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 = -½ vs. +½.
C
<lang 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;
} </lang>
- 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 <lang csharp> 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(); } }
} </lang>
- 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
<lang csharp> 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, 60) .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); } }
} </lang>
- 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
Crystal
<lang crystal>
- Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99
class Fraction
def initialize(n : Int64, d : Int64) @numerator = n @denominator = d end
def numerator @numerator end
def denominator @denominator end
def subtract(rhs_fraction) rhs_numerator = rhs_fraction.numerator * @denominator rhs_denominator = rhs_fraction.denominator * @denominator @numerator *= rhs_fraction.denominator @denominator *= rhs_fraction.denominator @numerator -= rhs_numerator self.reduce end
def multiply(value) @numerator *= value end
def reduce gcd = gcd(@numerator, @denominator) @numerator /= gcd @denominator /= gcd end
def to_s @numerator == 0 ? 0 : @numerator.to_s + '/' + @denominator.to_s end
end
def gcd(a, b)
# we need b>0 because b on its own isn't considered true b > 0 ? gcd(b, a % b) : a
end
def calculate_bernoulli(bern)
row = [] of Fraction 0_i64.step(bern) do |m| row << Fraction.new(1_i64, m + 1) m.step(1, -1) do |j| row[j - 1].subtract(row[j]) row[j - 1].multiply(j) row[j - 1].reduce end end
row[0]
end
1_i64.step(30_i64) do |bern|
puts "#{bern} : #{calculate_bernoulli(bern).to_s}"
end
</lang>
- Output:
1 : 1/2 2 : 1/6 3 : 0 4 : -1/30 5 : 0 6 : 1/42 7 : 0 8 : -1/30 9 : 0 10 : 5/66 11 : 0 12 : -691/2730 13 : 0 14 : 7/6 15 : 0 16 : -3617/510 17 : 0 18 : 43867/798 19 : 0 20 : -174611/330 21 : 0 22 : 854513/138 23 : 0 24 : -236364091/2730 25 : 0 26 : 8553103/6 27 : 0 28 : -23749461029/870 29 : 0 30 : 8615841276005/14322
Clojure
<lang 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)))
</lang>
- 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.
<lang lisp>(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)))))</lang>
- 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.
<lang d>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);
}</lang> The output is exactly the same as the Python entry.
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. <lang lisp> (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))))) </lang>
- Output:
<lang lisp>
- 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 </lang>
Elixir
<lang 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</lang>
- 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#
<lang fsharp> 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
</lang>
- 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: <lang> 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 </lang> 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. <lang>
- 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
</lang> It gives the same result as the native implementation, but is slightly faster. <lang> [ 30 bernoulli-numbers ] time ... Running time: 0.004331652 seconds </lang>
FreeBASIC
<lang 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</lang>
- 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
FunL
FunL has pre-defined function B
in module integers
, which is defined as:
<lang funl>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) )</lang>
- 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
GAP
<lang gap>for a in Filtered(List([1 .. 60], n -> [n, Bernoulli(n)]), x -> x[2] <> 0) do
Print(a, "\n");
od;
[ 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 ]</lang>
Go
<lang go>package main
import (
"fmt" "math/big" "strings"
)
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 align(b *big.Rat, w int) string {
s := b.String() return strings.Repeat(" ", w-strings.Index(s, "/")) + s
}
func main() {
for n := 0; n <= 60; n++ { if b := b(n); b.Num().BitLen() > 0 { fmt.Printf("B(%2d) =%s\n", n, align(b, 45)) } }
}</lang>
- 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.
