Bernoulli numbers: Difference between revisions
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b(60) -1215233140483755572040304994079820246041491 / 56786730 |
b(60) -1215233140483755572040304994079820246041491 / 56786730 |
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</pre> |
</pre> |
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=={{header|Scheme}}== |
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{{works with|Chez Scheme}} |
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<lang scheme>; Return the n'th Bernoulli number. |
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(define bernoulli |
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(lambda (n) |
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(let ((a (make-vector (1+ n)))) |
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(do ((m 0 (1+ m))) |
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((> m n)) |
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(vector-set! a m (/ 1 (1+ m))) |
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(do ((j m (1- j))) |
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((< j 1)) |
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(vector-set! a (1- j) (* j (- (vector-ref a (1- j)) (vector-ref a j)))))) |
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(vector-ref a 0)))) |
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; Convert a rational to a string. If an integer, ends with "/1". |
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(define rational->string |
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(lambda (rational) |
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(format "~a/~a" (numerator rational) (denominator rational)))) |
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; Returns the string length of the numerator of a rational. |
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(define rational-numerator-length |
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(lambda (rational) |
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(string-length (format "~a" (numerator rational))))) |
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; Formats a rational with left-padding such that total length to the slash is as given. |
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(define rational-padded |
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(lambda (rational total-length-to-slash) |
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(let* ((length-padding (- total-length-to-slash (rational-numerator-length rational))) |
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(padding-string (make-string length-padding #\ ))) |
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(string-append padding-string (rational->string rational))))) |
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; Return the Bernoulli numbers 0 through n in a list. |
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(define make-bernoulli-list |
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(lambda (n) |
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(if (= n 0) |
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(list (bernoulli n)) |
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(append (make-bernoulli-list (1- n)) (list (bernoulli n)))))) |
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; Print the non-zero Bernoulli numbers 0 through 60 aligning the slashes. |
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(let* ((bernoullis-list (make-bernoulli-list 60)) |
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(numerator-lengths (map rational-numerator-length bernoullis-list)) |
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(max-numerator-length (apply max numerator-lengths))) |
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(let print-bernoulli ((index 0) (numbers bernoullis-list)) |
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(cond |
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((null? numbers)) |
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((= 0 (car numbers)) |
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(print-bernoulli (1+ index) (cdr numbers))) |
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(else |
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(printf "B(~2@a) = ~a~%" index (rational-padded (car numbers) max-numerator-length)) |
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(print-bernoulli (1+ index) (cdr numbers))))))</lang> |
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{{out}} |
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<pre>$ scheme --script bernoulli.scm |
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B( 0) = 1/1 |
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B( 1) = 1/2 |
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B( 2) = 1/6 |
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B( 4) = -1/30 |
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B( 6) = 1/42 |
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B( 8) = -1/30 |
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B(10) = 5/66 |
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B(12) = -691/2730 |
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B(14) = 7/6 |
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B(16) = -3617/510 |
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B(18) = 43867/798 |
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B(20) = -174611/330 |
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B(22) = 854513/138 |
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B(24) = -236364091/2730 |
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B(26) = 8553103/6 |
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B(28) = -23749461029/870 |
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B(30) = 8615841276005/14322 |
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B(32) = -7709321041217/510 |
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B(34) = 2577687858367/6 |
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B(36) = -26315271553053477373/1919190 |
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B(38) = 2929993913841559/6 |
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B(40) = -261082718496449122051/13530 |
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B(42) = 1520097643918070802691/1806 |
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B(44) = -27833269579301024235023/690 |
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B(46) = 596451111593912163277961/282 |
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B(48) = -5609403368997817686249127547/46410 |
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B(50) = 495057205241079648212477525/66 |
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B(52) = -801165718135489957347924991853/1590 |
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B(54) = 29149963634884862421418123812691/798 |
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B(56) = -2479392929313226753685415739663229/870 |
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B(58) = 84483613348880041862046775994036021/354 |
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B(60) = -1215233140483755572040304994079820246041491/56786730</pre> |
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=={{header|Seed7}}== |
=={{header|Seed7}}== |
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Revision as of 13:29, 25 February 2022
You are encouraged to solve this task according to the task description, using any language you may know.
Bernoulli numbers are used in some series expansions of several functions (trigonometric, hyperbolic, gamma, etc.), and are extremely important in number theory and analysis.
Note that there are two definitions of Bernoulli numbers; this task will be using the modern usage (as per The National Institute of Standards and Technology convention).
The nth Bernoulli number is expressed as Bn.
- Task
- show the Bernoulli numbers B0 through B60.
- suppress the output of values which are equal to zero. (Other than B1 , all odd Bernoulli numbers have a value of zero.)
- express the Bernoulli numbers as fractions (most are improper fractions).
- the fractions should be reduced.
