Bernoulli numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Bernoulli numbers are used in some series expansions of several functions (trigonometric, hyperbolic, gamma, etc.), and are extremely important in number theory and analysis.
Note that there are two definitions of Bernoulli numbers; this task will be using the modern usage (as per The National Institute of Standards and Technology convention).
The nth Bernoulli number is expressed as Bn.
- Task
- show the Bernoulli numbers B0 through B60.
- suppress the output of values which are equal to zero. (Other than B1 , all odd Bernoulli numbers have a value of zero.)
- express the Bernoulli numbers as fractions (most are improper fractions).
- the fractions should be reduced.
- index each number in some way so that it can be discerned which Bernoulli number is being displayed.
- align the solidi (/) if used (extra credit).
- An algorithm
The Akiyama–Tanigawa algorithm for the "second Bernoulli numbers" as taken from wikipedia is as follows:
for m from 0 by 1 to n do A[m] ← 1/(m+1) for j from m by -1 to 1 do A[j-1] ← j×(A[j-1] - A[j]) return A[0] (which is Bn)
- See also
- Sequence A027641 Numerator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
- Sequence A027642 Denominator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
- Entry Bernoulli number on The Eric Weisstein's World of Mathematics (TM).
- Luschny's The Bernoulli Manifesto for a discussion on B1 = -½ versus +½.
Ada
Using a GMP thick binding available at http://www.codeforge.com/article/422541
WITH GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;
USE GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;
PROCEDURE Main IS
FUNCTION Bernoulli_Number (N : Natural) RETURN Unbounded_Fraction IS
FUNCTION "/" (Left, Right : Natural) RETURN Unbounded_Fraction IS
(To_Unbounded_Integer (Left) / To_Unbounded_Integer (Right));
A : ARRAY (0 .. N) OF Unbounded_Fraction;
BEGIN
FOR M IN 0 .. N LOOP
A (M) := 1 / (M + 1);
FOR J IN REVERSE 1 .. M LOOP
A (J - 1) := (J / 1 ) * (A (J - 1) - A (J));
END LOOP;
END LOOP;
RETURN A (0);
END Bernoulli_Number;
BEGIN
FOR I IN 0 .. 60 LOOP
IF I MOD 2 = 0 OR I = 1 THEN
DECLARE
B : Unbounded_Fraction := Bernoulli_Number (I);
S : String := Image (GMP.Rationals.Numerator (B));
BEGIN
Put_Line ("B (" & (IF I < 10 THEN " " ELSE "") & Trim (I'Img, Left)
& ")=" & (44 - S'Length) * " " & Image (B));
END;
END IF;
END LOOP;
END Main;
- Output:
B(0)= 1 / 1 B(1)= 1 / 2 B(2)= 1 / 6 B(4)= -1 / 30 B(6)= 1 / 42 B(8)= -1 / 30 B(10)= 5 / 66 B(12)= -691 / 2730 B(14)= 7 / 6 B(16)= -3617 / 510 B(18)= 43867 / 798 B(20)= -174611 / 330 B(22)= 854513 / 138 B(24)= -236364091 / 2730 B(26)= 8553103 / 6 B(28)= -23749461029 / 870 B(30)= 8615841276005 / 14322 B(32)= -7709321041217 / 510 B(34)= 2577687858367 / 6 B(36)= -26315271553053477373 / 1919190 B(38)= 2929993913841559 / 6 B(40)= -261082718496449122051 / 13530 B(42)= 1520097643918070802691 / 1806 B(44)= -27833269579301024235023 / 690 B(46)= 596451111593912163277961 / 282 B(48)= -5609403368997817686249127547 / 46410 B(50)= 495057205241079648212477525 / 66 B(52)= -801165718135489957347924991853 / 1590 B(54)= 29149963634884862421418123812691 / 798 B(56)= -2479392929313226753685415739663229 / 870 B(58)= 84483613348880041862046775994036021 / 354 B(60)=-1215233140483755572040304994079820246041491 / 56786730
ALGOL 68
Uses the LONG LONG INT mode of Algol 68G which allows large precision integers.
BEGIN
# Show Bernoulli numbers B0 to B60 as rational numbers #
# Uses code from the Arithmetic/Rational task modified to use #
# LONG LONG INT to allow for the large number of digits requried #
PR precision 100 PR # sets the precision of LONG LONG INT #
# Code from the Arithmetic/Rational task #
# ============================================================== #
MODE FRAC = STRUCT( LONG LONG INT num #erator#, den #ominator#);
PROC gcd = (LONG LONG INT a, b) LONG LONG INT: # greatest common divisor #
(a = 0 | b |: b = 0 | a |: ABS a > ABS b | gcd(b, a MOD b) | gcd(a, b MOD a));
PROC lcm = (LONG LONG INT a, b)LONG LONG INT: # least common multiple #
a OVER gcd(a, b) * b;
PRIO // = 9; # higher then the ** operator #
OP // = (LONG LONG INT num, den)FRAC: ( # initialise and normalise #
LONG LONG INT common = gcd(num, den);
IF den < 0 THEN
( -num OVER common, -den OVER common)
ELSE
( num OVER common, den OVER common)
FI
);
OP + = (FRAC a, b)FRAC: (
LONG LONG INT common = lcm(den OF a, den OF b);
FRAC result := ( common OVER den OF a * num OF a + common OVER den OF b * num OF b, common );
num OF result//den OF result
);
OP - = (FRAC a, b)FRAC: a + -b;
OP - = (FRAC frac)FRAC: (-num OF frac, den OF frac);
# ============================================================== #
# end code from the Arithmetic/Rational task #
# Additional FRACrelated operators #
OP * = ( INT a, FRAC b )FRAC: ( num OF b * a ) // den OF b;
OP // = ( INT a, INT b )FRAC: LONG LONG INT( a ) // LONG LONG INT( b );
# returns the nth Bernoulli number, n must be >= 0 #
# Uses the algorithm suggested by the task, so B(1) is +1/2 #
PROC bernoulli = ( INT n )FRAC:
IF n < 0
THEN # n is out of range # 0 // 1
ELSE # n is valid #
[ 0 : n ]FRAC a;
FOR i FROM LWB a TO UPB a DO a[ i ] := 0 // 1 OD;
FOR m FROM 0 TO n DO
a[ m ] := 1 // ( m + 1 );
FOR j FROM m BY -1 TO 1 DO
a[ j - 1 ] := j * ( a[ j - 1 ] - a[ j ] )
OD
OD;
a[ 0 ]
FI # bernoulli # ;
FOR n FROM 0 TO 60 DO
FRAC bn := bernoulli( n );
IF num OF bn /= 0 THEN
# have a non-0 Bn #
print( ( "B(", whole( n, -2 ), ") ", whole( num OF bn, -50 ) ) );
print( ( " / ", whole( den OF bn, 0 ), newline ) )
FI
OD
END
- Output:
B( 0) 1 / 1 B( 1) 1 / 2 B( 2) 1 / 6 B( 4) -1 / 30 B( 6) 1 / 42 B( 8) -1 / 30 B(10) 5 / 66 B(12) -691 / 2730 B(14) 7 / 6 B(16) -3617 / 510 B(18) 43867 / 798 B(20) -174611 / 330 B(22) 854513 / 138 B(24) -236364091 / 2730 B(26) 8553103 / 6 B(28) -23749461029 / 870 B(30) 8615841276005 / 14322 B(32) -7709321041217 / 510 B(34) 2577687858367 / 6 B(36) -26315271553053477373 / 1919190 B(38) 2929993913841559 / 6 B(40) -261082718496449122051 / 13530 B(42) 1520097643918070802691 / 1806 B(44) -27833269579301024235023 / 690 B(46) 596451111593912163277961 / 282 B(48) -5609403368997817686249127547 / 46410 B(50) 495057205241079648212477525 / 66 B(52) -801165718135489957347924991853 / 1590 B(54) 29149963634884862421418123812691 / 798 B(56) -2479392929313226753685415739663229 / 870 B(58) 84483613348880041862046775994036021 / 354 B(60) -1215233140483755572040304994079820246041491 / 56786730
AppleScript
To be able to handle numbers up to B(60) and beyond, this script represents the numbers with lists whose items are the digit values — which of course requires custom math routines.
on bernoullis(n) -- Return a list of "numerator / denominator" texts representing Bernoulli numbers B(0) to B(n).
set listMathScript to getListMathScript(10) -- Script object providing custom list math routines.
set output to {}
-- Akiyama–Tanigawa algorithm for the "second Bernoulli numbers".
-- List 'a' will contain {numerator, denominator} lists representing fractions.
-- The numerators and denominators will in turn be lists containing integers representing their (decimal) digits.
set a to {}
repeat with m from 0 to n
-- Append the structure for 1 / (m + 1) to the end of a.
set {numerator2, denominator2} to {{1}, listMathScript's intToList(m + 1)}
set a's end to result
repeat with j from m to 1 by -1
-- Retrieve the preceding numerator and denominator.
set {numerator1, denominator1} to a's item j
tell listMathScript
-- Get the two fractions' lowest common denominator and adjust the numerators accordingly.
set lcd to its lcm(denominator1, denominator2)
set numerator1 to its multiply(numerator1, its |div|(lcd, denominator1))
set numerator2 to its multiply(numerator2, its |div|(lcd, denominator2))
-- Subtract numerator2 from numerator1 and multiply the result by j.
-- Assign the results to numerator2 and denominator2 for the next iteration.
set numerator2 to its multiply(its subtract(numerator1, numerator2), its intToList(j))
set denominator2 to lcd
end tell
-- Also store them in a's slot j. No need to reduce them here.
set a's item j to {numerator2, denominator2}
end repeat
-- The fraction just stored in a's first slot is Bernoulli(m). Reduce it and append a text representation to the output.
tell listMathScript
set gcd to its hcf(numerator2, denominator2)
set numerator2 to its |div|(numerator2, gcd)
set denominator2 to its |div|(denominator2, gcd)
set end of output to its listToText(numerator2) & (" / " & its listToText(denominator2))
end tell
end repeat
return output
end bernoullis
on getListMathScript(base)
script
on multiply(lst1, lst2) -- Multiply lst1 by lst2.
set lst1Length to (count lst1)
set lst2Length to (count lst2)
set productLength to lst1Length + lst2Length - 1
set product to {}
repeat productLength times
set product's end to 0
end repeat
-- Long multiplication algorithm, updating product digits on the fly instead of summing rows at the end.
repeat with lst2Index from -1 to -lst2Length by -1
set lst2Digit to lst2's item lst2Index
if (lst2Digit is not 0) then
set carry to 0
set productIndex to lst2Index
repeat with lst1Index from lst1's length to 1 by -1
tell lst2Digit * (lst1's item lst1Index) + carry + (product's item productIndex)
set product's item productIndex to (it mod base)
set carry to (it div base)
end tell
set productIndex to productIndex - 1
end repeat
if (carry = 0) then
else if (productIndex < -productLength) then
set product's beginning to carry
else
set product's item productIndex to (product's item productIndex) + carry
end if
end if
end repeat
return product
end multiply
on subtract(lst1, lst2) -- Subtract lst2 from lst1.
set lst1Length to (count lst1)
set lst2Length to (count lst2)
-- Pad copies to equal lengths.
copy lst1 to lst1
repeat (lst2Length - lst1Length) times
set lst1's beginning to 0
end repeat
copy lst2 to lst2
repeat (lst1Length - lst2Length) times
set lst2's beginning to 0
end repeat
-- Is lst2's numeric value greater than lst1's?
set paddedLength to (count lst1)
repeat with i from 1 to paddedLength
set lst1Digit to lst1's item i
set lst2Digit to lst2's item i
set lst2Greater to (lst2Digit > lst1Digit)
if ((lst2Greater) or (lst1Digit > lst2Digit)) then exit repeat
end repeat
-- If so, set up to subtract lst1 from lst2 instead. We'll invert the result's sign at the end.
if (lst2Greater) then tell lst2
set lst2 to lst1
set lst1 to it
end tell
-- The subtraction at last!
set difference to {}
set borrow to 0
repeat with i from paddedLength to 1 by -1
tell (lst1's item i) + base - borrow - (lst2's item i)
set difference's beginning to (it mod base)
set borrow to 1 - (it div base)
end tell
end repeat
if (lst2Greater) then invert(difference)
return difference
end subtract
on |div|(lst1, lst2) -- List lst1 div lst2.
return divide(lst1, lst2)'s quotient
end |div|
on |mod|(lst1, lst2) -- List lst1 mod lst2.
return divide(lst1, lst2)'s remainder
end |mod|
on divide(lst1, lst2) -- Divide lst1 by lst2. Return a record containing separate lists for the quotient and remainder.
set dividend to trim(lst1)
set divisor to trim(lst2)
set dividendLength to (count dividend)
set divisorLength to (count divisor)
if (divisorLength > dividendLength) then return {quotient:{0}, remainder:dividend}
-- Note the dividend's and divisor's signs, but use absolute values in the division.
set dividendNegative to (dividend's beginning < 0)
if (dividendNegative) then invert(dividend)
set divisorNegative to (divisor's beginning < 0)
if (divisorNegative) then invert(divisor)
-- Long-division algorithm, but quotient digits are subtraction counts.
set quotient to {}
if (divisorLength > 1) then
set remainder to dividend's items 1 thru (divisorLength - 1)
else
set remainder to {}
end if
repeat with nextSlot from divisorLength to dividendLength
set remainder's end to dividend's item nextSlot
repeat with subtractionCount from 0 to base -- Only ever reaches base - 1.
set subtractionResult to trim(subtract(remainder, divisor))
if (subtractionResult's beginning < 0) then exit repeat
set remainder to subtractionResult
end repeat
set end of quotient to subtractionCount
end repeat
-- The quotient's negative if the input signs are different. Positive otherwise.
if (dividendNegative ≠ divisorNegative) then invert(quotient)
-- The remainder has the same sign as the dividend.
if (dividendNegative) then invert(remainder)
return {quotient:quotient, remainder:remainder}
end divide
on lcm(lst1, lst2) -- Lowest common multiple of lst1 and lst2.
return multiply(lst2, |div|(lst1, hcf(lst1, lst2)))
end lcm
on hcf(lst1, lst2) -- Highest common factor of lst1 and lst2.
set lst1 to trim(lst1)
set lst2 to trim(lst2)
repeat until (lst2 = {0})
set x to lst1
set lst1 to lst2
set lst2 to trim(|mod|(x, lst2))
end repeat
if (lst1's beginning < 0) then invert(lst1)
return lst1
end hcf
on invert(lst) -- Invert the sign of all lst's "digits".
repeat with thisDigit in lst
set thisDigit's contents to -thisDigit
end repeat
end invert
on trim(lst) -- Return a copy of lst with no leading zeros.
repeat with i from 1 to (count lst)
if (lst's item i is not 0) then exit repeat
end repeat
return lst's items i thru end
end trim
on intToList(n) -- Return a list of numbers representing n's digits.
set lst to {n mod base}
set n to n div base
repeat until (n = 0)
set beginning of lst to n mod base as integer
set n to n div base
end repeat
return lst
end intToList
on listToText(lst) -- Return the number represented by the input list as text.
-- This lazily assumes 2 <= base <= 10. :)
set lst to trim(lst)
if (lst's beginning < 0) then
invert(lst)
set lst's beginning to "-"
end if
return join(lst, "")
end listToText
end script
return result
end getListMathScript
on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join
on task()
set maxN to 60
set output to {""}
set padding to " = "
set bernoulliNumbers to bernoullis(maxN)
repeat with n from 0 to maxN
set bernie to bernoulliNumbers's item (n + 1)
if (bernie does not start with "0") then
set Bn to "B(" & n & ")"
set output's end to Bn & ¬
text ((count Bn) - 3) thru (50 - (offset of "/" in bernie)) of padding & ¬
bernie
end if
end repeat
return join(output, linefeed)
end task
task()
- Output:
"
B(0) = 1 / 1
B(1) = 1 / 2
B(2) = 1 / 6
B(4) = -1 / 30
B(6) = 1 / 42
B(8) = -1 / 30
B(10) = 5 / 66
B(12) = -691 / 2730
B(14) = 7 / 6
B(16) = -3617 / 510
B(18) = 43867 / 798
B(20) = -174611 / 330
B(22) = 854513 / 138
B(24) = -236364091 / 2730
B(26) = 8553103 / 6
B(28) = -23749461029 / 870
B(30) = 8615841276005 / 14322
B(32) = -7709321041217 / 510
B(34) = 2577687858367 / 6
B(36) = -26315271553053477373 / 1919190
B(38) = 2929993913841559 / 6
B(40) = -261082718496449122051 / 13530
B(42) = 1520097643918070802691 / 1806
B(44) = -27833269579301024235023 / 690
B(46) = 596451111593912163277961 / 282
B(48) = -5609403368997817686249127547 / 46410
B(50) = 495057205241079648212477525 / 66
B(52) = -801165718135489957347924991853 / 1590
B(54) = 29149963634884862421418123812691 / 798
B(56) = -2479392929313226753685415739663229 / 870
B(58) = 84483613348880041862046775994036021 / 354
B(60) = -1215233140483755572040304994079820246041491 / 56786730"
Bracmat
( BernoulliList
= B Bs answer indLn indexLen indexPadding
, n numberPadding p solPos solidusPos sp
. ( B
= m A a j b
. -1:?m
& :?A
& whl
' ( 1+!m:~>!arg:?m
& ((!m+1:?j)^-1:?a)
map
$ ( (
= .(-1+!j:?j)*(!arg+-1*!a):?a
)
. !A
)
: ?A
)
& !A:? @?b
& !b
)
& -1:?n
& :?Bs
& whl
' ( 1+!n:~>!arg:?n
& B$!n !Bs:?Bs
)
& @(!arg:? [?indexLen)
& 1+!indexLen:?indexLen
& !Bs:%@(?:? "/" [?solidusPos ?) ?
& 1+!solidusPos:?solidusPos:?p
& :?sp
& whl
' (!p+-1:~<0:?p&" " !sp:?sp)
& :?answer
& whl
' ( !Bs:%?B ?Bs
& ( !B:0
| (!B:/|str$(!B "/1"):?B)
& @(!B:? "/" [?solPos ?)
& @(!arg:? [?indLn)
& !sp
: ? [(-1*!indexLen+!indLn) ?indexPadding
: ? [(-1*!solidusPos+!solPos) ?numberPadding
& "B("
!arg
")="
!indexPadding
!numberPadding
(!B:>0&" "|)
!B
\n
!answer
: ?answer
)
& -1+!arg:?arg
)
& str$!answer
)
& BernoulliList$60;
B(0)= 1/1 B(1)= 1/2 B(2)= 1/6 B(4)= -1/30 B(6)= 1/42 B(8)= -1/30 B(10)= 5/66 B(12)= -691/2730 B(14)= 7/6 B(16)= -3617/510 B(18)= 43867/798 B(20)= -174611/330 B(22)= 854513/138 B(24)= -236364091/2730 B(26)= 8553103/6 B(28)= -23749461029/870 B(30)= 8615841276005/14322 B(32)= -7709321041217/510 B(34)= 2577687858367/6 B(36)= -26315271553053477373/1919190 B(38)= 2929993913841559/6 B(40)= -261082718496449122051/13530 B(42)= 1520097643918070802691/1806 B(44)= -27833269579301024235023/690 B(46)= 596451111593912163277961/282 B(48)= -5609403368997817686249127547/46410 B(50)= 495057205241079648212477525/66 B(52)= -801165718135489957347924991853/1590 B(54)= 29149963634884862421418123812691/798 B(56)= -2479392929313226753685415739663229/870 B(58)= 84483613348880041862046775994036021/354 B(60)=-1215233140483755572040304994079820246041491/56786730
C
#include <stdlib.h>
#include <gmp.h>
#define mpq_for(buf, op, n)\
do {\
size_t i;\
for (i = 0; i < (n); ++i)\
mpq_##op(buf[i]);\
} while (0)
void bernoulli(mpq_t rop, unsigned int n)
{
unsigned int m, j;
mpq_t *a = malloc(sizeof(mpq_t) * (n + 1));
mpq_for(a, init, n + 1);
for (m = 0; m <= n; ++m) {
mpq_set_ui(a[m], 1, m + 1);
for (j = m; j > 0; --j) {
mpq_sub(a[j-1], a[j], a[j-1]);
mpq_set_ui(rop, j, 1);
mpq_mul(a[j-1], a[j-1], rop);
}
}
mpq_set(rop, a[0]);
mpq_for(a, clear, n + 1);
free(a);
}
int main(void)
{
mpq_t rop;
mpz_t n, d;
mpq_init(rop);
mpz_inits(n, d, NULL);
unsigned int i;
for (i = 0; i <= 60; ++i) {
bernoulli(rop, i);
if (mpq_cmp_ui(rop, 0, 1)) {
mpq_get_num(n, rop);
mpq_get_den(d, rop);
gmp_printf("B(%-2u) = %44Zd / %Zd\n", i, n, d);
}
}
mpz_clears(n, d, NULL);
mpq_clear(rop);
return 0;
}
- Output:
B(0 ) = 1 / 1 B(1 ) = -1 / 2 B(2 ) = 1 / 6 B(4 ) = -1 / 30 B(6 ) = 1 / 42 B(8 ) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
C#
Using Mpir.NET
Translation of the C implementation
using Mpir.NET;
using System;
namespace Bernoulli
{
class Program
{
private static void bernoulli(mpq_t rop, uint n)
{
mpq_t[] a = new mpq_t[n + 1];
for (uint i = 0; i < n + 1; i++)
{
a[i] = new mpq_t();
}
for (uint m = 0; m <= n; ++m)
{
mpir.mpq_set_ui(a[m], 1, m + 1);
for (uint j = m; j > 0; --j)
{
mpir.mpq_sub(a[j - 1], a[j], a[j - 1]);
mpir.mpq_set_ui(rop, j, 1);
mpir.mpq_mul(a[j - 1], a[j - 1], rop);
}
mpir.mpq_set(rop, a[0]);
}
}
static void Main(string[] args)
{
mpq_t rop = new mpq_t();
mpz_t n = new mpz_t();
mpz_t d = new mpz_t();
for (uint i = 0; i <= 60; ++i)
{
bernoulli(rop, i);
if (mpir.mpq_cmp_ui(rop, 0, 1) != 0)
{
mpir.mpq_get_num(n, rop);
mpir.mpq_get_den(d, rop);
Console.WriteLine(string.Format("B({0, 2}) = {1, 44} / {2}", i, n, d));
}
}
Console.ReadKey();
}
}
}
- Output:
B(0 ) = 1 / 1 B(1 ) = -1 / 2 B(2 ) = 1 / 6 B(4 ) = -1 / 30 B(6 ) = 1 / 42 B(8 ) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Using Math.NET
using System;
using System.Console;
using System.Linq;
using MathNet.Numerics;
namespace Rosettacode.Rational.CS
{
class Program
{
private static readonly Func<int, BigRational> ℚ = BigRational.FromInt;
private static BigRational CalculateBernoulli(int n)
{
var a = InitializeArray(n);
foreach(var m in Enumerable.Range(1,n))
{
a[m] = ℚ(1) / (ℚ(m) + ℚ(1));
for (var j = m; j >= 1; j--)
{
a[j-1] = ℚ(j) * (a[j-1] - a[j]);
}
}
return a[0];
}
private static BigRational[] InitializeArray(int n)
{
var a = new BigRational[n + 1];
for (var x = 0; x < a.Length; x++)
{
a[x] = ℚ(x + 1);
}
return a;
}
static void Main()
{
Enumerable.Range(0, 61) // the second parameter is the number of range elements, and is not the final item of the range.
