Bernoulli numbers: Difference between revisions

 

:*   show the Bernoulli numbers   '''B'''₀   through   '''B'''₆₀.

:*   express the Bernoulli numbers as fractions  (most are improper fractions).

:*   the fractions should be reduced.

 

;See also

* Entry [http://mathworld.wolfram.com/BernoulliNumber.html Bernoulli number] on The Eric Weisstein's World of Mathematics (TM).

* Luschny's [http://luschny.de/math/zeta/The-Bernoulli-Manifesto.html The Bernoulli Manifesto] for a discussion on &nbsp; <big> '''B₁ &nbsp; = &nbsp; -&frac12;''' &nbsp; versus &nbsp; '''+&frac12;'''. </big>

<br><br>

=={{header|Ada}}==

Using a GMP thick binding available at http://www.codeforge.com/article/422541

 

USE GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;

 

END IF;

END LOOP;

{{out}}

<pre>

B(60)=-1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|ALGOL 68}}==

{{works with|ALGOL 68G|Any - tested with release 2.8.3.win32}}

Uses the LONG LONG INT mode of Algol 68G which allows large precision integers.

# Show Bernoulli numbers B0 to B60 as rational numbers #

 

OD

END

{{out}}

<pre>

B(60) -1215233140483755572040304994079820246041491 / 56786730

</pre>

 

=={{header|Bracmat}}==

= B Bs answer indLn indexLen indexPadding

, n numberPadding p solPos solidusPos sp

& str$!answer

)

<pre>B(0)= 1/1

B(1)= 1/2

B(58)= 84483613348880041862046775994036021/354

B(60)=-1215233140483755572040304994079820246041491/56786730</pre>

=={{header|C}}==

{{libheader|GMP}}

#include <stdlib.h>

#include <gmp.h>

return 0;

}

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|C sharp|C#}}==

 

{{libheader|Mpir.NET}}

Translation of the C implementation

using Mpir.NET;

using System;

}

}

{{out}}

<pre>

 

{{libheader|MathNet.Numerics}}

using System;

using System.Console;

}

}

{{out}}

<pre>

{{libheader|System.Numerics}}

Algo based on the example provided in the header of this RC page (the one from Wikipedia). <br/> Extra feature - one can override the default of 60 by supplying a suitable number on the command line. The column widths are not hard-coded, but will adapt to the widths of the items listed.

using System.Numerics;

using System.Collections.Generic;

}

}

{{out}}

Default (nothing entered on command line):

B(126) = 5556330281949274850616324408918951380525567307126747246796782304333594286400508981287241419934529638692081513802696639 / 4357878

</pre>

=={{header|C++}}==

{{Works with|C++11}}

{{libheader|boost}}

* Configured with: --prefix=/Library/Developer/CommandLineTools/usr --with-gxx-include-dir=/usr/include/c++/4.2.1

* Apple LLVM version 9.1.0 (clang-902.0.39.1)

 

return 0;

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|Clojure}}==

 

 

ns test-project-intellij.core

(println q ":" (format-ans ans)))

 

{{out}}

<pre>

60 : -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|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 [http://oeis.org/A027641 the numerator list].

 

(loop with a = (make-array (list (1+ n)))

for m from 0 to n do

n

(numerator r)

 

{{out}}

B(60): -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Crystal}}==

 

{{Trans|Ruby}}

 

 

class Bernoulli

puts "B(%2i) = %*i/%i" % [i, max_width, v.numerator, v.denominator]

end

 

{{Trans|Python}}

Version 1: compute each number separately.

 

def bernoulli(n)

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? }

 

 

{{Trans|Python}}

Version 2: create faster generator to compute array of numbers once.

 

def bernoulli2(limit)

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? }

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|D}}==

This uses the D module from the Arithmetic/Rational task.

{{trans|Python}}

 

auto bernoulli(in uint n) pure nothrow /*@safe*/ {

foreach (immutable b; berns)

writefln("B(%2d) = %*d/%d", b[0], width, b[1].tupleof);

The output is exactly the same as the Python entry.

 

=={{header|EchoLisp}}==

 

 

Only 'small' rationals are supported in EchoLisp, i.e numerator and demominator < 2^31. So, we create a class of 'large' rationals, supported by the bigint library, and then apply the magic formula.

