Riordan numbers: Difference between revisions
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1: { #1d #0d #1d #1d #3d #6d #15d #36d #91d #232d #603d #1585d #4213d 11298d #30537d #83097d #227475d #625992d #1730787d #4805595d #13393689d #37458330d #105089229d #295673994d #834086421d #2358641376d #6684761125d #18985057351d #54022715451d #154000562758d #439742222071d #1257643249140d } |
1: { #1d #0d #1d #1d #3d #6d #15d #36d #91d #232d #603d #1585d #4213d 11298d #30537d #83097d #227475d #625992d #1730787d #4805595d #13393689d #37458330d #105089229d #295673994d #834086421d #2358641376d #6684761125d #18985057351d #54022715451d #154000562758d #439742222071d #1257643249140d } |
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</pre> |
</pre> |
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=={{header|SETL}}== |
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<syntaxhighlight lang="setl">program riordan; |
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a := {[0, 1], [1, 0]}; |
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loop for n in [2..9999] do |
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a(n) := (n-1)*(2*a(n-1) + 3*a(n-2)) div (n+1); |
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end loop; |
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loop for n in [0..31] do |
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putchar(lpad(str a(n), 15)); |
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if n mod 4=3 then print; end if; |
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end loop; |
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loop for n in [999, 9999] do |
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print("The", str (n+1)+"th Riordan number has", #str a(n), "digits."); |
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end loop; |
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end program;</syntaxhighlight> |
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{{out}} |
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<pre> 1 0 1 1 |
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3 6 15 36 |
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91 232 603 1585 |
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4213 11298 30537 83097 |
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227475 625992 1730787 4805595 |
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13393689 37458330 105089229 295673994 |
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834086421 2358641376 6684761125 18985057351 |
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54022715451 154000562758 439742222071 1257643249140 |
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The 1000th Riordan number has 472 digits. |
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The 10000th Riordan number has 4765 digits.</pre> |
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=={{header|Wren}}== |
=={{header|Wren}}== |
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Revision as of 20:49, 7 November 2023
You are encouraged to solve this task according to the task description, using any language you may know.
Riordan numbers show up in several places in set theory. They are closely related to Motzkin numbers, and may be used to derive them.
Riordan numbers comprise the sequence a where:
a(0) = 1, a(1) = 0, for subsequent terms, a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1)
There are other generating functions, and you are free to use one most convenient for your language.
- Task
- Find and display the first 32 Riordan numbers.
- Stretch
- Find and display the digit count of the 1,000th Riordan number.
- Find and display the digit count of the 10,000th Riordan number.
- See also
11l
F riordan(nn)
V a = [BigInt(1), 0, 1]
L(n) 3 .< nn
a.append((n - 1) * (2 * a[n - 1] + 3 * a[n - 2]) I/ (n + 1))
R a
V rios = riordan(10'000)
L(i) 32
print(f:‘{commatize(rios[i]):18}’, end' I (i + 1) % 4 == 0 {"\n"} E ‘’)
print(‘The 1,000th Riordan has ’String(rios[999]).len‘ digits.’)
print(‘The 10,000th Rirdan has ’String(rios[9999]).len‘ digits.’)- Output:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The 1,000th Riordan has 472 digits.
The 10,000th Rirdan has 4765 digits.
Action!
Finds the first 13 Riordan numbers as Action! is limited to 16 bit integers (signed and unsiged).
;;; Find some Riordan numbers - limited to the first 13 as the largest integer
;;; Action! supports is unsigned 16-bit
;;; sets a to the riordan numbers 0 .. n, a must have n elements
PROC riordan( CARD n CARD ARRAY a )
CARD i
IF n >= 0 THEN
a( 0 ) = 1
IF n >= 1 THEN
a( 1 ) = 0
FOR i = 2 TO n DO
a( i ) = ( ( i - 1 )
* ( ( 2 * a( i - 1 ) )
+ ( 3 * a( i - 2 ) )
)
)
/ ( i + 1 )
OD
FI
FI
RETURN
PROC Main()
CARD ARRAY r( 13 )
CARD i
riordan( 13, r )
FOR i = 0 TO 12 DO
Put( ' )
PrintC( r( i ) )
OD
RETURN- Output:
1 0 1 1 3 6 15 36 91 232 603 1585 4213
ALGOL 68
Uses ALGOL 68G's LONG LONG INT which has programmer-specifiable precision. ALGOL 68G version 3 issues a warning that precision 5000 will impact performance but it still executes this program somewhat faster than version 2 does.