.Select(n => new {N = n, BernoulliNumber = CalculateBernoulli(n)})
.Where(b => !b.BernoulliNumber.Numerator.IsZero)
.Select(b => string.Format("B({0, 2}) = {1, 44} / {2}", b.N, b.BernoulliNumber.Numerator, b.BernoulliNumber.Denominator))
.ToList()
.ForEach(WriteLine);
}
}
}
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Using System.Numerics
Algo based on the example provided in the header of this RC page (the one from Wikipedia).
Extra feature - one can override the default of 60 by supplying a suitable number on the command line. The column widths are not hard-coded, but will adapt to the widths of the items listed.
using System;
using System.Numerics;
using System.Collections.Generic;
namespace bern
{
class Program
{
struct BerNum { public int index; public BigInteger Numer, Denomin; };
static int w1 = 1, w2 = 1; // widths for formatting output
static int max = 60; // default maximum, can override on command line
// returns nth Bernoulli number
static BerNum CalcBernoulli(int n)
{
BerNum res;
BigInteger f;
BigInteger[] nu = new BigInteger[n + 1],
de = new BigInteger[n + 1];
for (int m = 0; m <= n; m++)
{
nu[m] = 1; de[m] = m + 1;
for (int j = m; j > 0; j--)
if ((f = BigInteger.GreatestCommonDivisor(
nu[j - 1] = j * (de[j] * nu[j - 1] - de[j - 1] * nu[j]),
de[j - 1] *= de[j])) != BigInteger.One)
{ nu[j - 1] /= f; de[j - 1] /= f; }
}
res.index = n; res.Numer = nu[0]; res.Denomin = de[0];
w1 = Math.Max(n.ToString().Length, w1); // ratchet up widths
w2 = Math.Max(res.Numer.ToString().Length, w2);
if (max > 50) Console.Write("."); // progress dots appear for larger values
return res;
}
static void Main(string[] args)
{
List<BerNum> BNumbList = new List<BerNum>();
// defaults to 60 when no (or invalid) command line parameter is present
if (args.Length > 0) {
int.TryParse(args[0], out max);
if (max < 1 || max > Int16.MaxValue) max = 60;
if (args[0] == "0") max = 0;
}
for (int i = 0; i <= max; i++) // fill list with values
{
BerNum BNumb = CalcBernoulli(i);
if (BNumb.Numer != BigInteger.Zero) BNumbList.Add(BNumb);
}
if (max > 50) Console.WriteLine();
string strFmt = "B({0, " + w1.ToString() + "}) = {1, " + w2.ToString() + "} / {2}";
// display formatted list
foreach (BerNum bn in BNumbList)
Console.WriteLine(strFmt , bn.index, bn.Numer, bn.Denomin);
if (System.Diagnostics.Debugger.IsAttached) Console.Read();
}
}
}
- Output:
Default (nothing entered on command line):
............................................................. B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Output with "8" entered on command line:
B(0) = 1 / 1 B(1) = 1 / 2 B(2) = 1 / 6 B(4) = -1 / 30 B(6) = 1 / 42 B(8) = -1 / 30
Output with "126" entered on the command line:
............................................................................................................................... B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B( 10) = 5 / 66 B( 12) = -691 / 2730 B( 14) = 7 / 6 B( 16) = -3617 / 510 B( 18) = 43867 / 798 B( 20) = -174611 / 330 B( 22) = 854513 / 138 B( 24) = -236364091 / 2730 B( 26) = 8553103 / 6 B( 28) = -23749461029 / 870 B( 30) = 8615841276005 / 14322 B( 32) = -7709321041217 / 510 B( 34) = 2577687858367 / 6 B( 36) = -26315271553053477373 / 1919190 B( 38) = 2929993913841559 / 6 B( 40) = -261082718496449122051 / 13530 B( 42) = 1520097643918070802691 / 1806 B( 44) = -27833269579301024235023 / 690 B( 46) = 596451111593912163277961 / 282 B( 48) = -5609403368997817686249127547 / 46410 B( 50) = 495057205241079648212477525 / 66 B( 52) = -801165718135489957347924991853 / 1590 B( 54) = 29149963634884862421418123812691 / 798 B( 56) = -2479392929313226753685415739663229 / 870 B( 58) = 84483613348880041862046775994036021 / 354 B( 60) = -1215233140483755572040304994079820246041491 / 56786730 B( 62) = 12300585434086858541953039857403386151 / 6 B( 64) = -106783830147866529886385444979142647942017 / 510 B( 66) = 1472600022126335654051619428551932342241899101 / 64722 B( 68) = -78773130858718728141909149208474606244347001 / 30 B( 70) = 1505381347333367003803076567377857208511438160235 / 4686 B( 72) = -5827954961669944110438277244641067365282488301844260429 / 140100870 B( 74) = 34152417289221168014330073731472635186688307783087 / 6 B( 76) = -24655088825935372707687196040585199904365267828865801 / 30 B( 78) = 414846365575400828295179035549542073492199375372400483487 / 3318 B( 80) = -4603784299479457646935574969019046849794257872751288919656867 / 230010 B( 82) = 1677014149185145836823154509786269900207736027570253414881613 / 498 B( 84) = -2024576195935290360231131160111731009989917391198090877281083932477 / 3404310 B( 86) = 660714619417678653573847847426261496277830686653388931761996983 / 6 B( 88) = -1311426488674017507995511424019311843345750275572028644296919890574047 / 61410 B( 90) = 1179057279021082799884123351249215083775254949669647116231545215727922535 / 272118 B( 92) = -1295585948207537527989427828538576749659341483719435143023316326829946247 / 1410 B( 94) = 1220813806579744469607301679413201203958508415202696621436215105284649447 / 6 B( 96) = -211600449597266513097597728109824233673043954389060234150638733420050668349987259 / 4501770 B( 98) = 67908260672905495624051117546403605607342195728504487509073961249992947058239 / 6 B(100) = -94598037819122125295227433069493721872702841533066936133385696204311395415197247711 / 33330 B(102) = 3204019410860907078243020782116241775491817197152717450679002501086861530836678158791 / 4326 B(104) = -319533631363830011287103352796174274671189606078272738327103470162849568365549721224053 / 1590 B(106) = 36373903172617414408151820151593427169231298640581690038930816378281879873386202346572901 / 642 B(108) = -3469342247847828789552088659323852541399766785760491146870005891371501266319724897592306597338057 / 209191710 B(110) = 7645992940484742892248134246724347500528752413412307906683593870759797606269585779977930217515 / 1518 B(112) = -2650879602155099713352597214685162014443151499192509896451788427680966756514875515366781203552600109 / 1671270 B(114) = 21737832319369163333310761086652991475721156679090831360806110114933605484234593650904188618562649 / 42 B(116) = -309553916571842976912513458033841416869004128064329844245504045721008957524571968271388199595754752259 / 1770 B(118) = 366963119969713111534947151585585006684606361080699204301059440676414485045806461889371776354517095799 / 6 B(120) = -51507486535079109061843996857849983274095170353262675213092869167199297474922985358811329367077682677803282070131 / 2328255930 B(122) = 49633666079262581912532637475990757438722790311060139770309311793150683214100431329033113678098037968564431 / 6 B(124) = -95876775334247128750774903107542444620578830013297336819553512729358593354435944413631943610268472689094609001 / 30 B(126) = 5556330281949274850616324408918951380525567307126747246796782304333594286400508981287241419934529638692081513802696639 / 4357878
C++
/**
* Configured with: --prefix=/Library/Developer/CommandLineTools/usr --with-gxx-include-dir=/usr/include/c++/4.2.1
* Apple LLVM version 9.1.0 (clang-902.0.39.1)
* Target: x86_64-apple-darwin17.5.0
* Thread model: posix
*/
#include <boost/multiprecision/cpp_int.hpp> // 1024bit precision
#include <boost/rational.hpp> // Rationals
#include <iostream> // formatting with std::cout
#include <vector> // Container
typedef boost::rational<boost::multiprecision::int1024_t> rational; // reduce boilerplate
rational bernoulli(size_t n) {
auto out = std::vector<rational>();
for (size_t m = 0; m <= n; m++) {
out.emplace_back(1, (m + 1)); // automatically constructs object
for (size_t j = m; j >= 1; j--) {
out[j - 1] = rational(j) * (out[j - 1] - out[j]);
}
}
return out[0];
}
int main() {
for (size_t n = 0; n <= 60; n += n >= 2 ? 2 : 1) {
auto b = bernoulli(n);
std::cout << "B(" << std::right << std::setw(2) << n << ") = ";
std::cout << std::right << std::setw(44) << b.numerator();
std::cout << " / " << b.denominator() << std::endl;
}
return 0;
}
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Clojure
ns test-project-intellij.core
(:gen-class))
(defn a-t [n]
" Used Akiyama-Tanigawa algorithm with a single loop rather than double nested loop "
" Clojure does fractional arithmetic automatically so that part is easy "
(loop [m 0
j m
A (vec (map #(/ 1 %) (range 1 (+ n 2))))] ; Prefil A(m) with 1/(m+1), for m = 1 to n
(cond ; Three way conditional allows single loop
(>= j 1) (recur m (dec j) (assoc A (dec j) (* j (- (nth A (dec j)) (nth A j))))) ; A[j-1] ← j×(A[j-1] - A[j]) ;
(< m n) (recur (inc m) (inc m) A) ; increment m, reset j = m
:else (nth A 0))))
(defn format-ans [ans]
" Formats answer so that '/' is aligned for all answers "
(if (= ans 1)
(format "%50d / %8d" 1 1)
(format "%50d / %8d" (numerator ans) (denominator ans))))
;; Generate a set of results for [0 1 2 4 ... 60]
(doseq [q (flatten [0 1 (range 2 62 2)])
:let [ans (a-t q)]]
(println q ":" (format-ans ans)))
- Output:
0 : 1 / 1 1 : 1 / 2 2 : 1 / 6 4 : -1 / 30 6 : 1 / 42 8 : -1 / 30 10 : 5 / 66 12 : -691 / 2730 14 : 7 / 6 16 : -3617 / 510 18 : 43867 / 798 20 : -174611 / 330 22 : 854513 / 138 24 : -236364091 / 2730 26 : 8553103 / 6 28 : -23749461029 / 870 30 : 8615841276005 / 14322 32 : -7709321041217 / 510 34 : 2577687858367 / 6 36 : -26315271553053477373 / 1919190 38 : 2929993913841559 / 6 40 : -261082718496449122051 / 13530 42 : 1520097643918070802691 / 1806 44 : -27833269579301024235023 / 690 46 : 596451111593912163277961 / 282 48 : -5609403368997817686249127547 / 46410 50 : 495057205241079648212477525 / 66 52 : -801165718135489957347924991853 / 1590 54 : 29149963634884862421418123812691 / 798 56 : -2479392929313226753685415739663229 / 870 58 : 84483613348880041862046775994036021 / 354 60 : -1215233140483755572040304994079820246041491 / 56786730
Common Lisp
An implementation of the simple algorithm.
Be advised that the pseudocode algorithm specifies (j * (a[j-1] - a[j])) in the inner loop; implementing that as-is gives the wrong value (1/2) where n = 1, whereas subtracting a[j]-a[j-1] yields the correct value (B[1]=-1/2). See the numerator list.
(defun bernouilli (n)
(loop with a = (make-array (list (1+ n)))
for m from 0 to n do
(setf (aref a m) (/ 1 (+ m 1)))
(loop for j from m downto 1 do
(setf (aref a (- j 1))
(* j (- (aref a j) (aref a (- j 1))))))
finally (return (aref a 0))))
;;Print outputs to stdout:
(loop for n from 0 to 60 do
(let ((b (bernouilli n)))
(when (not (zerop b))
(format t "~a: ~a~%" n b))))
;;For the "extra credit" challenge, we need to align the slashes.
(let (results)
;;collect the results
(loop for n from 0 to 60 do
(let ((b (bernouilli n)))
(when (not (zerop b)) (push (cons b n) results))))
;;parse the numerators into strings; save the greatest length in max-length
(let ((max-length (apply #'max (mapcar (lambda (r)
(length (format nil "~a" (numerator r))))
(mapcar #'car results)))))
;;Print the numbers with using the fixed-width formatter: ~Nd, where N is
;;the number of leading spaces. We can't just pass in the width variable
;;but we can splice together a formatting string that includes it.
;;We also can't use the fixed-width formatter on a ratio, so we have to split
;;the ratio and splice it back together like idiots.
(loop for n in (mapcar #'cdr (reverse results))
for r in (mapcar #'car (reverse results)) do
(format t (concatenate 'string
"B(~2d): ~"
(format nil "~a" max-length)
"d/~a~%")
n
(numerator r)
(denominator r)))))
- Output:
B( 0): 1/1 B( 1): -1/2 B( 2): 1/6 B( 4): -1/30 B( 6): 1/42 B( 8): -1/30 B(10): 5/66 B(12): -691/2730 B(14): 7/6 B(16): -3617/510 B(18): 43867/798 B(20): -174611/330 B(22): 854513/138 B(24): -236364091/2730 B(26): 8553103/6 B(28): -23749461029/870 B(30): 8615841276005/14322 B(32): -7709321041217/510 B(34): 2577687858367/6 B(36): -26315271553053477373/1919190 B(38): 2929993913841559/6 B(40): -261082718496449122051/13530 B(42): 1520097643918070802691/1806 B(44): -27833269579301024235023/690 B(46): 596451111593912163277961/282 B(48): -5609403368997817686249127547/46410 B(50): 495057205241079648212477525/66 B(52): -801165718135489957347924991853/1590 B(54): 29149963634884862421418123812691/798 B(56): -2479392929313226753685415739663229/870 B(58): 84483613348880041862046775994036021/354 B(60): -1215233140483755572040304994079820246041491/56786730
Crystal
require "big"
class Bernoulli
include Iterator(Tuple(Int32, BigRational))
def initialize
@a = [] of BigRational
@m = 0
end
def next
@a << BigRational.new(1, @m+1)
@m.downto(1) { |j| @a[j-1] = j*(@a[j-1] - @a[j]) }
v = @m.odd? && @m != 1 ? BigRational.new(0, 1) : @a.first
return {@m, v}
ensure
@m += 1
end
end
b = Bernoulli.new
bn = b.first(61).to_a
max_width = bn.map { |_, v| v.numerator.to_s.size }.max
bn.reject { |i, v| v.zero? }.each do |i, v|
puts "B(%2i) = %*i/%i" % [i, max_width, v.numerator, v.denominator]
end
Version 1: compute each number separately.
require "big"
def bernoulli(n)
ar = [] of BigRational
(0..n).each do |m|
ar << BigRational.new(1, m+1)
m.downto(1) { |j| ar[j-1] = j * (ar[j-1] - ar[j]) }
end
ar[0] # (which is Bn)
end
b_nums = (0..61).map { |i| bernoulli(i) }
width = b_nums.map{ |b| b.numerator.to_s.size }.max
b_nums.each_with_index { |b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? }
Version 2: create faster generator to compute array of numbers once.
require "big"
def bernoulli2(limit)
ar = [] of BigRational
(0..limit).each do |m|
ar << BigRational.new(1, m+1)
m.downto(1) { |j| ar[j-1] = j * (ar[j-1] - ar[j]) }
yield ar[0] # use Bn value in required block
end
end
b_nums = [] of BigRational
bernoulli2(61){ |b| b_nums << b }
width = b_nums.map{ |b| b.numerator.to_s.size }.max
b_nums.each_with_index { |b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? }
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
D
This uses the D module from the Arithmetic/Rational task.
import std.stdio, std.range, std.algorithm, std.conv, arithmetic_rational;
auto bernoulli(in uint n) pure nothrow /*@safe*/ {
auto A = new Rational[n + 1];
foreach (immutable m; 0 .. n + 1) {
A[m] = Rational(1, m + 1);
foreach_reverse (immutable j; 1 .. m + 1)
A[j - 1] = j * (A[j - 1] - A[j]);
}
return A[0];
}
void main() {
immutable berns = 61.iota.map!bernoulli.enumerate.filter!(t => t[1]).array;
immutable width = berns.map!(b => b[1].numerator.text.length).reduce!max;
foreach (immutable b; berns)
writefln("B(%2d) = %*d/%d", b[0], width, b[1].tupleof);
}
The output is exactly the same as the Python entry.
Delphi
Thanks Rudy Velthuis for the Velthuis.BigRationals library.
program Bernoulli_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Velthuis.BigRationals;
function b(n: Integer): BigRational;
begin
var a: TArray<BigRational>;
SetLength(a, n + 1);
for var m := 0 to High(a) do
begin
a[m] := BigRational.Create(1, m + 1);
for var j := m downto 1 do
begin
a[j - 1] := (a[j - 1] - a[j]) * j;
end;
end;
Result := a[0];
end;
begin
for var n := 0 to 60 do
begin
var bb := b(n);
if bb.Numerator.BitLength > 0 then
writeln(format('B(%2d) =%45s/%s', [n, bb.Numerator.ToString, bb.Denominator.ToString]));
end;
readln;
end.
EchoLisp
Only 'small' rationals are supported in EchoLisp, i.e numerator and demominator < 2^31. So, we create a class of 'large' rationals, supported by the bigint library, and then apply the magic formula.