(lib 'bigint) ;; lerge numbers

(lib 'gloops) ;; classes

(define-method div (Rational Rational) (lambda (r q)

(normalize (Rational (* r.a q.b) (* r.b q.a)))))

{{Output}}

;; Bernoulli numbers

;; http://rosettacode.org/wiki/Bernoulli_numbers

 

(B 1) → 1 / 2

=={{header|Elixir}}==

defmodule Rational do

import Kernel, except: [div: 2]

end

 

 

{{out}}

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|F sharp|F#}}==

{{libheader|MathNet.Numerics.FSharp}}

open MathNet.Numerics

open System

printf ""

0

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|Factor}}==

One could use the "bernoulli" word from the math.extras vocabulary as follows:

[

0 1 1 "%2d : %d / %d\n" printf

58 : 84483613348880041862046775994036021 / 354

60 : -1215233140483755572040304994079820246041491 / 56786730

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.

n 1 + 0 <array> :> tab

1 1 tab set-nth

"%2d : %d / %d\n" printf

] each

It gives the same result as the native implementation, but is slightly faster.

...

=={{header|FreeBASIC}}==

{{libheader|GMP}}

' compile with: fbc -s console

' uses gmp

Print :Print "hit any key to end program"

Sleep

{{out}}

<pre>B( 0) = 1/1

B(58) = 84483613348880041862046775994036021/354

B(60) = -1215233140483755572040304994079820246041491/56786730</pre>

 

=={{header|FunL}}==

FunL has pre-defined function <code>B</code> in module <code>integers</code>, which is defined as:

 

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

 

{{out}}

B(60) = -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Fōrmulæ}}==

 

 

 

 

=={{header|GAP}}==

 

Print(a, "\n");

od;

[ 56, -2479392929313226753685415739663229/870 ]

[ 58, 84483613348880041862046775994036021/354 ]

=={{header|Go}}==

 

import (

}

}

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Haskell}}==

====Task algorithm====

The implementation of the algorithm is in the function bernoullis. The rest is for printing the results.

 

import System.Environment

 

berno i = 1 % (i + 1)

ulli _ [_] = []

{{Out}}

<pre>B(0) = 1/1

 

====Derivation from Faulhaber's triangle====

 

bernouillis :: Integer -> [Rational]

 

faulhaber :: [Ratio Integer] -> Integer -> [Ratio Integer]

 

main :: IO ()

main = do

let xs = bernouillis 60

putStrLn $

fTable

(filter ((0 /=) . snd) $ zip [0 ..] xs)

 

fTable s xShow fxShow f xs =

let w = maximum (length . xShow <$> xs)

 

showRatio :: Int -> Rational -> String

showRatio w r =

let d = denominator r

 

rjust :: Int -> a -> [a] -> [a]

{{Out}}

<pre>Bernouillis from Faulhaber triangle:

58 -> 84483613348880041862046775994036021 / 354

60 -> -1215233140483755572040304994079820246041491 / 56786730</pre>

=={{header|Icon}} and {{header|Unicon}}==

 

The following works in both languages:

 

procedure main(args)

procedure align(r,n)

return repl(" ",n-find("/",r))||r

 

Sample run:

->

</pre>

=={{header|J}}==

 

'''Implementation:'''

 

 

'''Task:'''

 

B 0 1

B 1 -1/2

B56 -2479392929313226753685415739663229/870

B58 84483613348880041862046775994036021/354

=={{header|Java}}==

 

public class BernoulliNumbers {

return A[0];

}

<pre>B(0 ) = 1

B(1 ) = 1 / 2

B(58) = 84483613348880041862046775994036021 / 354

B(60) = -1215233140483755572040304994079820246041491 / 56786730</pre>

=={{header|jq}}==

{{works with|jq|1.4}}

 

'''BigInt Stubs''':

# def lessOrEqual(x; y): # x <= y

# def long_add(x;y): # x+y

 

def long_mod(x;y):

 

# A fraction is represented by [numerator, denominator] in reduced form, with the sign on top

 

| if $a == $b then ["0", "1"]

else add($a; [ ($b[0]|negate), $b[1] ] )

 

'''Bernoulli Numbers''':

def bernoulli(n):

reduce range(0; n+1) as $m

.[$j-1] = multiply( [($j|tostring), "1"]; minus( .[$j-1] ; .[$j]) ) ))

| .[0] # (which is Bn)

 

'''The task''':

{{out}}

The following output was obtained using the previously mentioned BigInt library.