<lang Haskell>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)</lang>
- 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
<lang haskell>import Data.Ratio (Ratio, numerator, denominator, (%))
bernouilliNumbers :: Integer -> [Rational] bernouilliNumbers =
fmap head . tail . scanl (\rs n -> let xs = zipWith ((*) . (n %)) [2 ..] rs in 1 - sum xs : xs) [] . enumFromTo 0
main :: IO () main =
(putStrLn . unlines) (concat $ zipWith (\i x -> let n = numerator x in [ concat ["B(", show i, ") = ", show n, "/", show (denominator x)] | n /= 0 ]) [0 ..] (bernouilliNumbers 60))</lang>
- 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
Icon and Unicon
The following works in both languages: <lang unicon>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</lang>
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:
<lang J>B=:3 :0"0
+/,(<:*(_1^[)*!*(y^~1+[)%1+])"0/~i.1x+y
)</lang>
Task:
<lang J> require'strings'
'B',.rplc&'r/_-'"1": (#~ 0 ~: {:"1)(,.B) i.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</lang>
Java
<lang 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]; }
}</lang>
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: <lang jq># 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;</lang>
Fractions:<lang jq>
- 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 ; </lang>
Bernoulli Numbers: <lang jq># 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) ;</lang>
The task: <lang jq>range(0;61) | if . % 2 == 0 or . == 1 then "\(.): \(bernoulli(.) )" else empty end</lang>
- Output:
The following output was obtained using the previously mentioned BigInt library. <lang sh>$ 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"]</lang>
Julia
<lang Julia>function bernoulli(n)
A = Vector{Rational{BigInt}}(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(num(x)) + Int(x < 0), max, B) argdigits = ndigits(n) for i = 0 : n if num(B[i + 1]) & 1 == 1 println( "B(", lpad(i, argdigits), ") = ", lpad(num(B[i + 1]), pad), " / ", den(B[i + 1]) ) end end
end
display(60)</lang>
Produces virtually the same output as the Python version.
Kotlin
<lang scala>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))
}</lang>
- Output:
Produces virtually the same output as the Java version.
Maple
<lang Maple>print(select(n->n[2]<>0,[seq([n,bernoulli(n,1)],n=0..60)]));</lang>
- 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. <lang Mathematica>bernoulli[n_] := Module[{a = ConstantArray[0, n + 2]},
Do[ am = 1/m; If[m == 1 && a1 != 0, Print[{m - 1, a1}]]; Do[ aj - 1 = (j - 1)*(aj - 1 - aj); If[j == 2 && a1 != 0, Print[{m - 1, a1}]]; , {j, m, 2, -1}]; , {m, 1, n + 1}]; ]
bernoulli[60]</lang>
- 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.)
<lang Mathematica>Table[{i, BernoulliB[i]}, {i, 0, 60}]; Select[%, #2 != 0 &] // TableForm</lang>
- 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)
PARI/GP
<lang parigp>for(n=0,60,t=bernfrac(n);if(t,print(n" "t)))</lang>
- 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
<lang pascal> (* Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99 *)
program BernoulliForAda99;
type
Fraction = object private numerator, denominator: Int64; public procedure assign(n, d: Int64); procedure subtract(rhs: Fraction); procedure multiply(value: Int64); procedure reduce(); procedure writeOutput();
end;
function gcd(a, b: Int64):Int64;
begin
if (b = 0) then gcd := a else gcd := gcd(b, a mod b)
end;
procedure Fraction.writeOutput();
begin
write(numerator); if (numerator <> 0) then begin write('/'); write(denominator); end;
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);
begin
numerator := numerator * value;
end;
procedure Fraction.reduce();
var gcdResult: Int64;
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, n); 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...');
for b:= 1 to 25 do begin write(b); write(' : '); result := calculateBernoulli(b); result.writeOutput(); writeln; end;
end.
</lang>
- Output:
Calculating Bernoulli numbers... 1 : 1/2 2 : 1/6 3 : 0 4 : 1/-30 5 : 0 6 : 1/42 7 : 0 8 : 1/-30 9 : 0 10 : 5/66 11 : 0 12 : 691/-2730 13 : 0 14 : 7/6 15 : 0 16 : -3617/510 17 : 0 18 : 43867/798 19 : 0 20 : 174611/-330 21 : 0 22 : 854513/138 23 : 0 24 : 236364091/-2730 25 : 0
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.
<lang perl>#!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(); </lang> The output is exactly the same as the Python entry.
We can also use modules for faster results. E.g.
<lang perl>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;
}</lang> with identical output. Or: <lang perl>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 $num != 0;
}</lang> with the difference being that Pari chooses = -½.
Perl 6
Simple
First, a straighforward implementation of the naïve algorithm in the task description.
<lang perl6>sub bernoulli($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;</lang>
- 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 Perl 6 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.
<lang perl6>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;</lang>
- 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.