(lib 'bigint) ;; lerge numbers
(lib 'gloops) ;; classes
(define-class Rational null ((a :initform #0) (b :initform #1)))
(define-method tostring (Rational) (lambda (r) (format "%50d / %d" r.a r.b)))
(define-method normalize (Rational) (lambda (r) ;; divide a and b by gcd
(let ((g (gcd r.a r.b)))
(set! r.a (/ r.a g)) (set! r.b (/ r.b g))
(when (< r.b 0) (set! r.a ( - r.a)) (set! r.b (- r.b))) ;; denominator > 0
r)))
(define-method initialize (Rational) (lambda (r) (normalize r)))
(define-method add (Rational) (lambda (r n) ;; + Rational any number
(normalize (Rational (+ (* (+ #0 n) r.b) r.a) r.b))))
(define-method add (Rational Rational) (lambda (r q) ;;; + Rational Rational
(normalize (Rational (+ (* r.a q.b) (* r.b q.a)) (* r.b q.b)))))
(define-method sub (Rational Rational) (lambda (r q)
(normalize (Rational (- (* r.a q.b) (* r.b q.a)) (* r.b q.b)))))
(define-method mul (Rational Rational) (lambda (r q)
(normalize (Rational (* r.a q.a) (* r.b q.b)))))
(define-method mul (Rational) (lambda (r n)
(normalize (Rational (* r.a (+ #0 n)) r.b ))))
(define-method div (Rational Rational) (lambda (r q)
(normalize (Rational (* r.a q.b) (* r.b q.a)))))
- Output:
;; Bernoulli numbers
;; http://rosettacode.org/wiki/Bernoulli_numbers
(define A (make-vector 100 0))
(define (B n)
(for ((m (1+ n))) ;; #1 creates a large integer
(vector-set! A m (Rational #1 (+ #1 m)))
(for ((j (in-range m 0 -1)))
(vector-set! A (1- j)
(mul (sub (vector-ref A (1- j)) (vector-ref A j)) j))))
(vector-ref A 0))
(for ((b (in-range 0 62 2))) (writeln b (B b))) →
0 1 / 1
2 1 / 6
4 -1 / 30
6 1 / 42
8 -1 / 30
10 5 / 66
12 -691 / 2730
14 7 / 6
16 -3617 / 510
18 43867 / 798
20 -174611 / 330
22 854513 / 138
24 -236364091 / 2730
26 8553103 / 6
28 -23749461029 / 870
30 8615841276005 / 14322
32 -7709321041217 / 510
34 2577687858367 / 6
36 -26315271553053477373 / 1919190
38 2929993913841559 / 6
40 -261082718496449122051 / 13530
42 1520097643918070802691 / 1806
44 -27833269579301024235023 / 690
46 596451111593912163277961 / 282
48 -5609403368997817686249127547 / 46410
50 495057205241079648212477525 / 66
52 -801165718135489957347924991853 / 1590
54 29149963634884862421418123812691 / 798
56 -2479392929313226753685415739663229 / 870
58 84483613348880041862046775994036021 / 354
60 -1215233140483755572040304994079820246041491 / 56786730
(B 1) → 1 / 2
Elixir
defmodule Bernoulli do
defmodule Rational do
import Kernel, except: [div: 2]
defstruct numerator: 0, denominator: 1
def new(numerator, denominator\\1) do
sign = if numerator * denominator < 0, do: -1, else: 1
{numerator, denominator} = {abs(numerator), abs(denominator)}
gcd = gcd(numerator, denominator)
%Rational{numerator: sign * Kernel.div(numerator, gcd),
denominator: Kernel.div(denominator, gcd)}
end
def sub(a, b) do
new(a.numerator * b.denominator - b.numerator * a.denominator,
a.denominator * b.denominator)
end
def mul(a, b) when is_integer(a) do
new(a * b.numerator, b.denominator)
end
defp gcd(a,0), do: a
defp gcd(a,b), do: gcd(b, rem(a,b))
end
def numbers(n) do
Stream.transform(0..n, {}, fn m,acc ->
acc = Tuple.append(acc, Rational.new(1,m+1))
if m>0 do
new =
Enum.reduce(m..1, acc, fn j,ar ->
put_elem(ar, j-1, Rational.mul(j, Rational.sub(elem(ar,j-1), elem(ar,j))))
end)
{[elem(new,0)], new}
else
{[elem(acc,0)], acc}
end
end) |> Enum.to_list
end
def task(n \\ 61) do
b_nums = numbers(n)
width = Enum.map(b_nums, fn b -> b.numerator |> to_string |> String.length end)
|> Enum.max
format = 'B(~2w) = ~#{width}w / ~w~n'
Enum.with_index(b_nums)
|> Enum.each(fn {b,i} ->
if b.numerator != 0, do: :io.fwrite format, [i, b.numerator, b.denominator]
end)
end
end
Bernoulli.task
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
F#
open MathNet.Numerics
open System
open System.Collections.Generic
let calculateBernoulli n =
let ℚ(x) = BigRational.FromInt x
let A = Array.init<BigRational> (n+1) (fun x -> ℚ(x+1))
for m in [1..n] do
A.[m] <- ℚ(1) / (ℚ(m) + ℚ(1))
for j in [m..(-1)..1] do
A.[j-1] <- ℚ(j) * (A.[j-1] - A.[j])
A.[0]
[<EntryPoint>]
let main argv =
for n in [0..60] do
let bernoulliNumber = calculateBernoulli n
match bernoulliNumber.Numerator.IsZero with
| false ->
let formatedString = String.Format("B({0, 2}) = {1, 44} / {2}", n, bernoulliNumber.Numerator, bernoulliNumber.Denominator)
printfn "%s" formatedString
| true ->
printf ""
0
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Factor
One could use the "bernoulli" word from the math.extras vocabulary as follows:
IN: scratchpad
[
0 1 1 "%2d : %d / %d\n" printf
1 -1 2 "%2d : %d / %d\n" printf
30 iota [
1 + 2 * dup bernoulli [ numerator ] [ denominator ] bi
"%2d : %d / %d\n" printf
] each
] time
0 : 1 / 1
1 : -1 / 2
2 : 1 / 6
4 : -1 / 30
6 : 1 / 42
8 : -1 / 30
10 : 5 / 66
12 : -691 / 2730
14 : 7 / 6
16 : -3617 / 510
18 : 43867 / 798
20 : -174611 / 330
22 : 854513 / 138
24 : -236364091 / 2730
26 : 8553103 / 6
28 : -23749461029 / 870
30 : 8615841276005 / 14322
32 : -7709321041217 / 510
34 : 2577687858367 / 6
36 : -26315271553053477373 / 1919190
38 : 2929993913841559 / 6
40 : -261082718496449122051 / 13530
42 : 1520097643918070802691 / 1806
44 : -27833269579301024235023 / 690
46 : 596451111593912163277961 / 282
48 : -5609403368997817686249127547 / 46410
50 : 495057205241079648212477525 / 66
52 : -801165718135489957347924991853 / 1590
54 : 29149963634884862421418123812691 / 798
56 : -2479392929313226753685415739663229 / 870
58 : 84483613348880041862046775994036021 / 354
60 : -1215233140483755572040304994079820246041491 / 56786730
Running time: 0.00489444 seconds
Alternatively a method described by Brent and Harvey (2011) in "Fast computation of Bernoulli, Tangent and Secant numbers" https://arxiv.org/pdf/1108.0286.pdf is shown.
:: bernoulli-numbers ( n -- )
n 1 + 0 <array> :> tab
1 1 tab set-nth
2 n [a,b] [| k |
k 1 - dup
tab nth *
k tab set-nth
] each
2 n [a,b] [| k |
k n [a,b] [| j |
j tab nth
j k - 2 + *
j 1 - tab nth
j k - * +
j tab set-nth
] each
] each
1 :> s!
1 n [a,b] [| k |
k 2 * dup
2^ dup 1 - *
k tab nth
swap / *
s * k tab set-nth
s -1 * s!
] each
0 1 1 "%2d : %d / %d\n" printf
1 -1 2 "%2d : %d / %d\n" printf
1 n [a,b] [| k |
k 2 * k tab nth
[ numerator ] [ denominator ] bi
"%2d : %d / %d\n" printf
] each
;
It gives the same result as the native implementation, but is slightly faster.
[ 30 bernoulli-numbers ] time
...
Running time: 0.004331652 seconds
Fermat
Func Bern(m) = Sigma<k=0,m>[Sigma<v=0,k>[(-1)^v*Bin(k,v)*(v+1)^m/(k+1)]].;
for i=0, 60 do b:=Bern(i); if b<>0 then !!(i,b) fi od;
- Output:
0 1
1 1 / 2 2 1 / 6 4 -1 / 30 6 1 / 42 8 -1 / 30 10 5 / 66 12 -691 / 2730 14 7 / 6 16 -3617 / 510 18 43867 / 798 20 -174611 / 330 22 854513 / 138 24 -236364091 / 2730 26 8553103 / 6 28 -23749461029 / 870 30 8615841276005 / 14322 32 -7709321041217 / 510 34 2577687858367 / 6 36 -26315271553053477373 / 1919190 38 2929993913841559 / 6 40 -261082718496449122051 / 13530 42 1520097643918070802691 / 1806 44 -27833269579301024235023 / 690 46 596451111593912163277961 / 282 48 -5609403368997817686249127547 / 46410 50 495057205241079648212477525 / 66 52 -801165718135489957347924991853 / 1590 54 29149963634884862421418123812691 / 798 56 -2479392929313226753685415739663229 / 870 58 84483613348880041862046775994036021 / 35460 -1215233140483755572040304994079820246041491 / 56786730
FreeBASIC
' version 08-10-2016
' compile with: fbc -s console
' uses gmp
#Include Once "gmp.bi"
#Define max 60
Dim As Long n
Dim As ZString Ptr gmp_str :gmp_str = Allocate(1000) ' 1000 char
Dim Shared As Mpq_ptr tmp, big_j
tmp = Allocate(Len(__mpq_struct)) :Mpq_init(tmp)
big_j = Allocate(Len(__mpq_struct)) :Mpq_init(big_j)
Dim Shared As Mpq_ptr a(max), b(max)
For n = 0 To max
A(n) = Allocate(Len(__mpq_struct)) :Mpq_init(A(n))
B(n) = Allocate(Len(__mpq_struct)) :Mpq_init(B(n))
Next
Function Bernoulli(n As Integer) As Mpq_ptr
Dim As Long m, j
For m = 0 To n
Mpq_set_ui(A(m), 1, m + 1)
For j = m To 1 Step - 1
Mpq_sub(tmp, A(j - 1), A(j))
Mpq_set_ui(big_j, j, 1) 'big_j = j
Mpq_mul(A(j - 1), big_j, tmp)
Next
Next
Return A(0)
End Function
' ------=< MAIN >=------
For n = 0 To max
Mpq_set(B(n), Bernoulli(n))
Mpq_get_str(gmp_str, 10, B(n))
If *gmp_str <> "0" Then
If *gmp_str = "1" Then *gmp_str = "1/1"
Print Using "B(##) = "; n;
Print Space(45 - InStr(*gmp_str, "/")); *gmp_str
End If
Next
' empty keyboard buffer
While Inkey <> "" :Wend
Print :Print "hit any key to end program"
Sleep
End
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Frink
BernoulliNumber[n] :=
{
a = new array
for m = 0 to n
{
a@m = 1/(m+1)
for j = m to 1 step -1
a@(j-1) = j * (a@(j-1) - a@j)
}
return a@0
}
result = new array
for n=0 to 60
{
b = BernoulliNumber[n]
if b != 0
{
[num,den] = numeratorDenominator[b]
result.push[[n, num, "/", den]]
}
}
println[formatTable[result, "right"]]
- Output:
0 1 / 1 1 1 / 2 2 1 / 6 4 -1 / 30 6 1 / 42 8 -1 / 30 10 5 / 66 12 -691 / 2730 14 7 / 6 16 -3617 / 510 18 43867 / 798 20 -174611 / 330 22 854513 / 138 24 -236364091 / 2730 26 8553103 / 6 28 -23749461029 / 870 30 8615841276005 / 14322 32 -7709321041217 / 510 34 2577687858367 / 6 36 -26315271553053477373 / 1919190 38 2929993913841559 / 6 40 -261082718496449122051 / 13530 42 1520097643918070802691 / 1806 44 -27833269579301024235023 / 690 46 596451111593912163277961 / 282 48 -5609403368997817686249127547 / 46410 50 495057205241079648212477525 / 66 52 -801165718135489957347924991853 / 1590 54 29149963634884862421418123812691 / 798 56 -2479392929313226753685415739663229 / 870 58 84483613348880041862046775994036021 / 354 60 -1215233140483755572040304994079820246041491 / 56786730
FunL
FunL has pre-defined function B
in module integers
, which is defined as:
import integers.choose
def B( n ) = sum( 1/(k + 1)*sum((if 2|r then 1 else -1)*choose(k, r)*(r^n) | r <- 0..k) | k <- 0..n )
for i <- 0..60 if i == 1 or 2|i
printf( "B(%2d) = %s\n", i, B(i) )
- Output:
B( 0) = 1 B( 1) = -1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Solution. The following function reduces to the n-th Bernoulli number. It is a replica of the Akiyama–Tanigawa algorithm.
Test case. Showing the Bernoulli numbers B0 to B60
GAP
for a in Filtered(List([0 .. 60], n -> [n, Bernoulli(n)]), x -> x[2] <> 0) do
Print(a, "\n");
od;
[ 0, 1 ]
[ 1, -1/2 ]
[ 2, 1/6 ]
[ 4, -1/30 ]
[ 6, 1/42 ]
[ 8, -1/30 ]
[ 10, 5/66 ]
[ 12, -691/2730 ]
[ 14, 7/6 ]
[ 16, -3617/510 ]
[ 18, 43867/798 ]
[ 20, -174611/330 ]
[ 22, 854513/138 ]
[ 24, -236364091/2730 ]
[ 26, 8553103/6 ]
[ 28, -23749461029/870 ]
[ 30, 8615841276005/14322 ]
[ 32, -7709321041217/510 ]
[ 34, 2577687858367/6 ]
[ 36, -26315271553053477373/1919190 ]
[ 38, 2929993913841559/6 ]
[ 40, -261082718496449122051/13530 ]
[ 42, 1520097643918070802691/1806 ]
[ 44, -27833269579301024235023/690 ]
[ 46, 596451111593912163277961/282 ]
[ 48, -5609403368997817686249127547/46410 ]
[ 50, 495057205241079648212477525/66 ]
[ 52, -801165718135489957347924991853/1590 ]
[ 54, 29149963634884862421418123812691/798 ]
[ 56, -2479392929313226753685415739663229/870 ]
[ 58, 84483613348880041862046775994036021/354 ]
[ 60, -1215233140483755572040304994079820246041491/56786730 ]
Go
package main
import (
"fmt"
"math/big"
)
func b(n int) *big.Rat {
var f big.Rat
a := make([]big.Rat, n+1)
for m := range a {
a[m].SetFrac64(1, int64(m+1))
for j := m; j >= 1; j-- {
d := &a[j-1]
d.Mul(f.SetInt64(int64(j)), d.Sub(d, &a[j]))
}
}
return f.Set(&a[0])
}
func main() {
for n := 0; n <= 60; n++ {
if b := b(n); b.Num().BitLen() > 0 {
fmt.Printf("B(%2d) =%45s/%s\n", n, b.Num(), b.Denom())
}
}
}
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Haskell
Task algorithm
This program works as a command line utility, that reads from stdin the number of elements to compute (default 60) and prints them in stdout. The implementation of the algorithm is in the function bernoullis. The rest is for printing the results.
import Data.Ratio
import System.Environment
main = getArgs >>= printM . defaultArg
where
defaultArg as =
if null as
then 60
else read (head as)
printM m =
mapM_ (putStrLn . printP) .
takeWhile ((<= m) . fst) . filter (\(_, b) -> b /= 0 % 1) . zip [0 ..] $
bernoullis
printP (i, r) =
"B(" ++ show i ++ ") = " ++ show (numerator r) ++ "/" ++ show (denominator r)
bernoullis = map head . iterate (ulli 1) . map berno $ enumFrom 0
where
berno i = 1 % (i + 1)
ulli _ [_] = []
ulli i (x:y:xs) = (i % 1) * (x - y) : ulli (i + 1) (y : xs)
- Output:
B(0) = 1/1 B(1) = 1/2 B(2) = 1/6 B(4) = -1/30 B(6) = 1/42 B(8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Derivation from Faulhaber's triangle
import Data.Bool (bool)
import Data.Ratio (Ratio, denominator, numerator, (%))
-------------------- BERNOULLI NUMBERS -------------------
bernouillis :: Integer -> [Rational]
bernouillis =
fmap head
. tail
. scanl faulhaber []
. enumFromTo 0
faulhaber :: [Ratio Integer] -> Integer -> [Ratio Integer]
faulhaber rs n =
(:) =<< (-) 1 . sum $
zipWith ((*) . (n %)) [2 ..] rs
--------------------------- TEST -------------------------
main :: IO ()
main = do
let xs = bernouillis 60
w = length $ show (numerator (last xs))
putStrLn $
fTable
"Bernouillis from Faulhaber triangle:\n"
(show . fst)
(showRatio w . snd)
id
(filter ((0 /=) . snd) $ zip [0 ..] xs)
------------------------ FORMATTING ----------------------
fTable ::
String ->
(a -> String) ->
(b -> String) ->
(a -> b) ->
[a] ->
String
fTable s xShow fxShow f xs =
let w = maximum (length . xShow <$> xs)
in unlines $
s :
fmap
( ((<>) . rjust w ' ' . xShow)
<*> ((" -> " <>) . fxShow . f)
)
xs
showRatio :: Int -> Rational -> String
showRatio w r =
let d = denominator r
in rjust w ' ' $ show (numerator r)
<> bool [] (" / " <> show d) (1 /= d)
rjust :: Int -> a -> [a] -> [a]
rjust n c = drop . length <*> (replicate n c <>)
- Output:
Bernouillis from Faulhaber triangle: 0 -> 1 1 -> 1 / 2 2 -> 1 / 6 4 -> -1 / 30 6 -> 1 / 42 8 -> -1 / 30 10 -> 5 / 66 12 -> -691 / 2730 14 -> 7 / 6 16 -> -3617 / 510 18 -> 43867 / 798 20 -> -174611 / 330 22 -> 854513 / 138 24 -> -236364091 / 2730 26 -> 8553103 / 6 28 -> -23749461029 / 870 30 -> 8615841276005 / 14322 32 -> -7709321041217 / 510 34 -> 2577687858367 / 6 36 -> -26315271553053477373 / 1919190 38 -> 2929993913841559 / 6 40 -> -261082718496449122051 / 13530 42 -> 1520097643918070802691 / 1806 44 -> -27833269579301024235023 / 690 46 -> 596451111593912163277961 / 282 48 -> -5609403368997817686249127547 / 46410 50 -> 495057205241079648212477525 / 66 52 -> -801165718135489957347924991853 / 1590 54 -> 29149963634884862421418123812691 / 798 56 -> -2479392929313226753685415739663229 / 870 58 -> 84483613348880041862046775994036021 / 354 60 -> -1215233140483755572040304994079820246041491 / 56786730
Isabelle
Note that only the few invocations primcorec/definition in the last subsection are the implementation; the rest is about the correctness proof.
Adapted from the corresponding entry in the Archive of Formal Proofs.
Defines Bernoulli numbers as the sequence of numbers with the exponential generating function and then proves that the Akiyama–Tanigawa transform takes an ordinary generating function to the exponential generating function , from which it follows that applying it to the sequence with ordinary generating function gives exactly the Bernoulli numbers.
theory Mini_Bernoulli
imports Complex_Main
"HOL-Computational_Algebra.Computational_Algebra"
"HOL-Combinatorics.Stirling"
"HOL-Library.Stream"
"Coinductive.Coinductive_List"
"HOL-Library.Code_Target_Numeral"
"HOL-Library.Code_Lazy"
begin
subsection ‹Definition of Bernoulli numbers›
(* We define Bernoulli numbers in the standard fashion as those numbers B_n whose exponential
generating function function is X / (1 - e^(-X)): *)
definition bernoulli_fps :: "rat fps"
where "bernoulli_fps = fps_X / (1 - fps_exp (-1))"
definition bernoulli :: "nat ⇒ rat" where
"bernoulli n = fps_nth bernoulli_fps n * fact n"
subsection ‹Stirling numbers of the 2nd kind›
(* We use the Stirling numbers from the Isabelle library and prove a few additional things
about them. *)
lemma Stirling_n_0: "Stirling n 0 = (if n = 0 then 1 else 0)"
by (cases n) simp_all
lemma sum_Stirling_binomial:
"Stirling (Suc n) (Suc m) = (∑i=0..n. Stirling i m * (n choose i))"
proof -
have "real (Stirling (Suc n) (Suc m)) = real (∑i = 0..n. Stirling i m * (n choose i))"
proof (induction n arbitrary: m)
case (Suc n m)
have "real (∑i = 0..Suc n. Stirling i m * (Suc n choose i)) =
real (∑i = 0..n. Stirling (Suc i) m * (Suc n choose Suc i)) + real (Stirling 0 m)"
by (subst sum.atLeast0_atMost_Suc_shift) simp_all
also have "real (∑i = 0..n. Stirling (Suc i) m * (Suc n choose Suc i)) =
real (∑i = 0..n. (n choose i) * Stirling (Suc i) m) +
real (∑i = 0..n. (n choose Suc i) * Stirling (Suc i) m)"
by (simp add: algebra_simps sum.distrib)
also have "(∑i = 0..n. (n choose Suc i) * Stirling (Suc i) m) =
(∑i = Suc 0..Suc n. (n choose i) * Stirling i m)"
by (subst sum.shift_bounds_cl_Suc_ivl) simp_all
also have "… = (∑i = Suc 0..n. (n choose i) * Stirling i m)"
by (intro sum.mono_neutral_right) auto
also have "… = real (∑i = 0..n. Stirling i m * (n choose i)) - real (Stirling 0 m)"
by (simp add: sum.atLeast_Suc_atMost mult_ac)
also have "real (∑i = 0..n. Stirling i m * (n choose i)) = real (Stirling (Suc n) (Suc m))"
by (rule Suc.IH [symmetric])
also have "real (∑i = 0..n. (n choose i) * Stirling (Suc i) m) =
real m * real (Stirling (Suc n) (Suc m)) + real (Stirling (Suc n) m)"
by (cases m; (simp only: Suc.IH, simp add: algebra_simps sum.distrib
sum_distrib_left sum_distrib_right))
also have "… + (real (Stirling (Suc n) (Suc m)) - real (Stirling 0 m)) + real (Stirling 0 m) =
real (Suc m * Stirling (Suc n) (Suc m) + Stirling (Suc n) m)"
by (simp add: algebra_simps del: Stirling.simps)
also have "Suc m * Stirling (Suc n) (Suc m) + Stirling (Suc n) m =
Stirling (Suc (Suc n)) (Suc m)"
by (rule Stirling.simps(4) [symmetric])
finally show ?case ..
qed simp_all
thus ?thesis by (subst (asm) of_nat_eq_iff)
qed
(* The exponential generating function of the Stirling numbers (of the 2nd kind) with
fixed second argument $m$ is S_m(X) = (e^X - 1)^m / m!.