0: ["1","1"]

1: ["1","2"]

56: ["-2479392929313226753685415739663229","870"]

58: ["84483613348880041862046775994036021","354"]

=={{header|Julia}}==

A = Vector{Rational{BigInt}}(undef, n + 1)

for m = 0 : n

for (n, b) in enumerate(BernoulliList(60))

isodd(numerator(b)) && println("B($(n-1)) = $b")

 

Produces virtually the same output as the Python version.

=={{header|Kotlin}}==

{{trans|Java}}

{{works with|Commons Math|3.3.5}}

 

 

object Bernoulli {

if (n % 2 == 0 || n == 1)

System.out.printf("B(%-2d) = %-1s%n", n, Bernoulli(n))

{{out}}

Produces virtually the same output as the Java version.

=={{header|Lua}}==

LuaJIT version with FFI and GMP library

{{libheader|luagmp}}

{{works with|LuaJIT|2.0-2.1}}

local gmp = require 'gmp' ('libgmp')

local ffi = require'ffi'

gmp.z_clears(n,d)

gmp.q_clear(rop)

{{out}}

<pre>> time ./bernoulli_gmp.lua

./bernoulli_gmp.lua 0,02s user 0,00s system 97% cpu 0,022 total</pre>

Time compare: Python 0.591 sec, C 0.023 sec, Lua 0.022-0.025

=={{header|Maple}}==

{{out}}

<pre>[[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]]</pre>

=={{header|Mathematica}} / {{header|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.

Do[

a[[m]] = 1/m;

, {m, 1, n + 1}];

]

{{out}}

<pre>{0,1}

(Note from task's author: nobody is forced to use any specific algorithm, the one shown is just a suggestion.)

 

{{out}}

<pre>0 1

58 84483613348880041862046775994036021/354

60 -(1215233140483755572040304994079820246041491/56786730)</pre>

 

=={{header|Nim}}==

import strformat

 

a[m] = newRat(1, m + 1)

for j in countdown(m, 1):

result = a[0]

 

for (n, b) in values:

let s = fmt"{($b.num).alignString(maxLen, '>')} / {b.denom}"

 

{{out}}

<pre> 0: 1 / 1

2: 1 / 6

4: -1 / 30

58: 84483613348880041862046775994036021 / 354

60: -1215233140483755572040304994079820246041491 / 56786730</pre>

=={{header|PARI/GP}}==

{{out}}

<pre>0 1

58 84483613348880041862046775994036021/354

60 -1215233140483755572040304994079820246041491/56786730</pre>

=={{header|Pascal|FreePascal}}==

{{libheader|BigDecimalMath}}

Tested with fpc 3.0.4

(* Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99 *)

end;

end.

 

{{out}}

B(60) : -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|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.

 

use strict;

use warnings;

 

bernoulli_print();

The output is exactly the same as the Python entry.

 

We can also use modules for faster results. E.g.

{{libheader|ntheory}}

 

for my $n (0 .. 60) {

my($num,$den) = bernfrac($n);

printf "B(%2d) = %44s/%s\n", $n, $num, $den if $num != 0;

with identical output. Or:

 

for my $n (0 .. 60) {

my($num,$den) = split "/", bernfrac($n);

printf("B(%2d) = %44s/%s\n", $n, $num, $den||1) if $num != 0;

with the difference being that Pari chooses <math>B_1</math> = -&frac12;.

=={{header|Phix}}==

{{libheader|Phix/mpfr}}

{{trans|C}}

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|PicoLisp}}==

Brute force and method by Srinivasa Ramanujan.

 

(de fact (N)

(test (berno N) (berno-brute N)) )

 

=={{header|PL/I}}==

declare i fixed binary;

declare B complex fixed (31);

(3 A, column(10), F(32), 2 A);

end;

The above uses GCD (see Rosetta Code) extended for 31-digit working.