<lang perl6>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 ... *)
- </lang>
Phix
Using routines from Arithmetic/Rational, adapted to use bigatoms
Note: this exposed a bug in ba_trunc, so you may want to check for a mod dated 19/11/15 in that routine (in builtins\bigatom.e).
<lang Phix>include builtins\bigatom.e
constant NUM = 1, DEN = 2
type ba_frac(object r)
return sequence(r) and length(r)=2 and bigatom(r[NUM]) and bigatom(r[DEN])
end type
function ba_gcd(bigatom u, bigatom v) bigatom t
u = ba_floor(ba_abs(u)) v = ba_floor(ba_abs(v)) while v!=BA_ZERO do t = u u = v v = ba_remainder(t, v) end while return u
end function
function ba_frac_normalise(bigatom n, bigatom d) bigatom g
if ba_compare(d,BA_ZERO)<0 then n = ba_sub(0,n) d = ba_sub(0,d) end if g = ba_gcd(n,d) return {ba_idivide(n,g),ba_idivide(d,g)}
end function
function ba_frac_sub(ba_frac a, ba_frac b) bigatom {an,ad} = a,
{bn,bd} = b return ba_frac_normalise(ba_sub(ba_multiply(an,bd),ba_multiply(bn,ad)),ba_multiply(ad,bd))
end function
function ba_frac_mul(ba_frac a, ba_frac b) bigatom {an,ad} = a,
{bn,bd} = b return ba_frac_normalise(ba_multiply(an,bn),ba_multiply(ad,bd))
end function</lang> After which the code itself is pretty trivial <lang Phix>sequence a = {} for m=0 to 60 do
a = append(a,{ba_new(1),ba_new(m+1)}) for j=m to 1 by -1 do a[j] = ba_frac_mul({ba_new(j),ba_new(1)},ba_frac_sub(a[j+1],a[j])) end for if a[1][1]!=BA_ZERO then printf(1,"B(%2d) = %44s / %s\n",{m,ba_sprint(a[1][1]),ba_sprint(a[1][2])}) end if
end for</lang>
- 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. <lang PicoLisp>(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)</lang>
PL/I
<lang 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;</lang> 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
<lang python>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))</lang>
- 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(): <lang python>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))</lang>
Output is exactly the same as before.
R
R has the built-in function bernoulli(n), where n is the index, a whole number greater or equal to 0. It returns the first n+1 Bernoulli numbers, that are defined as a sequence of rational numbers.
<lang r>
- Bernoulli numbers. 12/8/16 aev
require(pracma) bernoulli(60) </lang>
- Output:
> require(pracma) Loading required package: pracma > bernoulli(60) [1] 1.000000e+00 -5.000000e-01 1.666667e-01 0.000000e+00 -3.333333e-02 [6] 0.000000e+00 2.380952e-02 0.000000e+00 -3.333333e-02 0.000000e+00 [11] 7.575758e-02 0.000000e+00 -2.531136e-01 0.000000e+00 1.166667e+00 [16] 0.000000e+00 -7.092157e+00 0.000000e+00 5.497118e+01 0.000000e+00 [21] -5.291242e+02 0.000000e+00 6.192123e+03 0.000000e+00 -8.658025e+04 [26] 0.000000e+00 1.425517e+06 0.000000e+00 -2.729823e+07 0.000000e+00 [31] 6.015809e+08 0.000000e+00 -1.511632e+10 0.000000e+00 4.296146e+11 [36] 0.000000e+00 -1.371166e+13 0.000000e+00 4.883323e+14 0.000000e+00 [41] -1.929658e+16 0.000000e+00 8.416930e+17 0.000000e+00 -4.033807e+19 [46] 0.000000e+00 2.115075e+21 0.000000e+00 -1.208663e+23 0.000000e+00 [51] 7.500867e+24 0.000000e+00 -5.038778e+26 0.000000e+00 3.652878e+28 [56] 0.000000e+00 -2.849877e+30 0.000000e+00 2.386543e+32 0.000000e+00 [61] -2.139995e+34 >
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>#lang racket
- 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))))</lang>
- 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
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.