We use this fact to derive a closed form for them. *)
lemma Stirling_fps_aux:
"(fps_exp 1 - 1) ^ m $ n * fact n = (fact m * of_nat (Stirling n m) :: 'a :: field_char_0)"
proof (induction m arbitrary: n)
case 0
thus ?case by (simp add: Stirling_n_0)
next
case (Suc m n)
show ?case
proof (cases n)
case 0
thus ?thesis by simp
next
case (Suc n')
hence "(fps_exp 1 - 1 :: 'a fps) ^ Suc m $ n * fact n =
fps_deriv ((fps_exp 1 - 1) ^ Suc m) $ n' * fact n'"
by (simp_all add: algebra_simps del: power_Suc)
also have "fps_deriv ((fps_exp 1 - 1 :: 'a fps) ^ Suc m) =
fps_const (of_nat (Suc m)) * ((fps_exp 1 - 1) ^ m * fps_exp 1)"
by (subst fps_deriv_power) simp_all
also have "… $ n' * fact n' =
of_nat (Suc m) * ((∑i = 0..n'. (fps_exp 1 - 1) ^ m $ i / fact (n' - i)) * fact n')"
unfolding fps_mult_left_const_nth
by (simp add: fps_mult_nth Suc.IH sum_distrib_right del: of_nat_Suc)
also have "(∑i = 0..n'. (fps_exp 1 - 1 :: 'a fps) ^ m $ i / fact (n' - i)) * fact n' =
(∑i = 0..n'. (fps_exp 1 - 1) ^ m $ i * fact n' / fact (n' - i))"
by (subst sum_distrib_right, rule sum.cong) (simp_all add: divide_simps)
also have "… = (∑i = 0..n'. (fps_exp 1 - 1) ^ m $ i * fact i * of_nat (n' choose i))"
by (intro sum.cong refl) (simp_all add: binomial_fact)
also have "… = (∑i = 0..n'. fact m * of_nat (Stirling i m) * of_nat (n' choose i))"
by (simp only: Suc.IH)
also have "of_nat (Suc m) * … = (fact (Suc m) :: 'a) *
(∑i = 0..n'. of_nat (Stirling i m) * of_nat (n' choose i))" (is "_ = _ * ?S")
by (simp add: sum_distrib_left sum_distrib_right mult_ac del: of_nat_Suc)
also have "?S = of_nat (Stirling (Suc n') (Suc m))"
by (subst sum_Stirling_binomial) simp
also have "Suc n' = n" by (simp add: Suc)
finally show ?thesis .
qed
qed
theorem Stirling_closed_form:
"(of_nat (Stirling n k) :: 'a :: field_char_0) =
(∑j≤k. (-1)^(k - j) * of_nat (k choose j) * of_nat j ^ n) / fact k"
proof -
have "(fps_exp 1 - 1 :: 'a fps) = (fps_exp 1 + (-1))" by simp
also have "… ^ k = (∑j≤k. of_nat (k choose j) * fps_exp 1 ^ j * (- 1) ^ (k - j))"
unfolding binomial_ring ..
also have "… = (∑j≤k. fps_const ((-1) ^ (k - j) * of_nat (k choose j)) * fps_exp (of_nat j))"
by (simp add: fps_const_mult [symmetric] fps_const_power [symmetric]
fps_const_neg [symmetric] mult_ac fps_of_nat fps_exp_power_mult
del: fps_const_mult fps_const_power fps_const_neg)
also have "fps_nth … n = (∑j≤k. (- 1) ^ (k - j) * of_nat (k choose j) * of_nat j ^ n) / fact n"
by (simp add: fps_sum_nth sum_divide_distrib)
also have "… * fact n = (∑j≤k. (- 1) ^ (k - j) * of_nat (k choose j) * of_nat j ^ n)"
by simp
also note Stirling_fps_aux[of k n]
finally show ?thesis by (simp add: atLeast0AtMost field_simps)
qed
(*
We now define a somewhat ad-hoc operator formal power series: XD' maps a formal power
series A(X) to the series (X - 1) A'(X).
The relevance of this operator to us is that the n-fold iteration of this operator
is related to Stirling numbers.
*)
definition fps_XD' :: "'a :: field_char_0 fps ⇒ 'a fps"
where "fps_XD' = (λb. (fps_X - 1) * fps_deriv b)"
lemma fps_XD'_0 [simp]: "fps_XD' 0 = 0"
and fps_XD'_1 [simp]: "fps_XD' 1 = 0"
and fps_XD'_add [simp]: "fps_XD' (b + c) = fps_XD' b + fps_XD' c"
and fps_XD'_mult: "fps_XD' (b * c) = fps_XD' b * c + b * fps_XD' c"
by (simp_all add: fps_XD'_def algebra_simps)
lemma fps_XD'_power: "fps_XD' (b ^ n) = of_nat n * b ^ (n - 1) * fps_XD' b"
by (induction n) (simp_all add: algebra_simps fps_XD'_mult power_eq_if)
lemma fps_XD'_sum: "fps_XD' (sum f A) = sum (λx. fps_XD' (f x)) A"
by (induction A rule: infinite_finite_induct) simp_all
lemma fps_XD'_funpow:
defines "S ≡ λn i. fps_const (of_nat (Stirling n i))"
shows "(fps_XD' ^^ n) H = (∑m≤n. S n m * (fps_X - 1) ^ m * (fps_deriv ^^ m) H)"
proof (induction n arbitrary: H)
case 0
thus ?case by (simp add: S_def)
next
case (Suc n H)
define G where "G = (fps_X - 1 :: 'a fps)"
have "(∑m≤Suc n. S (Suc n) m * G ^ m * (fps_deriv ^^ m) H) =
(∑i≤n. of_nat (Suc i) * S n (Suc i) * G ^ Suc i * (fps_deriv ^^ Suc i) H) +
(∑i≤n. S n i * G ^ Suc i * (fps_deriv ^^ Suc i) H)"
(is "_ = sum (λi. ?f (Suc i)) … + ?S2")
by (subst sum.atMost_Suc_shift) (simp_all add: sum.distrib algebra_simps fps_of_nat S_def
fps_const_add [symmetric] fps_const_mult [symmetric] del: fps_const_add fps_const_mult)
also have "sum (λi. ?f (Suc i)) {..n} = sum (λi. ?f (Suc i)) {..<n}"
by (intro sum.mono_neutral_right) (auto simp: S_def)
also have "… = ?f 0 + …" by simp
also have "… = sum ?f {..n}" by (subst sum.atMost_shift [symmetric]) simp_all
also have "… + ?S2 = (∑x≤n. fps_XD' (S n x * G ^ x * (fps_deriv ^^ x) H))"
unfolding sum.distrib [symmetric]
proof (rule sum.cong, goal_cases)
case (2 i)
thus ?case unfolding fps_XD'_mult fps_XD'_power
by (cases i) (auto simp: fps_XD'_mult algebra_simps of_nat_diff S_def fps_XD'_def G_def)
qed simp_all
also have "… = (fps_XD' ^^ Suc n) H" by (simp add: Suc.IH fps_XD'_sum G_def)
finally show ?case by (simp add: G_def)
qed
subsection ‹The Akiyama–Tanigawa transform›
(* a single step in the Akiyama–Tanigawa matrix, i.e. how to get the next
line from the current one *)
definition AT_step :: "(nat ⇒ 'a :: field_char_0) ⇒ nat ⇒ 'a"
where "AT_step f = (λk. of_nat (k+1) * (f k - f (k+1)))"
definition AT_matrix :: "(nat ⇒ 'a :: field_char_0) ⇒ nat ⇒ nat ⇒ 'a" where
"AT_matrix f n = (AT_step ^^ n) f"
(* The following describes the ordinary generating function of the n-th row in the AT matrix. *)
definition AT_fps :: "(nat ⇒ 'a :: field_char_0) ⇒ nat ⇒ 'a fps" where
"AT_fps f n = (fps_X - 1) * Abs_fps (AT_matrix f n)"
lemma AT_fps_Suc: "AT_fps f (Suc n) = (fps_X - 1) * fps_deriv (AT_fps f n)"
proof (rule fps_ext)
fix m :: nat
show "AT_fps f (Suc n) $ m = ((fps_X - 1) * fps_deriv (AT_fps f n)) $ m"
by (cases m) (simp_all add: AT_fps_def AT_matrix_def AT_step_def fps_deriv_def algebra_simps)
qed
lemma AT_fps_altdef:
"AT_fps f n =
(∑m≤n. fps_const (of_nat (Stirling n m)) * (fps_X - 1)^m * (fps_deriv ^^ m) (AT_fps f 0))"
proof -
have "AT_fps f n = (fps_XD' ^^ n) (AT_fps f 0)"
by (induction n) (simp_all add: AT_fps_Suc fps_XD'_def)
also have "… = (∑m≤n. fps_const (of_nat (Stirling n m)) * (fps_X - 1) ^ m *
(fps_deriv ^^ m) (AT_fps f 0))"
by (rule fps_XD'_funpow)
finally show ?thesis .
qed
lemma AT_fps_0_nth: "AT_fps f 0 $ n = (if n = 0 then -f 0 else f (n - 1) - f n)"
by (simp add: AT_fps_def AT_matrix_def algebra_simps)
(* The following gives a closed form for the first column of the AT matrix, i.e. the
result of the transform. *)
lemma AT_matrix_firstcol:
"AT_matrix f n 0 = (∑k≤n. (-1) ^ k * fact k * of_nat (Stirling (Suc n) (Suc k)) * f k)"
proof (cases "n = 0")
case False
have "AT_matrix f n 0 = -(AT_fps f n $ 0)" by (simp add: AT_fps_def)
also have "AT_fps f n $ 0 =
(∑k≤n. of_nat (Stirling n k) * (- 1) ^ k * (fact k * AT_fps f 0 $ k))"
by (subst AT_fps_altdef) (simp add: fps_sum_nth fps_nth_power_0 fps_0th_higher_deriv)
also have "… = (∑k≤n. of_nat (Stirling n k) * (- 1) ^ k * (fact k * (f (k - 1) - f k)))"
using False by (intro sum.cong refl) (auto simp: Stirling_n_0 AT_fps_0_nth)
also have "… = (∑k≤n. fact k * (of_nat (Stirling n k) * (- 1) ^ k) * f (k - 1)) -
(∑k≤n. fact k * (of_nat (Stirling n k) * (- 1) ^ k) * f k)"
(is "_ = sum ?f _ - ?S2") by (simp add: sum_subtractf algebra_simps)
also from False have "sum ?f {..n} = sum ?f {0<..n}"
by (intro sum.mono_neutral_right) (auto simp: Stirling_n_0)
also have "… = sum ?f {0<..Suc n}"
by (intro sum.mono_neutral_left) auto
also have "{0<..Suc n} = {Suc 0..Suc n}" by auto
also have "sum ?f … = sum (λn. ?f (Suc n)) {0..n}"
by (subst sum.atLeast_Suc_atMost_Suc_shift) simp_all
also have "{0..n} = {..n}" by auto
also have "sum (λn. ?f (Suc n)) … - ?S2 =
(∑k≤n. -((-1)^k * fact k * of_nat (Stirling (Suc n) (Suc k)) * f k))"
by (subst sum_subtractf [symmetric], intro sum.cong) (simp_all add: algebra_simps)
also have "-… = (∑k≤n. ((-1)^k * fact k * of_nat (Stirling (Suc n) (Suc k)) * f k))"
by (simp add: sum_negf)
finally show ?thesis .
qed (simp_all add: AT_matrix_def)
(* The following theorem relates the exponential generating function B(X) of the transformed
sequence to the ordinary generating function A(X) of the original sequence.
Namely: B(X) = e^X A(1 - e^X). *)
theorem AT_fps:
"Abs_fps (λn. AT_matrix f n 0 / fact n) = fps_exp 1 * fps_compose (Abs_fps f) (1 - fps_exp 1)"
proof (rule fps_ext)
fix n :: nat
have "(fps_const (fact n) *
(fps_compose (Abs_fps (λn. AT_matrix f 0 n)) (1 - fps_exp 1) * fps_exp 1)) $ n =
(∑m≤n. ∑k≤m. (1 - fps_exp 1) ^ k $ m * fact n / fact (n - m) * f k)"
unfolding fps_mult_left_const_nth
by (simp add: fps_times_def fps_compose_def AT_matrix_firstcol sum_Stirling_binomial
field_simps sum_distrib_left sum_distrib_right atLeast0AtMost AT_matrix_def
del: Stirling.simps of_nat_Suc)
also have "… = (∑m≤n. ∑k≤m. (-1)^k * fact k * of_nat (Stirling m k * (n choose m)) * f k)"
proof (intro sum.cong refl, goal_cases)
case (1 m k)
have "(1 - fps_exp 1 :: 'a fps) ^ k = (-fps_exp 1 + 1 :: 'a fps) ^ k" by simp
also have "… = (∑i≤k. of_nat (k choose i) * (-1) ^ i * fps_exp (of_nat i))"
by (subst binomial_ring) (simp add: atLeast0AtMost power_minus' fps_exp_power_mult mult.assoc)
also have "… = (∑i≤k. fps_const (of_nat (k choose i) * (-1) ^ i) * fps_exp (of_nat i))"
by (simp add: fps_const_mult [symmetric] fps_of_nat fps_const_power [symmetric]
fps_const_neg [symmetric] del: fps_const_mult fps_const_power fps_const_neg)
also have "… $ m = (∑i≤k. of_nat (k choose i) * (- 1) ^ i * of_nat i ^ m) / fact m"
(is "_ = ?S / _") by (simp add: fps_sum_nth sum_divide_distrib [symmetric])
also have "?S = (-1) ^ k * (∑i≤k. (-1) ^ (k - i) * of_nat (k choose i) * of_nat i ^ m)"
by (subst sum_distrib_left, intro sum.cong refl) (auto simp: minus_one_power_iff)
also have "(∑i≤k. (-1) ^ (k - i) * of_nat (k choose i) * of_nat i ^ m) =
of_nat (Stirling m k) * (fact k :: 'a)"
by (subst Stirling_closed_form) (simp_all add: field_simps)
finally have *: "(1 - fps_exp 1 :: 'a fps) ^ k $ m * fact n / fact (n - m) =
(- 1) ^ k * fact k * of_nat (Stirling m k) * of_nat (n choose m)"
using 1 by (simp add: binomial_fact del: of_nat_Suc)
show ?case using 1 by (subst *) simp
qed
also have "… = (∑m≤n. ∑k≤n. (- 1) ^ k * fact k *
of_nat (Stirling m k * (n choose m)) * f k)"
by (rule sum.cong[OF refl], rule sum.mono_neutral_left) auto
also have "… = (∑k≤n. ∑m≤n. (- 1) ^ k * fact k *
of_nat (Stirling m k * (n choose m)) * f k)"
by (rule sum.swap)
also have "… = AT_matrix f n 0"
by (simp add: AT_matrix_firstcol sum_Stirling_binomial sum_distrib_left sum_distrib_right
mult.assoc atLeast0AtMost del: Stirling.simps)
finally show "Abs_fps (λn. AT_matrix f n 0 / fact n) $ n =
(fps_exp 1 * (Abs_fps f oo 1 - fps_exp 1)) $ n"
by (subst (asm) fps_mult_left_const_nth) (simp add: field_simps AT_matrix_def del: of_nat_Suc)
qed
(* If we specialise this to the input sequence a(k) = 1 / (k+1), this has the ordinary
generating function -ln(1 - X) / X, so the exponential generating function of the
AT transform is e^X (-ln (1 - (1 - e^X)) / (1 - e^X)) = X / (1 - e^(-X)),
which is exactly the exponential generating function of the Bernoulli numbers. *)
corollary bernoulli_conv_AT: "bernoulli n = AT_matrix (λk. 1 / of_nat (k+1)) n 0"
proof -
define f :: "nat ⇒ rat" where "f = (λn. 1 / of_nat (Suc n))"
note AT_fps[of f]
also {
have "fps_ln 1 = fps_X * Abs_fps (λn. (-1)^n / of_nat (Suc n) :: rat)"
by (intro fps_ext) (simp del: of_nat_Suc add: fps_ln_def)
hence "fps_ln 1 / fps_X = Abs_fps (λn. (-1)^n / of_nat (Suc n) :: rat)"
by (metis fps_X_neq_zero nonzero_mult_div_cancel_left)
also have "fps_compose … (-fps_X) = Abs_fps f"
by (simp add: fps_compose_uminus' fps_eq_iff f_def)
finally have "Abs_fps f = fps_compose (fps_ln 1 / fps_X) (-fps_X)" ..
also have "fps_ln 1 / fps_X oo - fps_X oo 1 - fps_exp (1::rat) = fps_ln 1 / fps_X oo fps_exp 1 - 1"
by (subst fps_compose_assoc [symmetric])
(simp_all add: fps_compose_uminus)
also have "… = (fps_ln 1 oo fps_exp 1 - 1) / (fps_exp 1 - 1)"
by (subst fps_compose_divide_distrib) auto
also have "… = fps_X / (fps_exp 1 - 1)" by (simp add: fps_ln_fps_exp_inv fps_inv_fps_exp_compose)
finally have "Abs_fps f oo 1 - fps_exp 1 = fps_X / (fps_exp 1 - 1)" .