 

B(36)= -26315271553053477373/1919190

</pre>

=={{header|Python}}==

===Python: Using task algorithm===

 

def bernoulli(n):

width = max(len(str(b.numerator)) for i,b in bn)

for i,b in bn:

 

{{out}}

===Python: Optimised task algorithm===

Using the optimization mentioned in the Perl entry to reduce intermediate calculations we create and use the generator bernoulli2():

A, m = [], 0

while True:

width = max(len(str(b.numerator)) for i,b in bn2)

for i,b in bn2:

 

Output is exactly the same as before.

 

=={{header|R}}==

 

library(pracma)

cat("B(",idx,") = ",n,"/",d,"\n", sep = "")

}

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Racket}}==

 

use the same emmitter... it's just a matter of how long to wait for the emission.

 

;; For: http://rosettacode.org/wiki/Bernoulli_numbers

 

(list 1/1 (app abs 1/2) 1/6 -1/30 1/42 -1/30 _ ...))

; timing only ...

 

{{out}}

B(58) = 84483613348880041862046775994036021/354

B(60) = -1215233140483755572040304994079820246041491/56786730</pre>

=={{header|Raku}}==

(formerly Perl 6)

First, a straighforward implementation of the naïve algorithm in the task description.

{{works with|Rakudo|2015.12}}

my @a;

for 0..$n -> $m {

my $form = "B(%2d) = \%{$width}d/%d\n";

 

{{out}}

<pre>B( 0) = 1/1

 

{{works with|Rakudo|2015.12}}

my @a;

for 0..* -> $m {

my $form = "B(%d)\t= \%{$width}d/%d\n";

 

{{out}}

<pre>B(0) = 1/1

{{works with|Rakudo|2016.12}}

 

this.key => this.key * (this.value - prev.value)

}

my $form = "B(%d)\t= \%{$width}d/%d\n";

 

Same output as memoization example

<br>

:::::::::::: where &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <big><big> <math> \binom kr</math> </big></big> &nbsp; &nbsp; &nbsp; is a binomial coefficient. <br>

parse arg N .; if N=='' | N=="," then N= 60 /*Not specified? Then use the default.*/

numeric digits max(9, n*2) /*increase the decimal digits if needed*/

w= max(length(N), 4); Nw= N + w + N % 4 /*used for aligning (output) fractions.*/

say copies('─',w) copies("─", max(78-w,Nw+2*w)) /*display 2nd line of title, separators*/

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. */

/*──────────────────────────────────────────────────────────────────────────────────────*/

bern: parse arg x; if x==0 then return '1/1' /*handle the special case of zero. */

$lcm= LCM(sd, ad) /*use Least Common Denominator function*/

sn= $lcm % sd * sn; sd= $lcm /*calculate the current numerator. */

sn= sn + an /* " " current " */

end /*k*/ /* [↑] calculate the SN/SD sequence.*/

/*──────────────────────────────────────────────────────────────────────────────────────*/

comb: procedure expose !.; parse arg x,y; if x==y then return 1

if x-y < y then y= x-y /*x-y < y? Then use a new Y.*/

z= perm(x, y); do j=2 for y-1; z= z % j

end /*j*/

/*──────────────────────────────────────────────────────────────────────────────────────*/

GCD: procedure; parse arg x,y; x= abs(x)

do until y==0; parse value x//y y with y x; end; return x

/*──────────────────────────────────────────────────────────────────────────────────────*/

/*──────────────────────────────────────────────────────────────────────────────────────*/

{{out|output|text=&nbsp; when using the default input:}}

<pre>

──── ───────────────────────────────────────────────────────────────────────────────────────

0 1/1

60 -1215233140483755572040304994079820246041491/56786730

</pre>

 

=={{header|Ruby}}==

{{trans|Python}}

ar = []

0.step do |m|

b_nums.each_with_index {|b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? }

 

{{out}}

<pre>

=={{header|Rust}}==

 

// the optimized function. The speeds vary significantly: relative

// speeds of optimized:iterator:naive implementations is 625:25:1.