BEGIN # find some Riordan numbers #
# returns a table of the Riordan numbers 0 .. n #
OP RIORDAN = ( INT n )[]LONG INT:
BEGIN
[ 0 : n ]LONG INT a;
IF n >= 0 THEN
a[ 0 ] := 1;
IF n >= 1 THEN
a[ 1 ] := 0;
FOR i FROM 2 TO UPB a DO
a[ i ] := ( ( i - 1 )
* ( ( 2 * a[ i - 1 ] )
+ ( 3 * a[ i - 2 ] )
)
)
OVER ( i + 1 )
OD
FI
FI;
a
END # RIORDAN # ;
# returns a string representation of n with commas #
PROC commatise = ( STRING unformatted )STRING:
BEGIN
STRING result := "";
INT ch count := 0;
FOR c FROM UPB unformatted BY -1 TO LWB unformatted DO
IF ch count <= 2 THEN
ch count +:= 1
ELSE
ch count := 1;
IF unformatted[ c ] = " " THEN " " ELSE "," FI +=: result
FI;
unformatted[ c ] +=: result
OD;
result
END; # commatise #
# returns the length of s #
OP LENGTH = ( STRING s )INT: ( UPB s - LWB s ) + 1;
BEGIN # show some Riordann numbers #
[]LONG INT r = RIORDAN 31;
INT shown := 0;
FOR i FROM LWB r TO UPB r DO
print( ( commatise( whole( r[ i ], -15 ) ) ) );
IF ( shown +:= 1 ) = 4 THEN
print( ( newline ) );
shown := 0
FI
OD
END;
BEGIN # calculate the length of the 1 000th and 10 000th Riordan numbers #
PR precision 5000 PR # allow up to 5 000 digits for LONG LONG INT #
LONG LONG INT r2 := -1, r1 := 1, r := 0;
print( ( newline ) );
FOR i FROM 2 TO 9 999 DO
r2 := r1;
r1 := r;
r := ( ( i - 1 )
* ( ( 2 * r1 )
+ ( 3 * r2 )
)
)
OVER ( i + 1 );
IF i = 999 OR i = 9 999 THEN
STRING rs = whole( r, 0 )[ @ 1 ];
print( ( "The ", whole( i + 1, -6 ), "th number is: "
, rs[ 1 : 20 ], "...", rs[ LENGTH rs - 19 : ]
, " with ", whole( LENGTH rs, -5 ), " digits"
, newline
)
)
FI
OD
END
END- Output:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The 1000th number is: 51077756867821111314...79942013897484633052 with 472 digits
The 10000th number is: 19927418577260688844...71395322020211157137 with 4765 digits
ALGOL W
Finds the first 22 Riordan numbers as Algol W is limited to signed 32 bit integers.
begin % -- find some Riordan numbers %
% -- sets a to the Riordan numbers 0 .. n - a must have bounds 0 :: n %
procedure riordan ( integer value n; integer array a ( * ) ) ;
if n >= 0 then begin
a( 0 ) := 1;
if n >= 1 then begin
a( 1 ) := 0;
for i := 2 until n do begin
a( i ) := ( ( i - 1 )
* ( ( 2 * a( i - 1 ) )
+ ( 3 * a( i - 2 ) )
)
)
div ( i + 1 )
end for_i
end if_n_ge_1
end riordan ;
begin % -- show some Riordann numbers %
integer array r ( 0 :: 21 );
integer shown;
riordan( 21, r );
shown := 0;
for i := 0 until 21 do begin
writeon( i_w := 9, s_w := 0, " ", r( i ) );
shown := shown + 1;
if shown = 4 then begin
write();
shown := 0
end if_shown_eq_4
end for_i
end;
end.- Output:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330
AppleScript
on riordanNumbers(n)
set a to {1, 0}
repeat with n from 3 to n
set {an2, an1} to a's items -2 thru -1
set a's end to (n - 2) * (an1 + an1 + 3 * an2) div n
end repeat
return a
end riordanNumbers
on intToText(int, separator)
set groups to {}
repeat while (int > 999)
set groups's beginning to ((1000 + (int mod 1000 as integer)) as text)'s text 2 thru 4
set int to int div 1000
end repeat
set groups's beginning to int as integer
return join(groups, separator)
end intToText
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 riordans to riordanNumbers(32)
set columnWidth to (count intToText(riordans's item 32, ",")) + 2
set output to {"1st 32 Riordan numbers:"}
set row to {}
repeat with i from 1 to 32 by 4
repeat with j from i to (i + 3)
set end of row to text -columnWidth thru -1 of ¬
(" " & intToText(riordans's item j, ","))
end repeat
set end of output to join(row, "")
set row to {}
end repeat
return join(output, linefeed)
end task
task()
- Output:
"1st 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140"
Arturo
riordan: function [n].memoize[
if zero? n -> return 1
if one? n -> return 0
return ((n-1) * ((2*riordan n-1) + 3*riordan n-2)) / n+1
]
riordans: map 0..31 => riordan
loop split.every: 4 riordans 'x ->
print map x 's -> pad to :string s 13
- Output:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
BASIC
QBasic
CONST limit = 31
DIM r(0 TO limit)
PRINT "First 32 Riordan numbers:"
CALL Riordan(limit, r())
FOR i = 0 TO limit
PRINT USING " #############"; r(i);
cont = cont + 1
IF cont MOD 4 = 0 THEN PRINT
NEXT i
END
SUB Riordan (n, a())
IF n >= 0 THEN
a(0) = 1
IF n >= 1 THEN
a(1) = 0
FOR i = 2 TO n
a(i) = ((i - 1) * ((2 * a(i - 1)) + (3 * a(i - 2)))) / (i + 1)
NEXT i
END IF
END IF
END SUB
True BASIC
SUB riordan (n,a())
IF n >= 0 THEN
LET a(0) = 1
IF n >= 1 THEN
LET a(1) = 0
FOR i = 2 TO n
LET a(i) = ((i-1)*((2*a(i-1))+(3*a(i-2))))/(i+1)
NEXT i
END IF
END IF
END SUB
LET limit = 31
LET cont = 0
DIM r(0)
MAT REDIM r(0 TO limit)
PRINT "First 32 Riordan numbers:"
CALL riordan(limit, r())
FOR i = 0 TO limit
PRINT USING " #############": r(i);
LET count = count + 1
IF MOD(count, 4) = 0 THEN PRINT
NEXT i
END
- Output:
Same as FreeBASIC entry.