}
also have "fps_exp (1::rat) - 1 = (1 - fps_exp (-1)) * fps_exp 1"
by (simp add: algebra_simps fps_exp_add_mult [symmetric])
also have "fps_exp 1 * (fps_X / …) = bernoulli_fps" unfolding bernoulli_fps_def
by (subst dvd_div_mult2_eq) (auto simp: fps_dvd_iff intro!: subdegree_leI)
finally have "Abs_fps (λn. AT_matrix f n 0 / fact n) = bernoulli_fps" .
hence "fps_nth (Abs_fps (λn. AT_matrix f n 0 / fact n)) n = fps_nth bernoulli_fps n"
by (rule arg_cong)
thus ?thesis by (simp add: fps_eq_iff f_def bernoulli_def field_simps)
qed
subsection ‹Efficient code›
(* Next, we implement the AT transform using infinite streams. We define the infinite
AT matrix as a stream of streams of numbers and then define its leftmost column as the
result of the transform. *)
primcorec AT_impl_step :: "nat ⇒ 'a :: field_char_0 stream ⇒ 'a stream" where
"AT_impl_step m xs =
of_nat m * (xs !! 0 - xs !! 1) ## AT_impl_step (Suc m) (stl xs)"
primcorec AT_matrix_impl :: "'a :: field_char_0 stream ⇒ 'a stream stream" where
"AT_matrix_impl xs = xs ## AT_matrix_impl (AT_impl_step 1 xs)"
definition AT_impl :: "'a :: field_char_0 stream ⇒ 'a :: field_char_0 stream" where
"AT_impl xs = smap shd (AT_matrix_impl xs)"
lemma snth_AT_impl_step [simp]:
"AT_impl_step m xs !! n = of_nat (m + n) * (xs !! n - xs !! (n+1))"
by (induction n arbitrary: m xs; subst AT_impl_step.code) auto
lemma snth_AT_matrix_impl [simp]: "AT_matrix_impl xs !! n !! k = AT_matrix (snth xs) n k"
by (induction n arbitrary: xs k; subst AT_matrix_impl.code)
(auto simp: snth_AT_impl_step [abs_def] AT_matrix_def funpow_Suc_right AT_step_def
simp del: funpow.simps of_nat_Suc)
lemma snth_AT_impl [simp]: "AT_impl xs !! n = AT_matrix (snth xs) n 0"
by (simp add: AT_impl_def flip: snth.simps(1))
definition bernoulli_stream :: "rat stream" where
"bernoulli_stream = AT_impl (smap (λi. 1 / of_nat i) (fromN 1))"
theorem bernoulli_stream_correct: "bernoulli_stream !! n = bernoulli n"
by (simp add: bernoulli_stream_def bernoulli_conv_AT snth_smap [abs_def])
end
code_lazy_type stream
code_lazy_type llist
primcorec llist_of_stream :: "'a stream ⇒ 'a llist" where
"llist_of_stream xs = LCons (shd xs) (llist_of_stream (stl xs))"
value "list_of (ltakeWhile (λ(i,_). i ≤ 60)
(llist_of_stream (sfilter (λ(_, x). x ≠ 0)
(szip (fromN 0) bernoulli_stream))))"
(* Output
"[(0, 1),
(1, 1 / 2),
(2, 1 / 6),
(4, - (1 / 30)),
(6, 1 / 42),
(8, - (1 / 30)),
(10, 5 / 66),
(12, - (691 / 2730)),
(14, 7 / 6),
(16, - (3617 / 510)),
(18, 43867 / 798),
(20, - (174611 / 330)),
(22, 854513 / 138),
(24, - (236364091 / 2730)),
(26, 8553103 / 6),
(28, - (23749461029 / 870)),
(30, 8615841276005 / 14322),
(32, - (7709321041217 / 510)),
(34, 2577687858367 / 6),
(36, - (26315271553053477373 / 1919190)),
(38, 2929993913841559 / 6),
(40, - (261082718496449122051 / 13530)),
(42, 1520097643918070802691 / 1806),
(44, - (27833269579301024235023 / 690)),
(46, 596451111593912163277961 / 282),
(48, - (5609403368997817686249127547 / 46410)),
(50, 495057205241079648212477525 / 66),
(52, - (801165718135489957347924991853 / 1590)),
(54, 29149963634884862421418123812691 / 798),
(56, - (2479392929313226753685415739663229 / 870)),
(58, 84483613348880041862046775994036021 / 354),
(60, - (1215233140483755572040304994079820246041491 / 56786730))]"
:: "(nat × rat) list"
*)
Icon and Unicon
The following works in both languages:
link "rational"
procedure main(args)
limit := integer(!args) | 60
every b := bernoulli(i := 0 to limit) do
if b.numer > 0 then write(right(i,3),": ",align(rat2str(b),60))
end
procedure bernoulli(n)
(A := table(0))[0] := rational(1,1,1)
every m := 1 to n do {
A[m] := rational(1,m+1,1)
every j := m to 1 by -1 do A[j-1] := mpyrat(rational(j,1,1), subrat(A[j-1],A[j]))
}
return A[0]
end
procedure align(r,n)
return repl(" ",n-find("/",r))||r
end
Sample run:
->bernoulli 60 0: (1/1) 1: (1/2) 2: (1/6) 4: (-1/30) 6: (1/42) 8: (-1/30) 10: (5/66) 12: (-691/2730) 14: (7/6) 16: (-3617/510) 18: (43867/798) 20: (-174611/330) 22: (854513/138) 24: (-236364091/2730) 26: (8553103/6) 28: (-23749461029/870) 30: (8615841276005/14322) 32: (-7709321041217/510) 34: (2577687858367/6) 36: (-26315271553053477373/1919190) 38: (2929993913841559/6) 40: (-261082718496449122051/13530) 42: (1520097643918070802691/1806) 44: (-27833269579301024235023/690) 46: (596451111593912163277961/282) 48: (-5609403368997817686249127547/46410) 50: (495057205241079648212477525/66) 52: (-801165718135489957347924991853/1590) 54: (29149963634884862421418123812691/798) 56: (-2479392929313226753685415739663229/870) 58: (84483613348880041862046775994036021/354) 60: (-1215233140483755572040304994079820246041491/56786730) ->
J
Implementation:
See Bernoulli Numbers Essay on the J wiki.
B=: {.&1 %. (i. ! ])@>:@i.@x:
Task:
'B' ,. rplc&'r/_-'"1": (#~ 0 ~: {:"1)(i. ,. B) 61
B 0 1
B 1 -1/2
B 2 1/6
B 4 -1/30
B 6 1/42
B 8 -1/30
B10 5/66
B12 -691/2730
B14 7/6
B16 -3617/510
B18 43867/798
B20 -174611/330
B22 854513/138
B24 -236364091/2730
B26 8553103/6
B28 -23749461029/870
B30 8615841276005/14322
B32 -7709321041217/510
B34 2577687858367/6
B36 -26315271553053477373/1919190
B38 2929993913841559/6
B40 -261082718496449122051/13530
B42 1520097643918070802691/1806
B44 -27833269579301024235023/690
B46 596451111593912163277961/282
B48 -5609403368997817686249127547/46410
B50 495057205241079648212477525/66
B52 -801165718135489957347924991853/1590
B54 29149963634884862421418123812691/798
B56 -2479392929313226753685415739663229/870
B58 84483613348880041862046775994036021/354
B60 -1215233140483755572040304994079820246041491/56786730
Java
import org.apache.commons.math3.fraction.BigFraction;
public class BernoulliNumbers {
public static void main(String[] args) {
for (int n = 0; n <= 60; n++) {
BigFraction b = bernouilli(n);
if (!b.equals(BigFraction.ZERO))
System.out.printf("B(%-2d) = %-1s%n", n , b);
}
}
static BigFraction bernouilli(int n) {
BigFraction[] A = new BigFraction[n + 1];
for (int m = 0; m <= n; m++) {
A[m] = new BigFraction(1, (m + 1));
for (int j = m; j >= 1; j--)
A[j - 1] = (A[j - 1].subtract(A[j])).multiply(new BigFraction(j));
}
return A[0];
}
}
B(0 ) = 1 B(1 ) = 1 / 2 B(2 ) = 1 / 6 B(4 ) = -1 / 30 B(6 ) = 1 / 42 B(8 ) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
jq
This section uses the Akiyama–Tanigawa algorithm for the second Bernoulli numbers, Bn. Therefore, the sign of B(1) differs from the modern definition.
The implementation presented here is intended for use with a "BigInt" library that uses string representations of decimal integers. Such a library is at BigInt.jq. To make the code in this section self-contained, stubs for the "BigInt" operations are provided in the first subsection.
BigInt Stubs:
# def negate:
# def lessOrEqual(x; y): # x <= y
# def long_add(x;y): # x+y
# def long_minus(x;y): # x-y
# def long_multiply(x;y) # x*y
# def long_divide(x;y): # x/y => [q,r]
# def long_div(x;y) # integer division
# def long_mod(x;y) # %
# In all cases, x and y must be strings
def negate: (- tonumber) | tostring;
def lessOrEqual(num1; num2): (num1|tonumber) <= (num2|tonumber);
def long_add(num1; num2): ((num1|tonumber) + (num2|tonumber)) | tostring;
def long_minus(x;y): ((num1|tonumber) - (num2|tonumber)) | tostring;
# multiply two decimal strings, which may be signed (+ or -)
def long_multiply(num1; num2):
((num1|tonumber) * (num2|tonumber)) | tostring;
# return [quotient, remainder]
# 0/0 = 1; n/0 => error
def long_divide(xx;yy): # x/y => [q,r] imples x == (y * q) + r
def ld(x;y):
def abs: if . < 0 then -. else . end;
(x|abs) as $x | (y|abs) as $y
| (if (x >= 0 and y > 0) or (x < 0 and y < 0) then 1 else -1 end) as $sign
| (if x >= 0 then 1 else -1 end) as $sx
| [$sign * ($x / $y | floor), $sx * ($x % $y)];
ld( xx|tonumber; yy|tonumber) | map(tostring);
def long_div(x;y):
long_divide(x;y) | .[0];
def long_mod(x;y):
((x|tonumber) % (y|tonumber)) | tostring;
Fractions:
# A fraction is represented by [numerator, denominator] in reduced form, with the sign on top
# a and b should be BigInt; return a BigInt
def gcd(a; b):
def long_abs: . as $in | if lessOrEqual("0"; $in) then $in else negate end;
# subfunction rgcd expects [a,b] as input
# i.e. a ~ .[0] and b ~ .[1]
def rgcd:
.[0] as $a | .[1] as $b
| if $b == "0" then $a
else [$b, long_mod($a ; $b ) ] | rgcd
end;
a as $a | b as $b
| [$a,$b] | rgcd | long_abs ;
def normalize:
.[0] as $p | .[1] as $q
| if $p == "0" then ["0", "1"]
elif lessOrEqual($q ; "0") then [ ($p|negate), ($q|negate)] | normalize
else gcd($p; $q) as $g
| [ long_div($p;$g), long_div($q;$g) ]
end ;
# a and b should be fractions expressed in the form [p, q]
def add(a; b):
a as $a | b as $b
| if $a[1] == "1" and $b[1] == "1" then [ long_add($a[0]; $b[0]) , "1"]
elif $a[1] == $b[1] then [ long_add( $a[0]; $b[0]), $a[1] ] | normalize
elif $a[0] == "0" then $b
elif $b[0] == "0" then $a
else [ long_add( long_multiply($a[0]; $b[1]) ; long_multiply($b[0]; $a[1])),
long_multiply($a[1]; $b[1]) ]
| normalize
end ;
# a and/or b may be BigInts, or [p,q] fractions
def multiply(a; b):
a as $a | b as $b
| if ($a|type) == "string" and ($b|type) == "string" then [ long_multiply($a; $b), "1"]
else
if $a|type == "string" then [ long_multiply( $a; $b[0]), $b[1] ]
elif $b|type == "string" then [ long_multiply( $b; $a[0]), $a[1] ]
else [ long_multiply( $a[0]; $b[0]), long_multiply($a[1]; $b[1]) ]
end
| normalize
end ;
def minus(a; b):
a as $a | b as $b
| if $a == $b then ["0", "1"]
else add($a; [ ($b[0]|negate), $b[1] ] )
end ;
Bernoulli Numbers:
# Using the algorithm in the task description:
def bernoulli(n):
reduce range(0; n+1) as $m
( [];
.[$m] = ["1", long_add($m|tostring; "1")] # i.e. 1 / ($m+1)
| reduce ($m - range(0 ; $m)) as $j
(.;
.[$j-1] = multiply( [($j|tostring), "1"]; minus( .[$j-1] ; .[$j]) ) ))
| .[0] # (which is Bn)
;
The task:
range(0;61)
| if . % 2 == 0 or . == 1 then "\(.): \(bernoulli(.) )" else empty end
- Output:
The following output was obtained using the previously mentioned BigInt library.
$ jq -n -r -f Bernoulli.jq
0: ["1","1"]
1: ["1","2"]
2: ["1","6"]
4: ["-1","30"]
6: ["1","42"]
8: ["-1","30"]
10: ["5","66"]
12: ["-691","2730"]
14: ["7","6"]
16: ["-3617","510"]
18: ["43867","798"]
20: ["-174611","330"]
22: ["854513","138"]
24: ["-236364091","2730"]
26: ["8553103","6"]
28: ["-23749461029","870"]
30: ["8615841276005","14322"]
32: ["-7709321041217","510"]
34: ["2577687858367","6"]
36: ["-26315271553053477373","1919190"]
38: ["2929993913841559","6"]
40: ["-261082718496449122051","13530"]
42: ["1520097643918070802691","1806"]
44: ["-27833269579301024235023","690"]
46: ["596451111593912163277961","282"]
48: ["-5609403368997817686249127547","46410"]
50: ["495057205241079648212477525","66"]
52: ["-801165718135489957347924991853","1590"]
54: ["29149963634884862421418123812691","798"]
56: ["-2479392929313226753685415739663229","870"]
58: ["84483613348880041862046775994036021","354"]
60: ["-1215233140483755572040304994079820246041491","56786730"]
Julia
function bernoulli(n)
A = Vector{Rational{BigInt}}(undef, n + 1)
for m = 0 : n
A[m + 1] = 1 // (m + 1)
for j = m : -1 : 1
A[j] = j * (A[j] - A[j + 1])
end
end
return A[1]
end
function display(n)
B = map(bernoulli, 0 : n)
pad = mapreduce(x -> ndigits(numerator(x)) + Int(x < 0), max, B)
argdigits = ndigits(n)
for i = 0 : n
if numerator(B[i + 1]) & 1 == 1
println(
"B(", lpad(i, argdigits), ") = ",
lpad(numerator(B[i + 1]), pad), " / ", denominator(B[i + 1])
)
end
end
end
display(60)
# Alternative: Following the comment in the Perl section it is much more efficient
# to compute the list of numbers instead of one number after the other.
function BernoulliList(len)
A = Vector{Rational{BigInt}}(undef, len + 1)
B = similar(A)
for n in 0 : len
A[n + 1] = 1 // (n + 1)
for j = n : -1 : 1
A[j] = j * (A[j] - A[j + 1])
end
B[n + 1] = A[1]
end
return B
end
for (n, b) in enumerate(BernoulliList(60))
isodd(numerator(b)) && println("B($(n-1)) = $b")
end
Produces virtually the same output as the Python version.
Kotlin
import org.apache.commons.math3.fraction.BigFraction
object Bernoulli {
operator fun invoke(n: Int) : BigFraction {
val A = Array(n + 1, init)
for (m in 0..n)
for (j in m downTo 1)
A[j - 1] = A[j - 1].subtract(A[j]).multiply(integers[j])
return A.first()
}
val max = 60
private val init = { m: Int -> BigFraction(1, m + 1) }
private val integers = Array(max + 1, { m: Int -> BigFraction(m) } )
}
fun main(args: Array<String>) {
for (n in 0..Bernoulli.max)
if (n % 2 == 0 || n == 1)
System.out.printf("B(%-2d) = %-1s%n", n, Bernoulli(n))
}
- Output:
Produces virtually the same output as the Java version.
Lua
LuaJIT version with FFI and GMP library
#!/usr/bin/env luajit
local gmp = require 'gmp' ('libgmp')
local ffi = require'ffi'
local mpz, mpq = gmp.types.z, gmp.types.q
local function mpq_for(buf, op, n)
for i=0,n-1 do
op(buf[i])
end
end
local function bernoulli(rop, n)
local a=ffi.new("mpq_t[?]", n+1)
mpq_for(a, gmp.q_init, n+1)
for m=0,n do
gmp.q_set_ui(a[m],1, m+1)
for j=m,1,-1 do
gmp.q_sub(a[j-1], a[j], a[j-1])
gmp.q_set_ui(rop, j, 1)
gmp.q_mul(a[j-1], a[j-1], rop)
end
end
gmp.q_set(rop,a[0])
mpq_for(a, gmp.q_clear, n+1)
end
do --MAIN
local rop=mpq()
local n,d=mpz(),mpz()
gmp.q_init(rop)
gmp.z_inits(n, d)
local to=arg[1] and tonumber(arg[1]) or 60
local from=arg[2] and tonumber(arg[2]) or 0
if from~=0 then to,from=from,to end
for i=from,to do
bernoulli(rop, i)
if gmp.q_cmp_ui(rop, 0, 1)~=0 then
gmp.q_get_num(n, rop)
gmp.q_get_den(d, rop)
gmp.printf("B(%-2g) = %44Zd / %Zd\n", i, n, d)
end
end
gmp.z_clears(n,d)
gmp.q_clear(rop)
end
- Output:
> time ./bernoulli_gmp.lua B(0 ) = 1 / 1 B(1 ) = -1 / 2 B(2 ) = 1 / 6 B(4 ) = -1 / 30 B(6 ) = 1 / 42 B(8 ) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730 ./bernoulli_gmp.lua 0,02s user 0,00s system 97% cpu 0,022 total
Time compare: Python 0.591 sec, C 0.023 sec, Lua 0.022-0.025
M2000 Interpreter
Using a subset of Class Rational Maximum B(28) (if we use double and not decimal then we get lower maximum: B(20))
module Bernoulli_Numbers {
Class Rational {
numerator as decimal, denominator as decimal
gcd=lambda->0
lcm=lambda->0
operator "+" {
Read l
denom=.lcm(l.denominator, .denominator)
.numerator<=denom/l.denominator*l.numerator+denom/.denominator*.numerator
if .numerator==0 then denom=1
.denominator<=denom
}
Operator Unary {
.numerator-!
}
Operator "-" {
Read l
Call Operator "+", -l
}
Group Real {
value {
link parent numerator, denominator to n, d
=n/d
}
}
Group ToString$ {
value {
link parent numerator, denominator to n, d
=Str$(n)+"/"+Str$(d,"")
}
}
class:
Module Rational (.numerator, .denominator) {
if .denominator=0 then .denominator<=1
while frac(.numerator)<>0 {
.numerator*=10@
.denominator*=10@
}
sgn=Sgn(.numerator)*Sgn(.denominator)
.denominator<=abs(.denominator)
.numerator<=abs(.numerator)*sgn
gcd1=lambda (a as decimal, b as decimal) -> {
if a<b then swap a,b
g=a mod b
while g {
a=b:b=g: g=a mod b
}
=abs(b)
}
gdcval=gcd1(abs(.numerator), .denominator)
if gdcval<.denominator and gdcval<>0 then
.denominator/=gdcval
.numerator/=gdcval
end if
.gcd<=gcd1
.lcm<=lambda gcd=gcd1 (a as decimal, b as decimal) -> {
=a/gcd(a,b)*b
}
}
}
Open "out.txt" for output as #f
r1=Rational(1,1)
b=0
nextBern()
b=1
nextBern()
for b=2 to 28 step 2
nextBern()
next
close #f
sub nextBern()
local m=@bernoulli(b)
Print #F, format$("B({0::-2})={1}",b, m.tostring$)
end sub
function bernoulli(n as decimal)
Local a(1 to n+1)=Rational(1,1), z
local decimal m, j
for m=0 to n
a(m + 1) = Rational(1 , m + 1)
if m<1 then continue
for j = m to 1
z= a(J) - a(j+1) + r1
a(j)=z
nn=z.gcd(j, z.denominator)
a(j).numerator*=j/nn
a(j).denominator/=nn
next
for a(1) {
this=rational(.numerator, .denominator)
}
next
=a(1)
end function
}
Bernoulli_Numbers
- Output:
B( 0)= 1/1 B( 1)= 3/2 B( 2)= 1/6 B( 4)=-1/30 B( 6)= 1/42 B( 8)=-1/30 B(10)= 5/66 B(12)=-691/2730 B(14)= 7/6 B(16)=-3617/510 B(18)= 43867/798 B(20)=-174611/330 B(22)= 854513/138 B(24)=-236364091/2730 B(26)= 8553103/6 B(28)=-23749461029/870
Maple
print(select(n->n[2]<>0,[seq([n,bernoulli(n,1)],n=0..60)]));
- Output:
[[0, 1], [1, 1/2], [2, 1/6], [4, -1/30], [6, 1/42], [8, -1/30], [10, 5/66], [12, -691/2730], [14, 7/6], [16, -3617/510], [18, 43867/798], [20, -174611/330], [22, 854513/138], [24, -236364091/2730], [26, 8553103/6], [28, -23749461029/870], [30, 8615841276005/14322], [32, -7709321041217/510], [34, 2577687858367/6], [36, -26315271553053477373/1919190], [38, 2929993913841559/6], [40, -261082718496449122051/13530], [42, 1520097643918070802691/1806], [44, -27833269579301024235023/690], [46, 596451111593912163277961/282], [48, -5609403368997817686249127547/46410], [50, 495057205241079648212477525/66], [52, -801165718135489957347924991853/1590], [54, 29149963634884862421418123812691/798], [56, -2479392929313226753685415739663229/870], [58, 84483613348880041862046775994036021/354], [60, -1215233140483755572040304994079820246041491/56786730]]
Mathematica / Wolfram Language
Mathematica has no native way for starting an array at index 0. I therefore had to build the array from 1 to n+1 instead of from 0 to n, adjusting the formula accordingly.
bernoulli[n_] := Module[{a = ConstantArray[0, n + 2]},
Do[
a[[m]] = 1/m;
If[m == 1 && a[[1]] != 0, Print[{m - 1, a[[1]]}]];
Do[
a[[j - 1]] = (j - 1)*(a[[j - 1]] - a[[j]]);
If[j == 2 && a[[1]] != 0, Print[{m - 1, a[[1]]}]];
, {j, m, 2, -1}];
, {m, 1, n + 1}];
]
bernoulli[60]
- Output:
{0,1} {1,1/2} {2,1/6} {4,-(1/30)} {6,1/42} {8,-(1/30)} {10,5/66} {12,-(691/2730)} {14,7/6} {16,-(3617/510)} {18,43867/798} {20,-(174611/330)} {22,854513/138} {24,-(236364091/2730)} {26,8553103/6} {28,-(23749461029/870)} {30,8615841276005/14322} {32,-(7709321041217/510)} {34,2577687858367/6} {36,-(26315271553053477373/1919190)} {38,2929993913841559/6} {40,-(261082718496449122051/13530)} {42,1520097643918070802691/1806} {44,-(27833269579301024235023/690)} {46,596451111593912163277961/282} {48,-(5609403368997817686249127547/46410)} {50,495057205241079648212477525/66} {52,-(801165718135489957347924991853/1590)} {54,29149963634884862421418123812691/798} {56,-(2479392929313226753685415739663229/870)} {58,84483613348880041862046775994036021/354} {60,-(1215233140483755572040304994079820246041491/56786730)}
Or, it's permissible to use the native Bernoulli number function instead of being forced to use the specified algorithm, we very simply have:
(Note from task's author: nobody is forced to use any specific algorithm, the one shown is just a suggestion.)