<lang rexx>/*REXX program calculates N number of Bernoulli numbers expressed as fractions. */ parse arg N .; if N== then N=60 /*Not specified? Then use the default.*/ !.=0; w=max(length(N), 4); Nw=N + w + N % 4 /*used for aligning (output) fractions.*/ say 'B(n)' center("Bernoulli number expressed as a fraction", max(78-w, Nw)) /*title*/ 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 /*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. */ do j=2 to x by 2; jp=j+1; d=j+j /*process the positive integers up to X*/ if d>digits() then numeric digits d /*increase the decimal digits if needed*/ 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; ad=$lcm /* " " next " */ sn=sn+an /* " " current " */ end /*k*/ /* [↑] calculate the SN/SD sequence.*/ sn=-sn /*adjust 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\==0 then return !.c.x.y /*combination computed before?*/ if x-y<y then y=x-y; z=perm(x, y); do j=2 to y; 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); return x*y/gcd(x,y) /*──────────────────────────────────────────────────────────────────────────────────────*/ perm: procedure expose !.; parse arg x,y; if !.p.x.y\==0 then return !.p.x.y
z=1; do j=x-y+1 to x; z=z*j; end; !.p.x.y=z; return z</lang>
output when using the default input:
B(n) Bernoulli number expressed as a fraction ──── ─────────────────────────────────────────────────────────────────────────────────────── 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
Ruby
<lang ruby>bernoulli = Enumerator.new do |y|
ar, m = [], 0 loop do ar << Rational(1, m+1) m.downto(1){|j| ar[j-1] = j*(ar[j-1] - ar[j]) } y << ar.first # yield m += 1 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? }
</lang>
- 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
<lang 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.index % 2 == 0) { 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<_>>(); }); }
} </lang>
- Output:
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
Scala
With Custom Rational Number Class
(code will run in Scala REPL with a cut-and-paste without need for a third-party library)
<lang scala>/** 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}" )
}} </lang>
- 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
Sidef
Recursive solution (with auto-memoization): <lang ruby>func bernoulli_number{}
func bern_helper(n, k) {
binomial(n, k) * (bernoulli_number(k) / (n - k + 1))
}
func bern_diff(n, k, d) {
n < k ? d : bern_diff(n, k + 1, d - bern_helper(n + 1, k))
}
bernoulli_number = func(n) is cached {
n.is_one && return 1/2 n.is_odd && return 0
n > 0 ? bern_diff(n - 1, 0, 1) : 1
}
for i (0..60) {
var num = bernoulli_number(i) || next printf("B(%2d) = %44s / %s\n", i, num.nude)
}</lang>
Iterative solution: <lang ruby>func bernoulli_print {
var a = [] for m (0..60) { a.append(1/(m+1)) for j (flip(1..m)) { (a[j-1] -= a[j]) *= j } a[0] || next printf("B(%2d) = %44s / %s\n", m, a[0].nude) }
} bernoulli_print()</lang>
- 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
<lang SPAD> for n in 0..60 | (b:=bernoulli(n)$INTHEORY; b~=0) repeat print [n,b] </lang> 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
Tcl
<lang 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]
}</lang>
- 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
<lang vbnet>' Bernoulli numbers - vb.net - 06/03/2017 Imports System.Numerics 'BinInteger
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</lang>
- 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). <lang zkl>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
}</lang> <lang zkl>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]
}</lang> <lang zkl>foreach b in ([0..1].chain([2..60,2])){ println("B(%2d)%s".fmt(b,B(b))) }</lang>
- 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
- Programming Tasks
- Mathematics
- C
- GMP
- C sharp
- Mpir.NET
- Examples needing attention
- MathNet.Numerics
- Crystal
- Crystal examples needing attention
- Clojure
- Common Lisp
- D
- EchoLisp
- EchoLisp examples needing attention
- Elixir
- F Sharp
- MathNet.Numerics.FSharp
- Factor
- FreeBASIC
- FunL
- GAP
- GAP examples needing attention
- Go
- Haskell
- Icon
- Unicon
- J
- Java
- Jq
- Julia
- Kotlin
- Maple
- Mathematica
- Wolfram Language
- PARI/GP
- Pascal
- Pascal examples needing attention
- Perl
- Ntheory
- Perl 6
- Phix
- PicoLisp
- PL/I
- Python
- R
- Racket
- REXX
- Ruby
- Rust
- Rust examples needing attention
- Scala
- Sidef
- SPAD
- Tcl
- Visual Basic .NET
- System.Numerics
- Zkl