Yabasic
limit = 31
dim r(limit)
print "First 32 Riordan numbers:"
Riordan(limit, r())
for i = 0 to 23
print r(i) using("#############");
cont = cont + 1
if mod(cont, 4) = 0 print
next i
for i = 24 to limit
print " ", str$(r(i));
cont = cont + 1
if mod(cont, 4) = 0 print
next i
end
sub Riordan (n, a())
local i
if n >= 0 then
a(0) = 1
if n >= 1 then
a(1) = 0
for i = 2 to n
a(i) = ((i-1) * ((2 * a(i-1)) + (3 * a(i-2)))) / (i+1)
next i
fi
fi
end subC++
#include <iomanip>
#include <iostream>
#include <gmpxx.h>
using big_int = mpz_class;
class riordan_number_generator {
public:
big_int next();
private:
big_int a0_ = 1;
big_int a1_ = 0;
int n_ = 0;
};
big_int riordan_number_generator::next() {
int n = n_++;
if (n == 0)
return a0_;
if (n == 1)
return a1_;
big_int a = (n - 1) * (2 * a1_ + 3 * a0_) / (n + 1);
a0_ = a1_;
a1_ = a;
return a;
}
std::string to_string(const big_int& num, size_t max_digits) {
std::string str = num.get_str();
size_t len = str.size();
if (len > max_digits)
str.replace(max_digits / 2, len - max_digits, "...");
return str;
}
int main() {
riordan_number_generator rng;
std::cout << "First 32 Riordan numbers:\n";
int i = 1;
for (; i <= 32; ++i) {
std::cout << std::setw(14) << rng.next()
<< (i % 4 == 0 ? '\n' : ' ');
}
for (; i < 1000; ++i)
rng.next();
auto num = rng.next();
++i;
std::cout << "\nThe 1000th is " << to_string(num, 40) << " ("
<< num.get_str().size() << " digits).\n";
for (; i < 10000; ++i)
rng.next();
num = rng.next();
std::cout << "The 10000th is " << to_string(num, 40) << " ("
<< num.get_str().size() << " digits).\n";
}
- Output:
First 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
The 1000th is 51077756867821111314...79942013897484633052 (472 digits).
The 10000th is 19927418577260688844...71395322020211157137 (4765 digits).
EasyLang
cnt = 32
print "First " & cnt & " Riordan numbers:"
app = 1 ; ap = 0
print app ; print ap
for n = 2 to cnt - 1
a = (n - 1) * (2 * ap + 3 * app) / (n + 1)
print a
app = ap
ap = a
.
F#
// Riordan numbers. Nigel Galloway: August 19th., 2022
let r()=seq{yield 1I; yield 0I; yield! Seq.unfold(fun(n,n1,n2)->let r=(n-1I)*(2I*n1+3I*n2)/(n+1I) in Some(r,(n+1I,r,n1)))(2I,0I,1I)}
let n=r()|>Seq.take 10000|>Array.ofSeq in n|>Array.take 32|>Seq.iter(printf "%A "); printfn "\nr[999] has %d digits\nr[9999] has %d digits" ((string n.[999]).Length) ((string n.[9999]).Length)
- Output:
1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 95673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140 r[999] has 472 digits r[9999] has 4765 digits
FreeBASIC
Const limit = 31
Sub Riordan (n As Integer, a() As Integer)
If n >= 0 Then
a(0) = 1
If n >= 1 Then
a(1) = 0
For i As Integer = 2 To n
a(i) = ((i-1) * ((2 * a(i-1)) + (3 * a(i-2)))) / (i+1)
Next i
End If
End If
End Sub
Dim As Integer r(0 To limit)
Dim As Byte cont = 0
Print "First 32 Riordan numbers:"
Riordan(limit, r())
For i As Integer = 0 To limit
Print Using " #############"; r(i);
cont += 1
If cont Mod 4 = 0 Then Print
Next i
Sleep- Output:
First 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
FutureBasic
_limit = 31
local fn Riordan( n as long, a(_limit) as long )
long i
if ( n >= 0 )
a(0) = 1
if ( n >= 1 )
a(1) = 0
for i = 2 to n
a(i) = ( ( i - 1 ) * ( ( 2 * a(i-1) ) + ( 3 * a(i-2) ) ) ) / ( i + 1 )
next
end if
end if
end fn
long i, count = 0, r(_limit)
printf @"First 32 Riordan numbers:"
fn Riordan( _limit, r(0) )
for i = 0 to 23
printf @"%16ld\b", r(i)
count++
if count mod 4 == 0 then print
next
count = 0
for i = 24 to _limit
printf @"%16s\b", fn StringUTF8String( Str(r(i)) )
count++
if count mod 4 == 0 then print
next
HandleEvents- Output:
First 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
Haskell
--------------------- RIORDAN NUMBERS --------------------
riordans :: [Integer]
riordans =
1 :
0 :
zipWith
div
( zipWith
(*)
[1 ..]
( zipWith
(+)
((2 *) <$> tail riordans)
((3 *) <$> riordans)
)
)
[3 ..]
-------------------------- TESTS -------------------------
main :: IO ()
main =
putStrLn "First 32 Riordan terms:"
>> mapM_ print (take 32 riordans)
>> mapM_
( \x ->
putStrLn $
concat
[ "\nDigit count of ",
show x,
"th Riordan term:\n",
(show . length . show)
(riordans !! pred x)
]
)
[1000, 10000]
- Output:
First 32 Riordan terms: 1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 295673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140 Digit count of 1000th Riordan term: 472 Digit count of 10000th Riordan term: 4765
J
Sequence extender:
riordanext=: (, (<: % >:)@# * 3 2 +/ .* _2&{.)