Table[{i, BernoulliB[i]}, {i, 0, 60}];
Select[%, #[[2]] != 0 &] // TableForm
- Output:
0 1 1 -(1/2) 2 1/6 4 -(1/30) 6 1/42 8 -(1/30) 10 5/66 12 -(691/2730) 14 7/6 16 -(3617/510) 18 43867/798 20 -(174611/330) 22 854513/138 24 -(236364091/2730) 26 8553103/6 28 -(23749461029/870) 30 8615841276005/14322 32 -(7709321041217/510) 34 2577687858367/6 36 -(26315271553053477373/1919190) 38 2929993913841559/6 40 -(261082718496449122051/13530) 42 1520097643918070802691/1806 44 -(27833269579301024235023/690) 46 596451111593912163277961/282 48 -(5609403368997817686249127547/46410) 50 495057205241079648212477525/66 52 -(801165718135489957347924991853/1590) 54 29149963634884862421418123812691/798 56 -(2479392929313226753685415739663229/870) 58 84483613348880041862046775994036021/354 60 -(1215233140483755572040304994079820246041491/56786730)
Maxima
Using built-in function bern
block(makelist([sconcat("B","(",i,")","="),bern(i)],i,0,60),
sublist(%%,lambda([x],x[2]#0)),
table_form(%%))
- Output:
matrix( ["B(0)=", 1], ["B(1)=", -1/2], ["B(2)=", 1/6], ["B(4)=", -1/30], ["B(6)=", 1/42], ["B(8)=", -1/30], ["B(10)=", 5/66], ["B(12)=", -691/2730], ["B(14)=", 7/6], ["B(16)=", -3617/510], ["B(18)=", 43867/798], ["B(20)=", -174611/330], ["B(22)=", 854513/138], ["B(24)=", -236364091/2730], ["B(26)=", 8553103/6], ["B(28)=", -23749461029/870], ["B(30)=", 8615841276005/14322], ["B(32)=", -7709321041217/510], ["B(34)=", 2577687858367/6], ["B(36)=", -26315271553053477373/1919190], ["B(38)=", 2929993913841559/6], ["B(40)=", -261082718496449122051/13530], ["B(42)=", 1520097643918070802691/1806], ["B(44)=", -27833269579301024235023/690], ["B(46)=", 596451111593912163277961/282], ["B(48)=", -5609403368997817686249127547/46410], ["B(50)=", 495057205241079648212477525/66], ["B(52)=", -801165718135489957347924991853/1590], ["B(54)=", 29149963634884862421418123812691/798], ["B(56)=", -2479392929313226753685415739663229/870], ["B(58)=", 84483613348880041862046775994036021/354], ["B(60)=", -1215233140483755572040304994079820246041491/56786730] )
Nim
import bignum
import strformat
const Lim = 60
#---------------------------------------------------------------------------------------------------
proc bernoulli(n: Natural): Rat =
## Compute a Bernoulli number using Akiyama–Tanigawa algorithm.
var a = newSeq[Rat](n + 1)
for m in 0..n:
a[m] = newRat(1, m + 1)
for j in countdown(m, 1):
a[j-1] = j * (a[j] - a[j-1])
result = a[0]
#———————————————————————————————————————————————————————————————————————————————————————————————————
type Info = tuple
n: int # Number index in Bernoulli sequence.
val: Rat # Bernoulli number.
var values: seq[Info] # List of values as Info tuples.
var maxLen = -1 # Maximum length.
# First step: compute the values and prepare for display.
for n in 0..Lim:
# Compute value.
if n != 1 and (n and 1) == 1: continue # Ignore odd "n" except 1.
let b = bernoulli(n)
# Check numerator length.
let len = ($b.num).len
if len > maxLen: maxLen = len
# Store information for next step.
values.add((n, b))
# Second step: display the values with '/' aligned.
for (n, b) in values:
let s = fmt"{($b.num).alignString(maxLen, '>')} / {b.denom}"
echo fmt"{n:2}: {s}"
- Output:
0: 1 / 1 1: -1 / 2 2: 1 / 6 4: -1 / 30 6: 1 / 42 8: -1 / 30 10: 5 / 66 12: -691 / 2730 14: 7 / 6 16: -3617 / 510 18: 43867 / 798 20: -174611 / 330 22: 854513 / 138 24: -236364091 / 2730 26: 8553103 / 6 28: -23749461029 / 870 30: 8615841276005 / 14322 32: -7709321041217 / 510 34: 2577687858367 / 6 36: -26315271553053477373 / 1919190 38: 2929993913841559 / 6 40: -261082718496449122051 / 13530 42: 1520097643918070802691 / 1806 44: -27833269579301024235023 / 690 46: 596451111593912163277961 / 282 48: -5609403368997817686249127547 / 46410 50: 495057205241079648212477525 / 66 52: -801165718135489957347924991853 / 1590 54: 29149963634884862421418123812691 / 798 56: -2479392929313226753685415739663229 / 870 58: 84483613348880041862046775994036021 / 354 60: -1215233140483755572040304994079820246041491 / 56786730
PARI/GP
for(n=0,60,t=bernfrac(n);if(t,print(n" "t)))
- Output:
0 1 1 -1/2 2 1/6 4 -1/30 6 1/42 8 -1/30 10 5/66 12 -691/2730 14 7/6 16 -3617/510 18 43867/798 20 -174611/330 22 854513/138 24 -236364091/2730 26 8553103/6 28 -23749461029/870 30 8615841276005/14322 32 -7709321041217/510 34 2577687858367/6 36 -26315271553053477373/1919190 38 2929993913841559/6 40 -261082718496449122051/13530 42 1520097643918070802691/1806 44 -27833269579301024235023/690 46 596451111593912163277961/282 48 -5609403368997817686249127547/46410 50 495057205241079648212477525/66 52 -801165718135489957347924991853/1590 54 29149963634884862421418123812691/798 56 -2479392929313226753685415739663229/870 58 84483613348880041862046775994036021/354 60 -1215233140483755572040304994079820246041491/56786730
Pascal
Tested with fpc 3.0.4
(* Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99 *)
program BernoulliForAda99;
uses BigDecimalMath; {library for arbitary high precision BCD numbers}
type
Fraction = object
private
numerator, denominator: BigDecimal;
public
procedure assign(n, d: Int64);
procedure subtract(rhs: Fraction);
procedure multiply(value: Int64);
procedure reduce();
procedure writeOutput();
end;
function gcd(a, b: BigDecimal):BigDecimal;
begin
if (b = 0) then begin
gcd := a;
end
else begin
gcd := gcd(b, a mod b);
end;
end;
procedure Fraction.writeOutput();
var sign : char;
begin
sign := ' ';
if (numerator<0) then sign := '-';
if (denominator<0) then sign := '-';
write(sign + BigDecimalToStr(abs(numerator)):45);
write(' / ');
write(BigDecimalToStr(abs(denominator)));
end;
procedure Fraction.assign(n, d: Int64);
begin
numerator := n;
denominator := d;
end;
procedure Fraction.subtract(rhs: Fraction);
begin
numerator := numerator * rhs.denominator;
numerator := numerator - (rhs.numerator * denominator);
denominator := denominator * rhs.denominator;
end;
procedure Fraction.multiply(value: Int64);
var
temp :BigDecimal;
begin
temp := value;
numerator := numerator * temp;
end;
procedure Fraction.reduce();
var gcdResult: BigDecimal;
begin
gcdResult := gcd(numerator, denominator);
begin
numerator := numerator div gcdResult; (* div is Int64 division *)
denominator := denominator div gcdResult; (* could also use round(d/r) *)
end;
end;
function calculateBernoulli(n: Int64) : Fraction;
var
m, j: Int64;
results: array of Fraction;
begin
setlength(results, 60) ; {largest value 60}
for m:= 0 to n do
begin
results[m].assign(1, m+1);
for j:= m downto 1 do
begin
results[j-1].subtract(results[j]);
results[j-1].multiply(j);
results[j-1].reduce();
end;
end;
calculateBernoulli := results[0];
end;
(* Main program starts here *)
var
b: Int64;
result: Fraction;
begin
writeln('Calculating Bernoulli numbers...');
writeln('B( 0) : 1 / 1');
for b:= 1 to 60 do
begin
if (b<3) or ((b mod 2) = 0) then begin
result := calculateBernoulli(b);
write('B(',b:2,')');
write(' : ');
result.writeOutput();
writeln;
end;
end;
end.
- Output:
Calculating Bernoulli numbers... B( 0) : 1 / 1 B( 1) : 1 / 2 B( 2) : 1 / 6 B( 4) : -1 / 30 B( 6) : 1 / 42 B( 8) : -1 / 30 B(10) : 5 / 66 B(12) : -691 / 2730 B(14) : -7 / 6 B(16) : -3617 / 510 B(18) : 43867 / 798 B(20) : -174611 / 330 B(22) : 854513 / 138 B(24) : -236364091 / 2730 B(26) : 8553103 / 6 B(28) : -23749461029 / 870 B(30) : 8615841276005 / 14322 B(32) : -7709321041217 / 510 B(34) : 2577687858367 / 6 B(36) : -26315271553053477373 / 1919190 B(38) : 2929993913841559 / 6 B(40) : -261082718496449122051 / 13530 B(42) : 1520097643918070802691 / 1806 B(44) : -27833269579301024235023 / 690 B(46) : -596451111593912163277961 / 282 B(48) : -5609403368997817686249127547 / 46410 B(50) : 495057205241079648212477525 / 66 B(52) : -801165718135489957347924991853 / 1590 B(54) : 29149963634884862421418123812691 / 798 B(56) : -2479392929313226753685415739663229 / 870 B(58) : 84483613348880041862046775994036021 / 354 B(60) : -1215233140483755572040304994079820246041491 / 56786730
Perl
The only thing in the suggested algorithm which depends on N is the number of times through the inner block. This means that all but the last iteration through the loop produce the exact same values of A.
Instead of doing the same calculations over and over again, I retain the A array until the final Bernoulli number is produced.
#!perl
use strict;
use warnings;
use List::Util qw(max);
use Math::BigRat;
my $one = Math::BigRat->new(1);
sub bernoulli_print {
my @a;
for my $m ( 0 .. 60 ) {
push @a, $one / ($m + 1);
for my $j ( reverse 1 .. $m ) {
# This line:
( $a[$j-1] -= $a[$j] ) *= $j;
# is a faster version of the following line:
# $a[$j-1] = $j * ($a[$j-1] - $a[$j]);
# since it avoids unnecessary object creation.
}
next unless $a[0];
printf "B(%2d) = %44s/%s\n", $m, $a[0]->parts;
}
}
bernoulli_print();
The output is exactly the same as the Python entry.
We can also use modules for faster results. E.g.
use ntheory qw/bernfrac/;
for my $n (0 .. 60) {
my($num,$den) = bernfrac($n);
printf "B(%2d) = %44s/%s\n", $n, $num, $den if $num != 0;
}
with identical output. Or:
use Math::Pari qw/bernfrac/;
for my $n (0 .. 60) {
my($num,$den) = split "/", bernfrac($n);
printf("B(%2d) = %44s/%s\n", $n, $num, $den||1) if $num != 0;
}
with the difference being that Pari chooses = -½.
Phix
with javascript_semantics include builtins/mpfr.e procedure bernoulli(mpq rop, integer n) sequence a = mpq_inits(n+1) for m=1 to n+1 do mpq_set_si(a[m], 1, m) for j=m-1 to 1 by -1 do mpq_sub(a[j], a[j+1], a[j]) mpq_set_si(rop, j, 1) mpq_mul(a[j], a[j], rop) end for end for mpq_set(rop, a[1]) a = mpq_free(a) end procedure mpq rop = mpq_init() mpz n = mpz_init(), d = mpz_init() for i=0 to 60 do bernoulli(rop, i) if mpq_cmp_si(rop, 0, 1) then mpq_get_num(n, rop) mpq_get_den(d, rop) string ns = mpz_get_str(n), ds = mpz_get_str(d) printf(1,"B(%2d) = %44s / %s\n", {i,ns,ds}) end if end for {n,d} = mpz_free({n,d}) rop = mpq_free(rop)
- Output:
B( 0) = 1 / 1 B( 1) = -1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
PicoLisp
Brute force and method by Srinivasa Ramanujan.
(load "@lib/frac.l")
(de fact (N)
(cache '(NIL) N
(if (=0 N) 1 (apply * (range 1 N))) ) )
(de binomial (N K)
(frac
(/
(fact N)
(* (fact (- N K)) (fact K)) )
1 ) )
(de A (N M)
(let Sum (0 . 1)
(for X M
(setq Sum
(f+
Sum
(f*
(binomial (+ N 3) (- N (* X 6)))
(berno (- N (* X 6)) ) ) ) ) )
Sum ) )
(de berno (N)
(cache '(NIL) N
(cond
((=0 N) (1 . 1))
((= 1 N) (-1 . 2))
((bit? 1 N) (0 . 1))
(T
(case (% N 6)
(0
(f/
(f-
(frac (+ N 3) 3)
(A N (/ N 6)) )
(binomial (+ N 3) N) ) )
(2
(f/
(f-
(frac (+ N 3) 3)
(A N (/ (- N 2) 6)) )
(binomial (+ N 3) N) ) )
(4
(f/
(f-
(f* (-1 . 1) (frac (+ N 3) 6))
(A N (/ (- N 4) 6)) )
(binomial (+ N 3) N) ) ) ) ) ) ) )
(de berno-brute (N)
(cache '(NIL) N
(let Sum (0 . 1)
(cond
((=0 N) (1 . 1))
((= 1 N) (-1 . 2))
((bit? 1 N) (0 . 1))
(T
(for (X 0 (> N X) (inc X))
(setq Sum
(f+
Sum
(f* (binomial (inc N) X) (berno-brute X)) ) ) )
(f/ (f* (-1 . 1) Sum) (binomial (inc N) N)) ) ) ) ) )
(for (N 0 (> 62 N) (inc N))
(if (or (= N 1) (not (bit? 1 N)))
(tab (2 4 -60) N " => " (sym (berno N))) ) )
(for (N 0 (> 400 N) (inc N))
(test (berno N) (berno-brute N)) )
(bye)
PL/I
Bern: procedure options (main); /* 4 July 2014 */
declare i fixed binary;
declare B complex fixed (31);
Bernoulli: procedure (n) returns (complex fixed (31));
declare n fixed binary;
declare anum(0:n) fixed (31), aden(0:n) fixed (31);
declare (j, m) fixed;
declare F fixed (31);
do m = 0 to n;
anum(m) = 1;
aden(m) = m+1;
do j = m to 1 by -1;
anum(j-1) = j*( aden(j)*anum(j-1) - aden(j-1)*anum(j) );
aden(j-1) = ( aden(j-1) * aden(j) );
F = gcd(abs(anum(j-1)), abs(aden(j-1)) );
if F ^= 1 then
do;
anum(j-1) = anum(j-1) / F;
aden(j-1) = aden(j-1) / F;
end;
end;
end;
return ( complex(anum(0), aden(0)) );
end Bernoulli;
do i = 0, 1, 2 to 36 by 2; /* 36 is upper limit imposed by hardware. */
B = Bernoulli(i);
put skip edit ('B(' , trim(i) , ')=' , real(B) , '/' , trim(imag(B)) )
(3 A, column(10), F(32), 2 A);
end;
end Bern;
The above uses GCD (see Rosetta Code) extended for 31-digit working.
Results obtained by this program are limited to the entries shown below due to the restrictions imposed by storing numbers in fixed decimal (31 digits).
B(0)= 1/1 B(1)= 1/2 B(2)= 1/6 B(4)= -1/30 B(6)= 1/42 B(8)= -1/30 B(10)= 5/66 B(12)= -691/2730 B(14)= 7/6 B(16)= -3617/510 B(18)= 43867/798 B(20)= -174611/330 B(22)= 854513/138 B(24)= -236364091/2730 B(26)= 8553103/6 B(28)= -23749461029/870 B(30)= 8615841276005/14322 B(32)= -7709321041217/510 B(34)= 2577687858367/6 B(36)= -26315271553053477373/1919190
Python
Python: Using task algorithm
from fractions import Fraction as Fr
def bernoulli(n):
A = [0] * (n+1)
for m in range(n+1):
A[m] = Fr(1, m+1)
for j in range(m, 0, -1):
A[j-1] = j*(A[j-1] - A[j])
return A[0] # (which is Bn)
bn = [(i, bernoulli(i)) for i in range(61)]
bn = [(i, b) for i,b in bn if b]
width = max(len(str(b.numerator)) for i,b in bn)
for i,b in bn:
print('B(%2i) = %*i/%i' % (i, width, b.numerator, b.denominator))
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Python: Optimised task algorithm
Using the optimization mentioned in the Perl entry to reduce intermediate calculations we create and use the generator bernoulli2():
def bernoulli2():
A, m = [], 0
while True:
A.append(Fr(1, m+1))
for j in range(m, 0, -1):
A[j-1] = j*(A[j-1] - A[j])
yield A[0] # (which is Bm)
m += 1
bn2 = [ix for ix in zip(range(61), bernoulli2())]
bn2 = [(i, b) for i,b in bn2 if b]
width = max(len(str(b.numerator)) for i,b in bn2)
for i,b in bn2:
print('B(%2i) = %*i/%i' % (i, width, b.numerator, b.denominator))
Output is exactly the same as before.
Quackery
$ "bigrat.qky" loadfile
[ 1+
' [ [] ] over of swap
times
[ i^ 1+ n->v 1/v
join swap i^ poke
i^ times
[ dup i 1+ peek do
dip over swap i peek do
v- i 1+ n->v v*
join swap i poke ] ]
1 split drop do ] is bernoulli ( n --> n/d )
61 times
[ i^ bernoulli
2dup v0= iff
2drop
else
[ i^ 10 < if sp
i^ echo sp
vulgar$
char / over find
44 swap - times sp
echo$ cr ] ]
- Output:
0 1/1 1 -1/2 2 1/6 4 -1/30 6 1/42 8 -1/30 10 5/66 12 -691/2730 14 7/6 16 -3617/510 18 43867/798 20 -174611/330 22 854513/138 24 -236364091/2730 26 8553103/6 28 -23749461029/870 30 8615841276005/14322 32 -7709321041217/510 34 2577687858367/6 36 -26315271553053477373/1919190 38 2929993913841559/6 40 -261082718496449122051/13530 42 1520097643918070802691/1806 44 -27833269579301024235023/690 46 596451111593912163277961/282 48 -5609403368997817686249127547/46410 50 495057205241079648212477525/66 52 -801165718135489957347924991853/1590 54 29149963634884862421418123812691/798 56 -2479392929313226753685415739663229/870 58 84483613348880041862046775994036021/354 60 -1215233140483755572040304994079820246041491/56786730
R
library(pracma)
for (idx in c(1,2*0:30)) {
b <- bernoulli(idx)
d <- as.character(denominator(b))
n <- as.character(numerator(b))
cat("B(",idx,") = ",n,"/",d,"\n", sep = "")
}
- Output:
B(1) = 1/2 B(0) = 1/1 B(2) = 1/6 B(4) = -1/30 B(6) = 1/42 B(8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Racket
This implements, firstly, the algorithm specified with the task... then the better performing bernoulli.3, which uses the "double sum formula" listed under REXX. The number generators all (there is also a bernoulli.2) use the same emmitter... it's just a matter of how long to wait for the emission.
#lang racket
;; For: http://rosettacode.org/wiki/Bernoulli_numbers
;; As described in task...
(define (bernoulli.1 n)
(define A (make-vector (add1 n)))
(for ((m (in-range 0 (add1 n))))
(vector-set! A m (/ (add1 m)))
(for ((j (in-range m (sub1 1) -1)))
(define new-A_j-1 (* j (- (vector-ref A (sub1 j)) (vector-ref A j))))
(vector-set! A (sub1 j) new-A_j-1)))
(vector-ref A 0))
(define (non-zero-bernoulli-indices s)
(sequence-filter (λ (n) (or (even? n) (= n 1))) s))
(define (bernoulli_0..n B N)
(for/list ((n (non-zero-bernoulli-indices (in-range (add1 N))))) (B n)))
;; From REXX description / http://mathworld.wolfram.com/BernoulliNumber.html #33
;; ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
;; bernoulli.2 is for illustrative purposes, binomial is very costly if there is no memoisation
;; (which math/number-theory doesn't do)
(require (only-in math/number-theory binomial))
(define (bernoulli.2 n)
(for/sum ((k (in-range 0 (add1 n))))
(* (/ (add1 k))
(for/sum ((r (in-range 0 (add1 k))))
(* (expt -1 r) (binomial k r) (expt r n))))))
;; Three things to do:
;; 1. (expt -1 r): is 1 for even r, -1 for odd r... split the sum between those two.