}

}

{{out}}

<pre>

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|Scala}}==

'''With Custom Rational Number Class'''<br/>

(code will run in Scala REPL with a cut-and-paste without need for a third-party library)

case class BFraction( numerator:BigInt, denominator:BigInt ) {

require( denominator != BigInt(0), "Denominator cannot be zero" )

println( f"$label%-6s $num / ${b.den}" )

}}

{{out}}

<pre>b(0) 1 / 1

b(60) -1215233140483755572040304994079820246041491 / 56786730

</pre>

 

=={{header|Seed7}}==

The program below uses [http://seed7.sourceforge.net/manual/types.htm#bigRational bigRational]

function automatically writes repeating decimals in parentheses, when necessary.

 

include "bigrat.s7i";

 

end if;

end for;

 

{{out}}

B(60) = -1215233140483755572040304994079820246041491 / 56786730 -21399949257225333665810744765191097.3(926741511617238745742183076926598872659158222352299560126106)

</pre>

=={{header|Sidef}}==

Built-in:

 

Recursive solution (with auto-memoization):

 

n.is_one && return 1/2

var Bn = bernoulli_number(n) || next

printf("B(%2d) = %44s / %s\n", n, Bn.nude)

 

Using Ramanujan's congruences (pretty fast):

 

return 1/2 if n.is_one

binomial(n+3, n - 6*k) * __FUNC__(n - 6*k)

})) / binomial(n+3, n)

 

Using Euler's product formula for the Riemann zeta function and the Von Staudt–Clausen theorem (very fast):

 

n.is_zero && return 1

 

(-1)**(n/2 + 1) * int(ceil(d*K / z)) / d

 

The Akiyama–Tanigawa algorithm:

var a = []

for m in (0..60) {

}

 

 

{{out}}

B(60) = -1215233140483755572040304994079820246041491 / 56786730

</pre>

=={{header|SPAD}}==

{{works with|FriCAS, OpenAxiom, Axiom}}

for n in 0..60 | (b:=bernoulli(n)$INTHEORY; b~=0) repeat print [n,b]

Package:[http://fricas.github.io/api/IntegerNumberTheoryFunctions.html?highlight=bernoulli IntegerNumberTheoryFunctions]

 

Type: Void

</pre>

=={{header|Swift}}==

 

Uses the Frac type defined in the [http://rosettacode.org/wiki/Arithmetic/Rational#Swift Rational] task.

 

 

public func bernoulli<T: BinaryInteger & SignedNumeric>(n: Int) -> Frac<T> {

 

print("B(\(n)) = \(b)")

 

{{out}}

B(58) = Frac(84483613348880041862046775994036021 / 354)

B(60) = Frac(-1215233140483755572040304994079820246041491 / 56786730)</pre>

=={{header|Tcl}}==

for {set m 0} {$m <= $n} {incr m} {

lappend A [list 1 [expr {$m + 1}]]

foreach {n num denom} $result {

puts [format {B_%-2d = %*lld/%lld} $n $len $num $denom]

{{out}}

<pre>

B_60 = -1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Visual Basic .NET}}==

{{works with|Visual Basic .NET|2013}}

{{libheader|System.Numerics}}

Imports System.Numerics 'BigInteger

 

End Sub 'bernoulli_BigInt

{{out}}

<pre>

B(60)=-1215233140483755572040304994079820246041491/56786730

</pre>

=={{header|Wren}}==

{{libheader|Wren-fmt}}

{{libheader|Wren-big}}

 

var bernoulli = Fn.new { |n|

var b = bernoulli.call(n)

if (b != BigRat.zero) Fmt.print("B($2d) = $44i / $i", n, b.num, b.den)

 

{{out}}

{{trans|EchoLisp}}

Uses lib GMP (GNU MP Bignum Library).

fcn init(_a,_b){ var a=_a, b=_b; normalize(); }

fcn toString{ "%50d / %d".fmt(a,b) }

}

fcn __opDiv(n){ self(a*n.b,b*n.a) } // Rat / Rat

fcn B(N){ // calculate Bernoulli(n)

var A=List.createLong(100,0); // aka static aka not thread safe

}

A[0]

{{out}}

<pre>