Task example:
riordanext^:(30) 1x 0
1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 295673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140
Stretch:
#":(1e3-1){riordanext^:(1e3) x:1 0
472
#":(1e4-1){riordanext^:(1e4) x:1 0
4765
Java
import java.math.BigInteger;
import java.util.List;
public final class RiordanNumbers {
public static void main(String[] args) {
final int limit = 10_000;
final BigInteger THREE = BigInteger.valueOf(3);
BigInteger[] riordans = new BigInteger[limit];
riordans[0] = BigInteger.ONE;
riordans[1] = BigInteger.ZERO;
for ( int n = 2; n < limit; n++ ) {
BigInteger term = BigInteger.TWO.multiply(riordans[n - 1]).add(THREE.multiply(riordans[n - 2]));
riordans[n] = BigInteger.valueOf(n - 1).multiply(term).divide(BigInteger.valueOf(n + 1));
}
System.out.println("The first 32 Riordan numbers:");
for ( int i = 0; i < 32; i++ ) {
System.out.print(String.format("%14d%s", riordans[i], ( i % 4 == 3 ? "\n" :" " )));
}
System.out.println();
for ( int count : List.of( 1_000, 10_000 ) ) {
int length = riordans[count - 1].toString().length();
System.out.println("The " + count + "th Riordan number has " + length + " digits");
}
}
}
- Output:
The first 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
The 1000th Riordan number has 472 digits
The 10000th Riordan number has 4765 digits
jq
The C implementation of jq has sufficient arithmetic accuracy for the first task, but because of the stretch task, the Go implementation has been used as gojq's integer arithmetic has unbounded accuracy.
Using the program below to calculate the first 100,000 Riordan numbers, gojq takes about 4.7 seconds on a 3GHz machine.
def riordan:
{ai: 1, a1: 0}
| ., foreach range(1; infinite) as $i (.;
{ai: ( ($i-1) * (2*.ai + 3*.a1) / ($i+1)),
a1: .ai } )
| .ai ;
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
def snip($n):
tostring|lpad(6)
+ ($n | tostring | "th: \(.[:10]) .. \(.[-10:]) (\(length) digits)" );
"First 32 Riordan numbers:",
foreach limit(100000; riordan) as $riordan (0; .+1;
if . <= 32 then $riordan
elif . == 1000 or . == 10000 or . == 100000 then snip($riordan)
else empty end)Invocation: jq -nr -f riordan.jq
- Output:
First 32 Riordan numbers: 1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 295673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140 1000th: 5107775686 .. 7484633052 (472 digits) 10000th: 1992741857 .. 0211157137 (4765 digits) 100000th: 5156659846 .. 4709713332 (47704 digits)
Julia
""" julia example for rosettacode.org/wiki/Riordan_number """
using Formatting
const riordans = zeros(BigInt, 10000)
riordans[begin] = 1
for i in firstindex(riordans)+1:lastindex(riordans)-1
riordans[i + 1] = (i - 1) * (2 * riordans[i] + 3 * riordans[i - 1]) ÷ (i + 1)
end
for i in 0:31
print(rpad(format(riordans[begin+i], commas = true), 18), (i + 1) % 4 == 0 ? "\n" : "")
end
println("\nThe 1,000th Riordan has $(length(string(riordans[1000]))) digits.")
println("The 10,000th Riordan has $(length(string(riordans[10_000]))) digits.")
- Output:
1 0 1 1 3 6 15 36 91 232 603 1,585 4,213 11,298 30,537 83,097 227,475 625,992 1,730,787 4,805,595 13,393,689 37,458,330 105,089,229 295,673,994 834,086,421 2,358,641,376 6,684,761,125 18,985,057,351 54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140 The 1,000th Riordan has 472 digits. The 10,000th Riordan has 4765 digits.
Nim
Task
import std/strformat
iterator riordan(): int =
var prev = 1
var curr = 0
yield prev
var n = 1
while true:
yield curr
inc n
let next = (n - 1) * (2 * curr + 3 * prev) div (n + 1)
prev = curr
curr = next
echo &"First 32 Riordan numbers:"
var count = 0
for n in riordan():
inc count
stdout.write &"{n:<13}"
stdout.write if count mod 4 == 0: '\n' else: ' '
if count == 32: break
- Output:
First 32 Riordan numbers: 1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 295673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140
Stretch task
import std/strformat
import integers
iterator riordan(): Integer =
var prev = newInteger(1)
var curr = newInteger(0)
yield prev
var n = 1
while true:
yield curr
inc n
let next = (n - 1) * (2 * curr + 3 * prev) div (n + 1)
prev = curr
curr = next
var count = 0
for n in riordan():
inc count
if count in [1000, 10000]:
echo &"The {count}th Riordan number has {len($n)} digits."
if count == 10000: break
- Output:
The 1000th Riordan number has 472 digits. The 10000th Riordan number has 4765 digits.