;; 2. splitting the sum might has arithmetic advantages, too. We're using rationals, so the smaller
;; summations should require less normalisation of intermediate, fractional results
;; 3. a memoised binomial... although the one from math/number-theory is fast, it is (and its
;; factorials are) computed every time which is redundant
(define kCr-memo (make-hasheq))
(define !-memo (make-vector 1000 #f))
(vector-set! !-memo 0 1) ;; seed the memo
(define (! k)
(cond [(vector-ref !-memo k) => values]
[else (define k! (* k (! (- k 1)))) (vector-set! !-memo k k!) k!]))
(define (kCr k r)
; If we want (kCr ... r>1000000) we'll have to reconsider this. However, until then...
(define hash-key (+ (* 1000000 k) r))
(hash-ref! kCr-memo hash-key (λ () (/ (! k) (! r) (! (- k r))))))
(define (bernoulli.3 n)
(for/sum ((k (in-range 0 (add1 n))))
(define k+1 (add1 k))
(* (/ k+1)
(- (for/sum ((r (in-range 0 k+1 2))) (* (kCr k r) (expt r n)))
(for/sum ((r (in-range 1 k+1 2))) (* (kCr k r) (expt r n)))))))
(define (display/align-fractions caption/idx-fmt Bs)
;; widths are one more than the order of magnitude
(define oom+1 (compose add1 order-of-magnitude))
(define-values (I-width N-width D-width)
(for/fold ((I 0) (N 0) (D 0))
((b Bs) (n (non-zero-bernoulli-indices (in-naturals))))
(define +b (abs b))
(values (max I (oom+1 (max n 1)))
(max N (+ (oom+1 (numerator +b)) (if (negative? b) 1 0)))
(max D (oom+1 (denominator +b))))))
(define (~a/w/a n w a) (~a n #:width w #:align a))
(for ((n (non-zero-bernoulli-indices (in-naturals))) (b Bs))
(printf "~a ~a/~a~%"
(format caption/idx-fmt (~a/w/a n I-width 'right))
(~a/w/a (numerator b) N-width 'right)
(~a/w/a (denominator b) D-width 'left))))
(module+ main
(display/align-fractions "B(~a) =" (bernoulli_0..n bernoulli.3 60)))
(module+ test
(require rackunit)
; correctness and timing tests
(check-match (time (bernoulli_0..n bernoulli.1 60))
(list 1/1 (app abs 1/2) 1/6 -1/30 1/42 -1/30 _ ...))
(check-match (time (bernoulli_0..n bernoulli.2 60))
(list 1/1 (app abs 1/2) 1/6 -1/30 1/42 -1/30 _ ...))
(check-match (time (bernoulli_0..n bernoulli.3 60))
(list 1/1 (app abs 1/2) 1/6 -1/30 1/42 -1/30 _ ...))
; timing only ...
(void (time (bernoulli_0..n bernoulli.3 100))))
- Output:
B( 0) = 1/1 B( 1) = -1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Raku
(formerly Perl 6)
Simple
First, a straighforward implementation of the naïve algorithm in the task description.
sub bernoulli($n) {
my @a;
for 0..$n -> $m {
@a[$m] = FatRat.new(1, $m + 1);
for reverse 1..$m -> $j {
@a[$j - 1] = $j * (@a[$j - 1] - @a[$j]);
}
}
return @a[0];
}
constant @bpairs = grep *.value.so, ($_ => bernoulli($_) for 0..60);
my $width = max @bpairs.map: *.value.numerator.chars;
my $form = "B(%2d) = \%{$width}d/%d\n";
printf $form, .key, .value.nude for @bpairs;
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
With memoization
Here is a much faster way, following the Perl solution that avoids recalculating previous values each time through the function. We do this in Raku by not defining it as a function at all, but by defining it as an infinite sequence that we can read however many values we like from (52, in this case, to get up to B(100)). In this solution we've also avoided subscripting operations; rather we use a sequence operator (...) iterated over the list of the previous solution to find the next solution. We reverse the array in this case to make reference to the previous value in the list more natural, which means we take the last value of the list rather than the first value, and do so conditionally to avoid 0 values.
constant bernoulli = gather {
my @a;
for 0..* -> $m {
@a = FatRat.new(1, $m + 1),
-> $prev {
my $j = @a.elems;
$j * (@a.shift - $prev);
} ... { not @a.elems }
take $m => @a[*-1] if @a[*-1];
}
}
constant @bpairs = bernoulli[^52];
my $width = max @bpairs.map: *.value.numerator.chars;
my $form = "B(%d)\t= \%{$width}d/%d\n";
printf $form, .key, .value.nude for @bpairs;
- Output:
B(0) = 1/1 B(1) = 1/2 B(2) = 1/6 B(4) = -1/30 B(6) = 1/42 B(8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730 B(62) = 12300585434086858541953039857403386151/6 B(64) = -106783830147866529886385444979142647942017/510 B(66) = 1472600022126335654051619428551932342241899101/64722 B(68) = -78773130858718728141909149208474606244347001/30 B(70) = 1505381347333367003803076567377857208511438160235/4686 B(72) = -5827954961669944110438277244641067365282488301844260429/140100870 B(74) = 34152417289221168014330073731472635186688307783087/6 B(76) = -24655088825935372707687196040585199904365267828865801/30 B(78) = 414846365575400828295179035549542073492199375372400483487/3318 B(80) = -4603784299479457646935574969019046849794257872751288919656867/230010 B(82) = 1677014149185145836823154509786269900207736027570253414881613/498 B(84) = -2024576195935290360231131160111731009989917391198090877281083932477/3404310 B(86) = 660714619417678653573847847426261496277830686653388931761996983/6 B(88) = -1311426488674017507995511424019311843345750275572028644296919890574047/61410 B(90) = 1179057279021082799884123351249215083775254949669647116231545215727922535/272118 B(92) = -1295585948207537527989427828538576749659341483719435143023316326829946247/1410 B(94) = 1220813806579744469607301679413201203958508415202696621436215105284649447/6 B(96) = -211600449597266513097597728109824233673043954389060234150638733420050668349987259/4501770 B(98) = 67908260672905495624051117546403605607342195728504487509073961249992947058239/6 B(100) = -94598037819122125295227433069493721872702841533066936133385696204311395415197247711/33330
Functional
And if you're a pure enough FP programmer to dislike destroying and reconstructing the array each time, here's the same algorithm without side effects. We use zip with the pair constructor => to keep values associated with their indices. This provides sufficient local information that we can define our own binary operator "bop" to reduce between each two terms, using the "triangle" form (called "scan" in Haskell) to return the intermediate results that will be important to compute the next Bernoulli number.
sub infix:<bop>(\prev, \this) {
this.key => this.key * (this.value - prev.value)
}
sub next-bernoulli ( (:key($pm), :value(@pa)) ) {
$pm + 1 => [
map *.value,
[\bop] ($pm + 2 ... 1) Z=> FatRat.new(1, $pm + 2), |@pa
]
}
constant bernoulli =
grep *.value,
map { .key => .value[*-1] },
(0 => [FatRat.new(1,1)], &next-bernoulli ... *)
;
constant @bpairs = bernoulli[^52];
my $width = max @bpairs.map: *.value.numerator.chars;
my $form = "B(%d)\t= \%{$width}d/%d\n";
printf $form, .key, .value.nude for @bpairs;
Same output as memoization example
REXX
The double sum formula used is number (33) from the entry Bernoulli number on Wolfram MathWorldTM.
- where is a binomial coefficient.
- where is a binomial coefficient.
/*REXX program calculates N number of Bernoulli numbers expressed as vulgar fractions.*/
parse arg N .; if N=='' | N=="," then N= 60 /*Not specified? Then use the default.*/
numeric digits max(9, n*2) /*increase the decimal digits if needed*/
w= max(length(N), 4); Nw= N + w + N % 4 /*used for aligning (output) fractions.*/
say 'B(n)' center("Bernoulli numbers expressed as vulgar fractions", max(78-w, Nw) )
say copies('─',w) copies("─", max(78-w,Nw+2*w)) /*display 2nd line of title, separators*/
!.= .; do #=0 to N /*process the numbers from 0 ──► N. */
b= bern(#); if b==0 then iterate /*calculate Bernoulli number, skip if 0*/
indent= max(0, nW - pos('/', b) ) /*calculate the alignment (indentation)*/
say right(#, w) left('', indent) b /*display the indented Bernoulli number*/
end /*#*/ /* [↑] align the Bernoulli fractions. */
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
bern: parse arg x; if x==0 then return '1/1' /*handle the special case of zero. */
if x==1 then return '-1/2' /* " " " " " one. */
if x//2 then return 0 /* " " " " " odds > 1.*/
do j=2 to x by 2; jp= j+1 /*process the positive integers up to X*/
sn= 1 - j /*define the numerator. */
sd= 2 /* " " denominator. */
do k=2 to j-1 by 2 /*calculate a SN/SD sequence. */
parse var @.k bn '/' ad /*get a previously calculated fraction.*/
an= comb(jp, k) * bn /*use COMBination for the next term. */
$lcm= LCM(sd, ad) /*use Least Common Denominator function*/
sn= $lcm % sd * sn; sd= $lcm /*calculate the current numerator. */
an= $lcm % ad * an /* " " next " */
sn= sn + an /* " " current " */
end /*k*/ /* [↑] calculate the SN/SD sequence.*/
sn= -sn /*flip the sign for the numerator. */
sd= sd * jp /*calculate the denominator. */
if sn\==1 then do; _= GCD(sn, sd) /*get the Greatest Common Denominator.*/
sn= sn%_; sd= sd%_ /*reduce the numerator and denominator.*/
end /* [↑] done with the reduction(s). */
@.j= sn'/'sd /*save the result for the next round. */
end /*j*/ /* [↑] done calculating Bernoulli #'s.*/
return sn'/'sd
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure expose !.; parse arg x,y; if x==y then return 1
if !.C.x.y\==. then return !.C.x.y /*combination computed before?*/
if x-y < y then y= x-y /*x-y < y? Then use a new Y.*/
z= perm(x, y); do j=2 for y-1; z= z % j
end /*j*/
!.C.x.y= z; return z /*assign memoization & return.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
GCD: procedure; parse arg x,y; x= abs(x)
do until y==0; parse value x//y y with y x; end; return x
/*──────────────────────────────────────────────────────────────────────────────────────*/
LCM: procedure; parse arg x,y /*X=ABS(X); Y=ABS(Y) not needed for Bernoulli #s.*/
/*IF Y==0 THEN RETURN 0 " " " " " */
$= x * y /*calculate part of the LCM here. */
do until y==0; parse value x//y y with y x
end /*until*/ /* [↑] this is a short & fast GCD*/
return $ % x /*divide the pre─calculated value.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
perm: procedure expose !.; parse arg x,y; if !.P.x.y\==. then return !.P.x.y
z= 1; do j=x-y+1 to x; z= z*j; end; !.P.x.y= z; return z
- output when using the default input:
B(n) Bernoulli numbers expressed as vulgar fractions ──── ─────────────────────────────────────────────────────────────────────────────────────── 0 1/1 1 -1/2 2 1/6 4 -1/30 6 1/42 8 -1/30 10 5/66 12 -691/2730 14 7/6 16 -3617/510 18 43867/798 20 -174611/330 22 854513/138 24 -236364091/2730 26 8553103/6 28 -23749461029/870 30 8615841276005/14322 32 -7709321041217/510 34 2577687858367/6 36 -26315271553053477373/1919190 38 2929993913841559/6 40 -261082718496449122051/13530 42 1520097643918070802691/1806 44 -27833269579301024235023/690 46 596451111593912163277961/282 48 -5609403368997817686249127547/46410 50 495057205241079648212477525/66 52 -801165718135489957347924991853/1590 54 29149963634884862421418123812691/798 56 -2479392929313226753685415739663229/870 58 84483613348880041862046775994036021/354 60 -1215233140483755572040304994079820246041491/56786730
Output notes: This version of REXX can compute and display all values up to B110 in sub─second.
RPL
Fractions such as a/b are here handled through the complex number data structure (a,b)
.
2 local words support the algorithm suggested by the task: FracSub
substracts 2 fractions and FracSimp
make a fraction irreducible
Unfortunately, floating-point precision prevents from going beyond B22.
RPL code | Comment |
---|---|
≪ DUP2 IM * ROT IM ROT RE * - ≫ 'FracSub' STO ≪ DUP C→R ABS SWAP ABS DUP2 < ≪ SWAP ≫ IFT WHILE DUP REPEAT SWAP OVER MOD END DROP / ≫ 'FracSimp' STO ≪ { } 1 ROT 1 + FOR m 1 m R→C + IF m 2 ≥ THEN m 2 FOR j DUP j 1 - GET OVER j GET FracSub C→R SWAP j 1 - * SWAP R→C FracSimp j 1 - SWAP PUT -1 STEP END NEXT 1 GET DUP RE SWAP 0 IFTE ≫ 'BRNOU' STO |
( (a,b) (c,d) -- (e,f) ) with e/f = a/b - c/d ( (a,b) -- (c,d) ) with c/d simplified fraction of a/b get GCD of a and b divide (a,b) by GCD ( n -- (a,b) ) with a/b = B(n) For m from 1 to n+1 do A[m] ← 1/m for j from m to 2 do (A[j-1] - A[j]) . * (j-1) A[j-1] ← return A[1] as a fraction or zero |
5 BRNOU 22 BRNOU
- Output:
2: 0 1: (854513,138)
HP-49+ version
Latest RPL implementations can natively handle long fractions and generate Bernoulli numbers.
≪ { }
0 ROT FOR n
IF n 2 > LASTARG MOD AND NOT THEN n IBERNOULLI + END
NEXT
≫ 'TASK' STO
60 TASK
- Output:
1: {1 -1/2 1/6 -1/30 1/42 -1/30 5/66 -691/2730 7/6 -3617/510 43867/798 -174611/330 854513/138 -236364091/2730 8553103/6 -23749461029/870 8615841276005/14322 -7709321041217/510 2577687858367/6 -26315271553053477373/1919190 2929993913841559/6 -261082718496449122051/13530 1520097643918070802691/1806 -27833269579301024235023/690 596451111593912163277961/282 -5609403368997817686249127547/46410 495057205241079648212477525/66 -801165718135489957347924991853/1590 9149963634884862421418123812691/798 -2479392929313226753685415739663229/870 84483613348880041862046775994036021/354 -1215233140483755572040304994079820246041491/56786730}
Runs in 3 minutes 40 on a HP-50g, against 1 hour and 30 minutes if calculating Bernoulli numbers with the above function.
Ruby
bernoulli = Enumerator.new do |y|
ar = []
0.step do |m|
ar << Rational(1, m+1)
m.downto(1){|j| ar[j-1] = j*(ar[j-1] - ar[j]) }
y << ar.first # yield
end
end
b_nums = bernoulli.take(61)
width = b_nums.map{|b| b.numerator.to_s.size}.max
b_nums.each_with_index {|b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? }
- Output:
B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Rust
// 2.5 implementations presented here: naive, optimized, and an iterator using
// the optimized function. The speeds vary significantly: relative
// speeds of optimized:iterator:naive implementations is 625:25:1.
#![feature(test)]
extern crate num;
extern crate test;
use num::bigint::{BigInt, ToBigInt};
use num::rational::{BigRational};
use std::cmp::max;
use std::env;
use std::ops::{Mul, Sub};
use std::process;
struct Bn {
value: BigRational,
index: i32
}
struct Context {
bigone_const: BigInt,
a: Vec<BigRational>,
index: i32 // Counter for iterator implementation
}
impl Context {
pub fn new() -> Context {
let bigone = 1.to_bigint().unwrap();
let a_vec: Vec<BigRational> = vec![];
Context {
bigone_const: bigone,
a: a_vec,
index: -1
}
}
}
impl Iterator for Context {
type Item = Bn;
fn next(&mut self) -> Option<Bn> {
self.index += 1;
Some(Bn { value: bernoulli(self.index as usize, self), index: self.index })
}
}
fn help() {
println!("Usage: bernoulli_numbers <up_to>");
}
fn main() {
let args: Vec<String> = env::args().collect();
let mut up_to: usize = 60;
match args.len() {
1 => {},
2 => {
up_to = args[1].parse::<usize>().unwrap();
},
_ => {
help();
process::exit(0);
}
}
let context = Context::new();
// Collect the solutions by using the Context iterator
// (this is not as fast as calling the optimized function directly).
let res = context.take(up_to + 1).collect::<Vec<_>>();
let width = res.iter().fold(0, |a, r| max(a, r.value.numer().to_string().len()));
for r in res.iter().filter(|r| *r.value.numer() != ToBigInt::to_bigint(&0).unwrap()) {
println!("B({:>2}) = {:>2$} / {denom}", r.index, r.value.numer(), width,
denom = r.value.denom());
}
}
// Implementation with no reused calculations.
fn _bernoulli_naive(n: usize, c: &mut Context) -> BigRational {
for m in 0..n + 1 {
c.a.push(BigRational::new(c.bigone_const.clone(), (m + 1).to_bigint().unwrap()));
for j in (1..m + 1).rev() {
c.a[j - 1] = (c.a[j - 1].clone().sub(c.a[j].clone())).mul(
BigRational::new(j.to_bigint().unwrap(), c.bigone_const.clone())
);
}
}
c.a[0].reduced()
}
// Implementation with reused calculations (does not require sequential calls).
fn bernoulli(n: usize, c: &mut Context) -> BigRational {
for i in 0..n + 1 {
if i >= c.a.len() {
c.a.push(BigRational::new(c.bigone_const.clone(), (i + 1).to_bigint().unwrap()));
for j in (1..i + 1).rev() {
c.a[j - 1] = (c.a[j - 1].clone().sub(c.a[j].clone())).mul(
BigRational::new(j.to_bigint().unwrap(), c.bigone_const.clone())
);
}
}
}
c.a[0].reduced()
}
#[cfg(test)]
mod tests {
use super::{Bn, Context, bernoulli, _bernoulli_naive};
use num::rational::{BigRational};
use std::str::FromStr;
use test::Bencher;
// [tests elided]
#[bench]
fn bench_bernoulli_naive(b: &mut Bencher) {
let mut context = Context::new();
b.iter(|| {
let mut res: Vec<Bn> = vec![];
for n in 0..30 + 1 {
let b = _bernoulli_naive(n, &mut context);
res.push(Bn { value:b.clone(), index: n as i32});
}
});
}
#[bench]
fn bench_bernoulli(b: &mut Bencher) {
let mut context = Context::new();
b.iter(|| {
let mut res: Vec<Bn> = vec![];
for n in 0..30 + 1 {
let b = bernoulli(n, &mut context);
res.push(Bn { value:b.clone(), index: n as i32});
}
});
}
#[bench]
fn bench_bernoulli_iter(b: &mut Bencher) {
b.iter(|| {
let context = Context::new();
let _res = context.take(30 + 1).collect::<Vec<_>>();
});
}
}
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Scala
With Custom Rational Number Class
(code will run in Scala REPL with a cut-and-paste without need for a third-party library)
/** Roll our own pared-down BigFraction class just for these Bernoulli Numbers */
case class BFraction( numerator:BigInt, denominator:BigInt ) {
require( denominator != BigInt(0), "Denominator cannot be zero" )
val gcd = numerator.gcd(denominator)
val num = numerator / gcd
val den = denominator / gcd
def unary_- = BFraction(-num, den)
def -( that:BFraction ) = that match {
case f if f.num == BigInt(0) => this
case f if f.den == this.den => BFraction(this.num - f.num, this.den)
case f => BFraction(((this.num * f.den) - (f.num * this.den)), this.den * f.den )
}
def *( that:Int ) = BFraction( num * that, den )
override def toString = num + " / " + den
}
def bernoulliB( n:Int ) : BFraction = {
val aa : Array[BFraction] = Array.ofDim(n+1)
for( m <- 0 to n ) {
aa(m) = BFraction(1,(m+1))
for( n <- m to 1 by -1 ) {
aa(n-1) = (aa(n-1) - aa(n)) * n
}
}
aa(0)
}
assert( {val b12 = bernoulliB(12); b12.num == -691 && b12.den == 2730 } )
val r = for( n <- 0 to 60; b = bernoulliB(n) if b.num != 0 ) yield (n, b)
val numeratorSize = r.map(_._2.num.toString.length).max
// Print the results
r foreach{ case (i,b) => {
val label = f"b($i)"
val num = (" " * (numeratorSize - b.num.toString.length)) + b.num
println( f"$label%-6s $num / ${b.den}" )
}}
- Output:
b(0) 1 / 1 b(1) 1 / 2 b(2) 1 / 6 b(4) -1 / 30 b(6) 1 / 42 b(8) -1 / 30 b(10) 5 / 66 b(12) -691 / 2730 b(14) 7 / 6 b(16) -3617 / 510 b(18) 43867 / 798 b(20) -174611 / 330 b(22) 854513 / 138 b(24) -236364091 / 2730 b(26) 8553103 / 6 b(28) -23749461029 / 870 b(30) 8615841276005 / 14322 b(32) -7709321041217 / 510 b(34) 2577687858367 / 6 b(36) -26315271553053477373 / 1919190 b(38) 2929993913841559 / 6 b(40) -261082718496449122051 / 13530 b(42) 1520097643918070802691 / 1806 b(44) -27833269579301024235023 / 690 b(46) 596451111593912163277961 / 282 b(48) -5609403368997817686249127547 / 46410 b(50) 495057205241079648212477525 / 66 b(52) -801165718135489957347924991853 / 1590 b(54) 29149963634884862421418123812691 / 798 b(56) -2479392929313226753685415739663229 / 870 b(58) 84483613348880041862046775994036021 / 354 b(60) -1215233140483755572040304994079820246041491 / 56786730
Scheme
; Return the n'th Bernoulli number.