Perl
use v5.36;
use bigint;
use experimental <builtin for_list>;
use List::Util 'max';
use List::Lazy 'lazy_list';
use Lingua::EN::Numbers qw(num2en_ordinal);
sub abbr ($d) { my $l = length $d; $l < 41 ? $d : substr($d,0,20) . '..' . substr($d,-20) . " ($l digits)" }
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub table ($c, @V) { my $t = $c * (my $w = 2 + max map { length } @V); ( sprintf( ('%'.$w.'s')x@V, @V) ) =~ s/.{1,$t}\K/\n/gr }
my @riordan;
my $riordan_lazy = lazy_list { state @r = (1,0); state $n = 1; $n++; push @r, ($n-1) * (2*$r[1] + 3*$r[0]) / ($n+1) ; shift @r };
push @riordan, $riordan_lazy->next() for 1..1e4;
say 'First thirty-two Riordan numbers:';
say table 4, map { comma $_ } @riordan[0..31];
say 'The ' . num2en_ordinal($_) . ': ' . abbr $riordan[$_ - 1] for 1e3, 1e4;
- Output:
First thirty-two Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The one thousandth: 51077756867821111314..79942013897484633052 (472 digits)
The ten thousandth: 19927418577260688844..71395322020211157137 (4765 digits)
Phix
with javascript_semantics requires("1.0.2") -- mpz_get_str(comma_fill) was not working [!!!] include mpfr.e constant limit = 10000 sequence a = {mpz_init(1),mpz_init(0)} for n=2 to limit do mpz an = mpz_init() mpz_mul_si(an,a[n],2) mpz_addmul_si(an,a[n-1],3) mpz_mul_si(an,an,n-1) assert(mpz_fdiv_q_ui(an,an,n+1)=0) a &= an end for printf(1,"First 32 Riordan numbers:\n%s\n", {join_by(apply(true,mpz_get_str,{a[1..32],10,true}),1,4," ",fmt:="%17s")}) for i in {1e3, 1e4} do printf(1,"The %6s: %s\n", {ordinal(i), mpz_get_short_str(a[i])}) end for
- Output:
First 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The one thousandth: 51077756867821111314...79942013897484633052 (472 digits)
The ten thousandth: 19927418577260688844...71395322020211157137 (4,765 digits)
PL/I
showRiordan: procedure options( main ); /* find some Riordan numbers */
%replace maxRiordan by 32;
/* sets a to the first n riordan numbers a must have bounds 1 : n */
/* so the first number has index 1, not 0 */
riordan: procedure( n, a );
declare n binary( 15 )fixed;
declare a ( maxRiordan )decimal( 14 );
declare ( r2, r1, ri, i ) decimal( 14 );
if n >= 1 then do;
r2 = 1;
a( 1 ) = r2;
if n >= 2 then do;
r1 = 0;
a( 2 ) = r1;
do i = 2 to n;
ri = ( ( i - 1 )
* ( ( 2 * r1 )
+ ( 3 * r2 )
)
)
/ ( i + 1 );
a( i + 1 ) = ri;
r2 = r1;
r1 = ri;
end;
end;
end;
end riordan ;
declare r ( maxRiordan )decimal( 14 ), i binary( 15 )fixed;
call riordan( maxRiordan, r );
do i = 1 to maxRiordan;
put list( ' ', r( i ) );
if mod( i, 4 ) = 0 then put skip;
end;
end showRiordan;- Output:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
PL/M
... under CP/M (or an emulator)
PL/M only handles 8 and 16 bit unsigned integers but also provides two-digit BCD addition/subtraction with carry.
This sample uses the BCD facility to implement 16-digit arithmetic and solve the basic task. Ethiopian multiplication and Egyptian division are used, hence the length of the sample.
100H: /* FIND SOME RIORDAN NUMBERS */
DECLARE FALSE LITERALLY '0';
DECLARE TRUE LITERALLY '0FFH';
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
DECLARE DEC$LAST LITERALLY '7'; /* SUBSCRIPT OF LAST DIGIT PAIR */
DECLARE DEC$LEN LITERALLY '8'; /* LENGTH OF A 16-DIGIT BCD NUMBER */
DECLARE DEC$16 LITERALLY '( DEC$LEN )BYTE'; /* TYPE DECLARATION OF A */
/* 16-DIGIT BCD NUMBER - 8 BYTES */
PR$DEC: PROCEDURE( A$PTR ); /* PRINT AN UNSIGNED 16-DIGIT BCD NUMBER */
DECLARE A$PTR ADDRESS;
DECLARE A BASED A$PTR DEC$16;
DECLARE ( D, ZERO$CHAR, I, V ) BYTE;
ZERO$CHAR = ' ';
DO I = 0 TO DEC$LAST - 1;
V = A( I );
D = SHR( V AND 0F0H, 4 );
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + ( ZERO$CHAR := '0' ) );
D = V AND 0FH;
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + ( ZERO$CHAR := '0' ) );
END;
V = A( DEC$LAST );
D = SHR( V AND 0F0H, 4 );
IF D = 0 THEN CALL PR$CHAR( ZERO$CHAR );
ELSE CALL PR$CHAR( D + '0' );
D = V AND 0FH;
CALL PR$CHAR( D + '0' );
END PR$DEC ;
/* SETS THE 16-DIGIT BCD VALUE IN A TO 0 */
INIT$DEC: PROCEDURE( A$PTR );
DECLARE A$PTR ADDRESS;
DECLARE A BASED A$PTR ( 0 )BYTE;
DECLARE I BYTE;
DO I = 0 TO DEC$LAST;
A( I ) = 0;
END;
END INIT$DEC ;
/* SETS THE 16-DIGIT BCD VALUE IN A TO B */
SET$DEC: PROCEDURE( A$PTR, B );
DECLARE A$PTR ADDRESS, B BYTE;
DECLARE A BASED A$PTR DEC$16;
DECLARE ( I, P, V, D1, D2 ) BYTE;
V = B;
P = DEC$LAST;
DO I = 0 TO DEC$LAST;
IF V = 0
THEN A( P ) = 0;
ELSE DO;
D1 = V MOD 10;
D2 = ( V := V / 10 ) MOD 10;
A( P ) = SHL( D2, 4 ) OR D1;
V = V / 10;
END;
P = P - 1;
END;
END SET$DEC ;
/* ASSIGN THE 16-DIGIT BCD VALUD IN B TO A */
MOV$DEC: PROCEDURE( A$PTR, B$PTR );
DECLARE ( A$PTR, B$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$16, B BASED B$PTR DEC$16;
DECLARE I BYTE;
DO I = 0 TO DEC$LAST;
A( I ) = B( I );
END;
END MOV$DEC ;
/* BCD ADDITION - ADDS B TO A, STORING THE RESULT IN A */
/* A AND B MUST HAVE 16 DIGITS */
ADD$DEC: PROCEDURE( A$PTR, B$PTR );
DECLARE ( A$PTR, B$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$16, B BASED B$PTR DEC$16;
DECLARE ( A0, A1, A2, A3, A4, A5, A6, A7 ) BYTE;
DECLARE ( B0, B1, B2, B3, B4, B5, B6, B7 ) BYTE;
/* SEPARATE THE DIGIT PAIRS */
A0 = A( 0 ); A1 = A( 1 ); A2 = A( 2 ); A3 = A( 3 );
A4 = A( 4 ); A5 = A( 5 ); A6 = A( 6 ); A7 = A( 7 );
B0 = B( 0 ); B1 = B( 1 ); B2 = B( 2 ); B3 = B( 3 );
B4 = B( 4 ); B5 = B( 5 ); B6 = B( 6 ); B7 = B( 7 );
/* DO THE ADDITIONS */
A7 = DEC( A7 + B7 );
A6 = DEC( A6 PLUS B6 );
A5 = DEC( A5 PLUS B5 );
A4 = DEC( A4 PLUS B4 );
A3 = DEC( A3 PLUS B3 );
A2 = DEC( A2 PLUS B2 );
A1 = DEC( A1 PLUS B1 );
A0 = DEC( A0 PLUS B0 );
/* RETURN THE RESULT */
A( 0 ) = A0; A( 1 ) = A1; A( 2 ) = A2; A( 3 ) = A3;
A( 4 ) = A4; A( 5 ) = A5; A( 6 ) = A6; A( 7 ) = A7;
END ADD$DEC;
/* RETURNS TRUE IF THE 16-DIGIT BCD NUMBER A IS <= B */
/* USING BCD SUBTRACTION WITH CARRY - SUBTRACTS A FROM B DISCARDING */
/* THE RESULT ABD RETURNING TRUE IF THE CARRY FLAG IS CLEAR */
DEC$LE: PROCEDURE( A$PTR, B$PTR )BYTE;
DECLARE ( A$PTR, B$PTR,C$PTR ) ADDRESS;
DECLARE A BASED A$PTR DEC$16, B BASED B$PTR DEC$16, C BASED C$PTR DEC$16;
DECLARE ( A0, A1, A2, A3, A4, A5, A6, A7 ) BYTE;
DECLARE ( B0, B1, B2, B3, B4, B5, B6, B7 ) BYTE;
DECLARE ( CFLAG, I ) BYTE;
/* SEPARATE THE DIGIT PAIRS */
A0 = A( 0 ); A1 = A( 1 ); A2 = A( 2 ); A3 = A( 3 );
A4 = A( 4 ); A5 = A( 5 ); A6 = A( 6 ); A7 = A( 7 );
B0 = B( 0 ); B1 = B( 1 ); B2 = B( 2 ); B3 = B( 3 );
B4 = B( 4 ); B5 = B( 5 ); B6 = B( 6 ); B7 = B( 7 );
/* SUBTRACTION A FROM B */
CFLAG = DEC( B7 - A7 );
CFLAG = DEC( B6 MINUS A6 );
CFLAG = DEC( B5 MINUS A5 );
CFLAG = DEC( B4 MINUS A4 );
CFLAG = DEC( B3 MINUS A3 );
CFLAG = DEC( B2 MINUS A2 );
CFLAG = DEC( B1 MINUS A1 );
CFLAG = DEC( B0 MINUS A0 );
CFLAG = CARRY; /* IF THERE'S NO CARRY, B IS > A AND SO A <= B */
RETURN CFLAG = 0;
END DEC$LE;
/* BCD MULTIPLICATION BY AN UNSIGNED INTEGER VIA ETHIOPIAN MULTIPLICATION */
/* MULTIPLIES A BY B, STORES THE RESULT IN A - A MUST HAVE 16 DIGITS */
MUL$DEC: PROCEDURE( A$PTR, B );
DECLARE A$PTR ADDRESS, B BYTE;
DECLARE V BYTE, R DEC$16, ACCUMULATOR DEC$16;
CALL MOV$DEC( .R, A$PTR );
V = B;
CALL INIT$DEC( .ACCUMULATOR );
DO WHILE( V > 0 );
IF ( V AND 1 ) = 1 THEN DO;
CALL ADD$DEC( .ACCUMULATOR, .R );
END;
V = SHR( V, 1 );
CALL ADD$DEC( .R, .R );
END;
CALL MOV$DEC( A$PTR, .ACCUMULATOR );
END MUL$DEC ;
/* POWERS OF 2 TABLE FOR THE DIVISION ROUTINE */
/* 2^54 IS LARGER THAN A 10^16 */
DECLARE POWERS$OF$2 ( 54 /* 16 POINTERS TO 16 DIGIT BCD NUMBERS */
)ADDRESS;
DECLARE POWER$DATA ( 864 /* 54 16-DIGIT BCD NUMBERS */ )BYTE;
DO;
DECLARE ( P, P$POS ) ADDRESS;
DO P = 0 TO LAST( POWER$DATA ); POWER$DATA( P ) = 0; END;