(define bernoulli
(lambda (n)
(let ((a (make-vector (1+ n))))
(do ((m 0 (1+ m)))
((> m n))
(vector-set! a m (/ 1 (1+ m)))
(do ((j m (1- j)))
((< j 1))
(vector-set! a (1- j) (* j (- (vector-ref a (1- j)) (vector-ref a j))))))
(vector-ref a 0))))
; Convert a rational to a string. If an integer, ends with "/1".
(define rational->string
(lambda (rational)
(format "~a/~a" (numerator rational) (denominator rational))))
; Returns the string length of the numerator of a rational.
(define rational-numerator-length
(lambda (rational)
(string-length (format "~a" (numerator rational)))))
; Formats a rational with left-padding such that total length to the slash is as given.
(define rational-padded
(lambda (rational total-length-to-slash)
(let* ((length-padding (- total-length-to-slash (rational-numerator-length rational)))
(padding-string (make-string length-padding #\ )))
(string-append padding-string (rational->string rational)))))
; Return the Bernoulli numbers 0 through n in a list.
(define make-bernoulli-list
(lambda (n)
(if (= n 0)
(list (bernoulli n))
(append (make-bernoulli-list (1- n)) (list (bernoulli n))))))
; Print the non-zero Bernoulli numbers 0 through 60 aligning the slashes.
(let* ((bernoullis-list (make-bernoulli-list 60))
(numerator-lengths (map rational-numerator-length bernoullis-list))
(max-numerator-length (apply max numerator-lengths)))
(let print-bernoulli ((index 0) (numbers bernoullis-list))
(cond
((null? numbers))
((= 0 (car numbers))
(print-bernoulli (1+ index) (cdr numbers)))
(else
(printf "B(~2@a) = ~a~%" index (rational-padded (car numbers) max-numerator-length))
(print-bernoulli (1+ index) (cdr numbers))))))
- Output:
$ scheme --script bernoulli.scm B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Seed7
The program below uses bigRational numbers. The Bernoulli numbers are written as fraction and as decimal number, with possible repeating decimals. The conversion of a bigRational number to string is done with the function str. This function automatically writes repeating decimals in parentheses, when necessary.
$ include "seed7_05.s7i";
include "bigrat.s7i";
const func bigRational: bernoulli (in integer: n) is func
result
var bigRational: bernoulli is bigRational.value;
local
var integer: m is 0;
var integer: j is 0;
var array bigRational: a is 0 times bigRational.value;
begin
a := [0 .. n] times bigRational.value;
for m range 0 to n do
a[m] := 1_ / bigInteger(succ(m));
for j range m downto 1 do
a[pred(j)] := bigRational(j) * (a[j] - a[pred(j)]);
end for;
end for;
bernoulli := a[0];
end func;
const proc: main is func
local
var bigRational: bernoulli is bigRational.value;
var integer: i is 0;
begin
for i range 0 to 60 do
bernoulli := bernoulli(i);
if bernoulli <> bigRational.value then
writeln("B(" <& i lpad 2 <& ") = " <& bernoulli.numerator lpad 44 <&
" / " <& bernoulli.denominator rpad 8 <& " " <& bernoulli);
end if;
end for;
end func;
- Output:
B( 0) = 1 / 1 1.0 B( 1) = -1 / 2 -0.5 B( 2) = 1 / 6 0.1(6) B( 4) = -1 / 30 -0.0(3) B( 6) = 1 / 42 0.0(238095) B( 8) = -1 / 30 -0.0(3) B(10) = 5 / 66 0.0(75) B(12) = -691 / 2730 -0.2(531135) B(14) = 7 / 6 1.1(6) B(16) = -3617 / 510 -7.0(9215686274509803) B(18) = 43867 / 798 54.9(711779448621553884) B(20) = -174611 / 330 -529.1(24) B(22) = 854513 / 138 6192.1(2318840579710144927536) B(24) = -236364091 / 2730 -86580.2(531135) B(26) = 8553103 / 6 1425517.1(6) B(28) = -23749461029 / 870 -27298231.0(6781609195402298850574712643) B(30) = 8615841276005 / 14322 601580873.9(006423683843038681748359167714) B(32) = -7709321041217 / 510 -15116315767.0(9215686274509803) B(34) = 2577687858367 / 6 429614643061.1(6) B(36) = -26315271553053477373 / 1919190 -13711655205088.3(327721590879485616) B(38) = 2929993913841559 / 6 488332318973593.1(6) B(40) = -261082718496449122051 / 13530 -19296579341940068.1(4863266814) B(42) = 1520097643918070802691 / 1806 841693047573682615.0(005537098560354374307862679955703211517165) B(44) = -27833269579301024235023 / 690 -40338071854059455413.0(7681159420289855072463) B(46) = 596451111593912163277961 / 282 2115074863808199160560.1(4539007092198581560283687943262411347517730496) B(48) = -5609403368997817686249127547 / 46410 -120866265222965259346027.3(119370825253178194354664942900237017884076707606) B(50) = 495057205241079648212477525 / 66 7500866746076964366855720.0(75) B(52) = -801165718135489957347924991853 / 1590 -503877810148106891413789303.0(5220125786163) B(54) = 29149963634884862421418123812691 / 798 36528776484818123335110430842.9(711779448621553884) B(56) = -2479392929313226753685415739663229 / 870 -2849876930245088222626914643291.0(6781609195402298850574712643) B(58) = 84483613348880041862046775994036021 / 354 238654274996836276446459819192192.1(4971751412429378531073446327683615819209039548022598870056) B(60) = -1215233140483755572040304994079820246041491 / 56786730 -21399949257225333665810744765191097.3(926741511617238745742183076926598872659158222352299560126106)
Sidef
Built-in:
say bernoulli(42).as_frac #=> 1520097643918070802691/1806
Recursive solution (with auto-memoization):
func bernoulli_number(n) is cached {
n.is_one && return 1/2
n.is_odd && return 0
1 - sum(^n, {|k|
binomial(n,k) * __FUNC__(k) / (n - k + 1)
})
}
for n in (0..60) {
var Bn = bernoulli_number(n) || next
printf("B(%2d) = %44s / %s\n", n, Bn.nude)
}
Using Ramanujan's congruences (pretty fast):
func ramanujan_bernoulli_number(n) is cached {
return 1/2 if n.is_one
return 0 if n.is_odd
((n%6 == 4 ? -1/2 : 1) * (n+3)/3 - sum(1 .. (n - n%6)/6, {|k|
binomial(n+3, n - 6*k) * __FUNC__(n - 6*k)
})) / binomial(n+3, n)
}
Using Euler's product formula for the Riemann zeta function and the Von Staudt–Clausen theorem (very fast):
func bernoulli_number_from_zeta(n) {
n.is_zero && return 1
n.is_one && return 1/2
n.is_odd && return 0
var log2B = (log(4*Num.tau*n)/2 + n*log(n) - n*log(Num.tau) - n)/log(2)
local Num!PREC = *(int(n + log2B) + (n <= 90 ? 18 : 0))
var K = 2*(n! / Num.tau**n)
var d = n.divisors.grep {|k| is_prime(k+1) }.prod {|k| k+1 }
var z = ceil((K*d).root(n-1)).primes.prod {|p| 1 - p.float**(-n) }
(-1)**(n/2 + 1) * int(ceil(d*K / z)) / d
}
The Akiyama–Tanigawa algorithm:
func bernoulli_print {
var a = []
for m in (0..60) {
a << 1/(m+1)
for j in (1..m -> flip) {
(a[j-1] -= a[j]) *= j
}
a[0] || next
printf("B(%2d) = %44s / %s\n", m, a[0].nude)
}
}
bernoulli_print()
- Output:
B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
SPAD
for n in 0..60 | (b:=bernoulli(n)$INTHEORY; b~=0) repeat print [n,b]
Package:IntegerNumberTheoryFunctions
- Output:
=============== Format: [n,B_n] =============== [0,1] 1 [1,- -] 2 1 [2,-] 6 1 [4,- --] 30 1 [6,--] 42 1 [8,- --] 30 5 [10,--] 66 691 [12,- ----] 2730 7 [14,-] 6 3617 [16,- ----] 510 43867 [18,-----] 798 174611 [20,- ------] 330 854513 [22,------] 138 236364091 [24,- ---------] 2730 8553103 [26,-------] 6 23749461029 [28,- -----------] 870 8615841276005 [30,-------------] 14322 7709321041217 [32,- -------------] 510 2577687858367 [34,-------------] 6 26315271553053477373 [36,- --------------------] 1919190 2929993913841559 [38,----------------] 6 261082718496449122051 [40,- ---------------------] 13530 1520097643918070802691 [42,----------------------] 1806 27833269579301024235023 [44,- -----------------------] 690 596451111593912163277961 [46,------------------------] 282 5609403368997817686249127547 [48,- ----------------------------] 46410 495057205241079648212477525 [50,---------------------------] 66 801165718135489957347924991853 [52,- ------------------------------] 1590 29149963634884862421418123812691 [54,--------------------------------] 798 2479392929313226753685415739663229 [56,- ----------------------------------] 870 84483613348880041862046775994036021 [58,-----------------------------------] 354 1215233140483755572040304994079820246041491 [60,- -------------------------------------------] 56786730 Type: Void
Swift
Uses the Frac type defined in the Rational task.
import BigInt
public func bernoulli<T: BinaryInteger & SignedNumeric>(n: Int) -> Frac<T> {
guard n != 0 else {
return 1
}
var arr = [Frac<T>]()
for m in 0...n {
arr.append(Frac(numerator: 1, denominator: T(m) + 1))
for j in stride(from: m, through: 1, by: -1) {
arr[j-1] = (arr[j-1] - arr[j]) * Frac(numerator: T(j), denominator: 1)
}
}
return arr[0]
}
for n in 0...60 {
let b = bernoulli(n: n) as Frac<BigInt>
guard b != 0 else {
continue
}
print("B(\(n)) = \(b)")
}
- Output:
B(0) = Frac(1 / 1) B(1) = Frac(1 / 2) B(2) = Frac(1 / 6) B(4) = Frac(-1 / 30) B(6) = Frac(1 / 42) B(8) = Frac(-1 / 30) B(10) = Frac(5 / 66) B(12) = Frac(-691 / 2730) B(14) = Frac(7 / 6) B(16) = Frac(-3617 / 510) B(18) = Frac(43867 / 798) B(20) = Frac(-174611 / 330) B(22) = Frac(854513 / 138) B(24) = Frac(-236364091 / 2730) B(26) = Frac(8553103 / 6) B(28) = Frac(-23749461029 / 870) B(30) = Frac(8615841276005 / 14322) B(32) = Frac(-7709321041217 / 510) B(34) = Frac(2577687858367 / 6) B(36) = Frac(-26315271553053477373 / 1919190) B(38) = Frac(2929993913841559 / 6) B(40) = Frac(-261082718496449122051 / 13530) B(42) = Frac(1520097643918070802691 / 1806) B(44) = Frac(-27833269579301024235023 / 690) B(46) = Frac(596451111593912163277961 / 282) B(48) = Frac(-5609403368997817686249127547 / 46410) B(50) = Frac(495057205241079648212477525 / 66) B(52) = Frac(-801165718135489957347924991853 / 1590) B(54) = Frac(29149963634884862421418123812691 / 798) B(56) = Frac(-2479392929313226753685415739663229 / 870) B(58) = Frac(84483613348880041862046775994036021 / 354) B(60) = Frac(-1215233140483755572040304994079820246041491 / 56786730)
Tcl
proc bernoulli {n} {
for {set m 0} {$m <= $n} {incr m} {
lappend A [list 1 [expr {$m + 1}]]
for {set j $m} {[set i $j] >= 1} {} {
lassign [lindex $A [incr j -1]] a1 b1
lassign [lindex $A $i] a2 b2
set x [set p [expr {$i * ($a1*$b2 - $a2*$b1)}]]
set y [set q [expr {$b1 * $b2}]]
while {$q} {set q [expr {$p % [set p $q]}]}
lset A $j [list [expr {$x/$p}] [expr {$y/$p}]]
}
}
return [lindex $A 0]
}
set len 0
for {set n 0} {$n <= 60} {incr n} {
set b [bernoulli $n]
if {[lindex $b 0]} {
lappend result $n {*}$b
set len [expr {max($len, [string length [lindex $b 0]])}]
}
}
foreach {n num denom} $result {
puts [format {B_%-2d = %*lld/%lld} $n $len $num $denom]
}
- Output:
B_0 = 1/1 B_1 = 1/2 B_2 = 1/6 B_4 = -1/30 B_6 = 1/42 B_8 = -1/30 B_10 = 5/66 B_12 = -691/2730 B_14 = 7/6 B_16 = -3617/510 B_18 = 43867/798 B_20 = -174611/330 B_22 = 854513/138 B_24 = -236364091/2730 B_26 = 8553103/6 B_28 = -23749461029/870 B_30 = 8615841276005/14322 B_32 = -7709321041217/510 B_34 = 2577687858367/6 B_36 = -26315271553053477373/1919190 B_38 = 2929993913841559/6 B_40 = -261082718496449122051/13530 B_42 = 1520097643918070802691/1806 B_44 = -27833269579301024235023/690 B_46 = 596451111593912163277961/282 B_48 = -5609403368997817686249127547/46410 B_50 = 495057205241079648212477525/66 B_52 = -801165718135489957347924991853/1590 B_54 = 29149963634884862421418123812691/798 B_56 = -2479392929313226753685415739663229/870 B_58 = 84483613348880041862046775994036021/354 B_60 = -1215233140483755572040304994079820246041491/56786730
Visual Basic .NET
' Bernoulli numbers - vb.net - 06/03/2017
Imports System.Numerics 'BigInteger
Module Bernoulli_numbers
Function gcd_BigInt(ByVal x As BigInteger, ByVal y As BigInteger) As BigInteger
Dim y2 As BigInteger
x = BigInteger.Abs(x)
Do
y2 = BigInteger.Remainder(x, y)
x = y
y = y2
Loop Until y = 0
Return x
End Function 'gcd_BigInt
Sub bernoul_BigInt(n As Integer, ByRef bnum As BigInteger, ByRef bden As BigInteger)
Dim j, m As Integer
Dim f As BigInteger
Dim anum(), aden() As BigInteger
ReDim anum(n + 1), aden(n + 1)
For m = 0 To n
anum(m + 1) = 1
aden(m + 1) = m + 1
For j = m To 1 Step -1
anum(j) = j * (aden(j + 1) * anum(j) - aden(j) * anum(j + 1))
aden(j) = aden(j) * aden(j + 1)
f = gcd_BigInt(BigInteger.Abs(anum(j)), BigInteger.Abs(aden(j)))
If f <> 1 Then
anum(j) = anum(j) / f
aden(j) = aden(j) / f
End If
Next
Next
bnum = anum(1) : bden = aden(1)
End Sub 'bernoul_BigInt
Sub bernoulli_BigInt()
Dim i As Integer
Dim bnum, bden As BigInteger
bnum = 0 : bden = 0
For i = 0 To 60
bernoul_BigInt(i, bnum, bden)
If bnum <> 0 Then
Console.WriteLine("B(" & i & ")=" & bnum.ToString("D") & "/" & bden.ToString("D"))
End If
Next i
End Sub 'bernoulli_BigInt
End Module 'Bernoulli_numbers
- Output:
B(0)=1/1 B(1)=1/2 B(2)=1/6 B(4)=-1/30 B(6)=1/42 B(8)=-1/30 B(10)=5/66 B(12)=-691/2730 B(14)=7/6 B(16)=-3617/510 B(18)=43867/798 B(20)=-174611/330 B(22)=854513/138 B(24)=-236364091/2730 B(26)=8553103/6 B(28)=-23749461029/870 B(30)=8615841276005/14322 B(32)=-7709321041217/510 B(34)=2577687858367/6 B(36)=-26315271553053477373/1919190 B(38)=2929993913841559/6 B(40)=-261082718496449122051/13530 B(42)=1520097643918070802691/1806 B(44)=-27833269579301024235023/690 B(46)=596451111593912163277961/282 B(48)=-5609403368997817686249127547/46410 B(50)=495057205241079648212477525/66 B(52)=-801165718135489957347924991853/1590 B(54)=29149963634884862421418123812691/798 B(56)=-2479392929313226753685415739663229/870 B(58)=84483613348880041862046775994036021/354 B(60)=-1215233140483755572040304994079820246041491/56786730
Wren
import "./fmt" for Fmt
import "./big" for BigRat
var bernoulli = Fn.new { |n|
if (n < 0) Fiber.abort("Argument must be non-negative")
var a = List.filled(n+1, null)
for (m in 0..n) {
a[m] = BigRat.new(1, m+1)
var j = m
while (j >= 1) {
a[j-1] = (a[j-1] - a[j]) * BigRat.new(j, 1)
j = j - 1
}
}
return (n != 1) ? a[0] : -a[0] // 'first' Bernoulli number
}
for (n in 0..60) {
var b = bernoulli.call(n)
if (b != BigRat.zero) Fmt.print("B($2d) = $44i / $i", n, b.num, b.den)
}
- Output:
B( 0) = 1 / 1 B( 1) = -1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
zkl
Uses lib GMP (GNU MP Bignum Library).
class Rational{ // Weenie Rational class, can handle BigInts
fcn init(_a,_b){ var a=_a, b=_b; normalize(); }
fcn toString{ "%50d / %d".fmt(a,b) }
fcn normalize{ // divide a and b by gcd
g:= a.gcd(b);
a/=g; b/=g;
if(b<0){ a=-a; b=-b; } // denominator > 0
self
}
fcn __opAdd(n){
if(Rational.isChildOf(n)) self(a*n.b + b*n.a, b*n.b); // Rat + Rat
else self(b*n + a, b); // Rat + Int
}
fcn __opSub(n){ self(a*n.b - b*n.a, b*n.b) } // Rat - Rat
fcn __opMul(n){
if(Rational.isChildOf(n)) self(a*n.a, b*n.b); // Rat * Rat
else self(a*n, b); // Rat * Int
}
fcn __opDiv(n){ self(a*n.b,b*n.a) } // Rat / Rat
}
var [const] BN=Import.lib("zklBigNum"); // libGMP (GNU MP Bignum Library)
fcn B(N){ // calculate Bernoulli(n)
var A=List.createLong(100,0); // aka static aka not thread safe
foreach m in (N+1){
A[m]=Rational(BN(1),BN(m+1));
foreach j in ([m..1, -1]){ A[j-1]= (A[j-1] - A[j])*j; }
}
A[0]
}
foreach b in ([0..1].chain([2..60,2])){ println("B(%2d)%s".fmt(b,B(b))) }
- Output:
B( 0) 1 / 1 B( 1) 1 / 2 B( 2) 1 / 6 B( 4) -1 / 30 B( 6) 1 / 42 B( 8) -1 / 30 B(10) 5 / 66 B(12) -691 / 2730 B(14) 7 / 6 B(16) -3617 / 510 B(18) 43867 / 798 B(20) -174611 / 330 B(22) 854513 / 138 B(24) -236364091 / 2730 B(26) 8553103 / 6 B(28) -23749461029 / 870 B(30) 8615841276005 / 14322 B(32) -7709321041217 / 510 B(34) 2577687858367 / 6 B(36) -26315271553053477373 / 1919190 B(38) 2929993913841559 / 6 B(40) -261082718496449122051 / 13530 B(42) 1520097643918070802691 / 1806 B(44) -27833269579301024235023 / 690 B(46) 596451111593912163277961 / 282 B(48) -5609403368997817686249127547 / 46410 B(50) 495057205241079648212477525 / 66 B(52) -801165718135489957347924991853 / 1590 B(54) 29149963634884862421418123812691 / 798 B(56) -2479392929313226753685415739663229 / 870 B(58) 84483613348880041862046775994036021 / 354 B(60) -1215233140483755572040304994079820246041491 / 56786730
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