POWER$DATA( DEC$LAST ) = 01H; /* SET LAST DIGIT OF THE 1ST POWER TO 1 */
P$POS = 0;
DO P = 0 TO LAST( POWERS$OF$2 );
POWERS$OF$2( P ) = .POWER$DATA( P$POS );
P$POS = P$POS + DEC$LEN;
END;
DO P = 1 TO LAST( POWERS$OF$2 );
CALL MOV$DEC( POWERS$OF$2( P ), POWERS$OF$2( P - 1 ) ); /* NEXT... */
CALL ADD$DEC( POWERS$OF$2( P ), POWERS$OF$2( P ) ); /* POWER */
END;
END;
/* BCD DIVISION BY AN UNSIGNED INTEGER VIA EGYPTIAN DIVISION */
/* DIVIDES A BY B, STORES THE RESULT IN A - A MUST HAVE 16 DIGITS */
DIV$DEC: PROCEDURE( A$PTR, B );
DECLARE A$PTR ADDRESS, B BYTE;
DECLARE DOUBLINGS ( 54 /* 16 POINTERS TO 16 DIGIT BCD NUMBERS */
)ADDRESS;
DECLARE DOUBLE$DATA ( 864 /* 54 16-DIGIT BCD NUMBERS */ )BYTE;
DECLARE ( D, D$POS ) ADDRESS;
DECLARE ACCUMULATOR DEC$16, QUOTIENT DEC$16, ACC$PLUS$$DOUBLING DEC$16;
DECLARE MORE$DOUBLINGS BYTE;
/* CONSTRUCT THE DOUBLINGS TABLE - A*1, A*2, A*3, ETC. */
CALL SET$DEC( .DOUBLE$DATA, B );
DOUBLINGS( 0 ) = .DOUBLE$DATA( 0 );
D$POS = 0; /* START OF THE FIRST DOUBLINGS ELEMENT */
D = 0;
MORE$DOUBLINGS = TRUE;
DO WHILE( MORE$DOUBLINGS );
D = D + 1;
D$POS = D$POS + DEC$LEN; /* POSITION TO THE NEXT ELEMENT */
DOUBLINGS( D ) = .DOUBLE$DATA( D$POS );
CALL MOV$DEC( DOUBLINGS( D ), DOUBLINGS( D - 1 ) );
CALL ADD$DEC( DOUBLINGS( D ), DOUBLINGS( D ) );
MORE$DOUBLINGS = DEC$LE( DOUBLINGS( D ), A$PTR )
AND D < LAST( DOUBLINGS );
END;
/* CONSTRUCT THE ACCUMULATOR AND QUOTIEMT */
CALL INIT$DEC( .ACCUMULATOR );
CALL INIT$DEC( .QUOTIENT );
D = D + 1;
DO WHILE( D >= 1 );
D = D - 1;
CALL MOV$DEC( .ACC$PLUS$DOUBLING, .ACCUMULATOR );
CALL ADD$DEC( .ACC$PLUS$DOUBLING, DOUBLINGS( D ) );
IF DEC$LE( .ACC$PLUS$DOUBLING, A$PTR ) THEN DO;
CALL MOV$DEC( .ACCUMULATOR, .ACC$PLUS$DOUBLING );
CALL ADD$DEC( .QUOTIENT, POWERS$OF$2( D ) );
END;
END;
CALL MOV$DEC( A$PTR, .QUOTIENT );
END DIV$DEC ;
/* TASK */
/* SETS A TO THE RIORDAN NUMBERS 0 .. N - LAST(A) MUST BE N */
RIORDAN: PROCEDURE( N, A$PTR );
DECLARE ( N, A$PTR ) ADDRESS;
DECLARE A BASED A$PTR ( 0 )ADDRESS;
DECLARE I ADDRESS;
DECLARE R2 DEC$16, R1 DEC$16;
DECLARE TWO$R1 DEC$16, THREE$R2 DEC$16;
CALL INIT$DEC( .R2 );
CALL INIT$DEC( .R1 );
IF N >= 0 THEN DO;
R2( LAST( R2 ) ) = 01H; /* SET LAST DIGIT OF R2 TO 1, I.E., R2 = 1 */
CALL MOV$DEC( A( 0 ), .R2 );
IF N >= 1 THEN DO;
CALL MOV$DEC( A( 1 ), .R1 );
DO I = 2 TO N;
CALL MOV$DEC( .TWO$R1, .R1 ); /* TWO$R1 = R1 ... */
CALL ADD$DEC( .TWO$R1, .R1 ); /* * 2 */
CALL MOV$DEC( .THREE$R2, .R2 ); /* THREE$R2 = R2 ... */
CALL ADD$DEC( .THREE$R2, .R2 ); /* THREE$R2 += R2 ... */
CALL ADD$DEC( .THREE$R2, .R2 ); /* THREE$R2 += R2 ... */
CALL ADD$DEC( .TWO$R1, .THREE$R2 ); /* TWO$R2 += THREE$R2 */
CALL MUL$DEC( .TWO$R1, I - 1 ); /* TWO$R2 *= ( I - 1 ) */
CALL DIV$DEC( .TWO$R1, I + 1 ); /* TWO$R1 /= ( I + 1 ) */
CALL MOV$DEC( A( I ), .TWO$R1 ); /* A( I ) = TWO$R1 */
CALL MOV$DEC( .R2, .R1 ); /* R2 = R1 */
CALL MOV$DEC( .R1, A( I ) ); /* R1 = A( I ) */
END;
END;
END;
END RIORDAN ;
/* CONSTRUCT AN ARRAY OF 16 DIGIT BCD NUMBERS */
DECLARE R ( 32 )ADDRESS; /* THE ARRAY OF RIORDAN NUMBERS */
DECLARE R$DATA ( 256 /* 32 * 8 */ )BYTE; /* THE RIORDAN NUMBER'S DIGITS */
DECLARE ( I, D$POS ) ADDRESS;
D$POS = 0;
DO I = 0 TO LAST( R );
R( I ) = .R$DATA( D$POS );
D$POS = D$POS + DEC$LEN;
END;
DO I = 0 TO LAST( R$DATA ); R$DATA( I ) = 0; END;
/* GET AND PRINT THE RIORDAN NUMBERS */
CALL RIORDAN( LAST( R ), .R );
DO I = 0 TO LAST( R );
CALL PR$CHAR( ' ' );
CALL PR$DEC( R( I ) );
IF ( I + 1 ) MOD 4 = 0 THEN CALL PR$NL;
END;
EOF- Output:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
Python
def Riordan(N):
a = [1, 0, 1]
for n in range(3, N):
a.append((n - 1) * (2 * a[n - 1] + 3 * a[n - 2]) // (n + 1))
return a
rios = Riordan(10_000)
for i in range(32):
print(f'{rios[i] : 18,}', end='\n' if (i + 1) % 4 == 0 else '')
print(f'The 1,000th Riordan has {len(str(rios[999]))} digits.')
print(f'The 10,000th Rirdan has {len(str(rios[9999]))} digits.')
- Output:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The 1,000th Riordan has 472 digits.
The 10,000th Rirdan has 4765 digits.
Quackery
[ dup -1 peek 2 *
over -2 peek 3 * +
over size tuck 1 - *
swap 1+ /
join ] is nextterm ( [ --> [ )
say "first 32 Riordan numbers: "
' [ 1 0 ]
30 times nextterm
witheach [ echo sp ]
cr cr
say "1000th Riordan number has "
' [ 1 0 ]
1000 2 - times nextterm
-1 peek number$ size echo
say " digits"
cr cr
say "10000th Riordan number has "
' [ 1 0 ]
10000 2 - times nextterm
-1 peek number$ size echo
say " digits"- Output:
first 32 Riordan numbers: 1 0 1 1 3 6 15 36 91 232 603 1585 4213 11298 30537 83097 227475 625992 1730787 4805595 13393689 37458330 105089229 295673994 834086421 2358641376 6684761125 18985057351 54022715451 154000562758 439742222071 1257643249140 1000th Riordan number has 472 digits 10000th Riordan number has 4765 digits
Raku
use Lingua::EN::Numbers;
my @riordan = 1, 0, { state $n = 1; (++$n - 1) / ($n + 1) × (3 × $^a + 2 × $^b) } … *;
my $upto = 32;
say "First {$upto.&cardinal} Riordan numbers:\n" ~ @riordan[^$upto]».&comma».fmt("%17s").batch(4).join("\n") ~ "\n";
sub abr ($_) { .chars < 41 ?? $_ !! .substr(0,20) ~ '..' ~ .substr(*-20) ~ " ({.chars} digits)" }
say "The {.Int.&ordinal}: " ~ abr @riordan[$_ - 1] for 1e3, 1e4
- Output:
First thirty-two Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
The one thousandth: 51077756867821111314..79942013897484633052 (472 digits)
The ten thousandth: 19927418577260688844..71395322020211157137 (4765 digits)
RPL
Needed to use unsigned integers to avoid the loss of the last digit of a(31).
≪ { #1 #0 }
WHILE DUP2 SIZE 1 - > REPEAT
DUP SIZE 1 -
DUP2 GETI 3 * 4 ROLLD
GET 2 * ROT +
OVER * SWAP 2 + /
+
END SWAP DROP
≫ 'RIORDAN' STO
31 RIORDAN
- Output:
1: { #1d #0d #1d #1d #3d #6d #15d #36d #91d #232d #603d #1585d #4213d 11298d #30537d #83097d #227475d #625992d #1730787d #4805595d #13393689d #37458330d #105089229d #295673994d #834086421d #2358641376d #6684761125d #18985057351d #54022715451d #154000562758d #439742222071d #1257643249140d }
SETL
program riordan;
a := {[0, 1], [1, 0]};
loop for n in [2..9999] do
a(n) := (n-1)*(2*a(n-1) + 3*a(n-2)) div (n+1);
end loop;
loop for n in [0..31] do
putchar(lpad(str a(n), 15));
if n mod 4=3 then print; end if;
end loop;
loop for n in [999, 9999] do
print("The", str (n+1)+"th Riordan number has", #str a(n), "digits.");
end loop;
end program;- Output:
1 0 1 1
3 6 15 36
91 232 603 1585
4213 11298 30537 83097
227475 625992 1730787 4805595
13393689 37458330 105089229 295673994
834086421 2358641376 6684761125 18985057351
54022715451 154000562758 439742222071 1257643249140
The 1000th Riordan number has 472 digits.
The 10000th Riordan number has 4765 digits.
Wren
import "./gmp" for Mpz
import "./fmt" for Fmt
var limit = 10000
var a = List.filled(limit, null)
a[0] = Mpz.one
a[1] = Mpz.zero
for (n in 2...limit) {
a[n] = (a[n-1] * 2 + a[n-2] * 3) * (n-1) / (n+1)
}
System.print("First 32 Riordan numbers:")
Fmt.tprint("$,17i", a[0..31], 4)
System.print()
for (i in [1e3, 1e4]) {
Fmt.print("$,8r: $20a ($,d digits)", i, a[i-1], a[i-1].toString.count)
}- Output:
First 32 Riordan numbers:
1 0 1 1
3 6 15 36
91 232 603 1,585
4,213 11,298 30,537 83,097
227,475 625,992 1,730,787 4,805,595
13,393,689 37,458,330 105,089,229 295,673,994
834,086,421 2,358,641,376 6,684,761,125 18,985,057,351
54,022,715,451 154,000,562,758 439,742,222,071 1,257,643,249,140
1,000th: 51077756867821111314...79942013897484633052 (472 digits)
10,000th: 19927418577260688844...71395322020211157137 (4,765 digits)