- index each number in some way so that it can be discerned which Bernoulli number is being displayed.
- align the solidi (/) if used (extra credit).
- An algorithm
The Akiyama–Tanigawa algorithm for the "second Bernoulli numbers" as taken from wikipedia is as follows:
for m from 0 by 1 to n do
A[m] ← 1/(m+1)
for j from m by -1 to 1 do
A[j-1] ← j×(A[j-1] - A[j])
return A[0] (which is Bn)
- See also
- Sequence A027641 Numerator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
- Sequence A027642 Denominator of Bernoulli number B_n on The On-Line Encyclopedia of Integer Sequences.
- Entry Bernoulli number on The Eric Weisstein's World of Mathematics (TM).
- Luschny's The Bernoulli Manifesto for a discussion on B1 = -½ versus +½.
Ada
Using a GMP thick binding available at http://www.codeforge.com/article/422541
<lang Ada>WITH GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings; USE GMP.Rationals, GMP.Integers, Ada.Text_IO, Ada.Strings.Fixed, Ada.Strings;
PROCEDURE Main IS
FUNCTION Bernoulli_Number (N : Natural) RETURN Unbounded_Fraction IS
FUNCTION "/" (Left, Right : Natural) RETURN Unbounded_Fraction IS
(To_Unbounded_Integer (Left) / To_Unbounded_Integer (Right));
A : ARRAY (0 .. N) OF Unbounded_Fraction;
BEGIN
FOR M IN 0 .. N LOOP
A (M) := 1 / (M + 1);
FOR J IN REVERSE 1 .. M LOOP
A (J - 1) := (J / 1 ) * (A (J - 1) - A (J));
END LOOP;
END LOOP;
RETURN A (0);
END Bernoulli_Number;
BEGIN
FOR I IN 0 .. 60 LOOP
IF I MOD 2 = 0 OR I = 1 THEN
DECLARE
B : Unbounded_Fraction := Bernoulli_Number (I);
S : String := Image (GMP.Rationals.Numerator (B));
BEGIN
Put_Line ("B (" & (IF I < 10 THEN " " ELSE "") & Trim (I'Img, Left)
& ")=" & (44 - S'Length) * " " & Image (B));
END;
END IF;
END LOOP;
END Main;</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
ALGOL 68
Uses the LONG LONG INT mode of Algol 68G which allows large precision integers. <lang algol68>BEGIN
# Show Bernoulli numbers B0 to B60 as rational numbers #
# Uses code from the Arithmetic/Rational task modified to use # # LONG LONG INT to allow for the large number of digits requried #
PR precision 100 PR # sets the precision of LONG LONG INT #
# Code from the Arithmetic/Rational task # # ============================================================== #
MODE FRAC = STRUCT( LONG LONG INT num #erator#, den #ominator#);
PROC gcd = (LONG LONG INT a, b) LONG LONG INT: # greatest common divisor #
(a = 0 | b |: b = 0 | a |: ABS a > ABS b | gcd(b, a MOD b) | gcd(a, b MOD a));
PROC lcm = (LONG LONG INT a, b)LONG LONG INT: # least common multiple #
a OVER gcd(a, b) * b;
PRIO // = 9; # higher then the ** operator #
OP // = (LONG LONG INT num, den)FRAC: ( # initialise and normalise #
LONG LONG INT common = gcd(num, den);
IF den < 0 THEN
( -num OVER common, -den OVER common)
ELSE
( num OVER common, den OVER common)
FI
);
OP + = (FRAC a, b)FRAC: (
LONG LONG INT common = lcm(den OF a, den OF b);
FRAC result := ( common OVER den OF a * num OF a + common OVER den OF b * num OF b, common );
num OF result//den OF result
);
OP - = (FRAC a, b)FRAC: a + -b,
* = (FRAC a, b)FRAC: (
LONG LONG INT num = num OF a * num OF b,
den = den OF a * den OF b;
LONG LONG INT common = gcd(num, den);
(num OVER common) // (den OVER common)
);
OP - = (FRAC frac)FRAC: (-num OF frac, den OF frac);
# ============================================================== #
# end code from the Arithmetic/Rational task #
# Additional FRACrelated operators # OP * = ( INT a, FRAC b )FRAC: ( num OF b * a ) // den OF b; OP // = ( INT a, INT b )FRAC: LONG LONG INT( a ) // LONG LONG INT( b );
# returns the nth Bernoulli number, n must be >= 0 #
# Uses the algorithm suggested by the task, so B(1) is +1/2 #
PROC bernoulli = ( INT n )FRAC:
IF n < 0
THEN # n is out of range # 0 // 1
ELSE # n is valid #
[ 0 : n ]FRAC a;
FOR i FROM LWB a TO UPB a DO a[ i ] := 0 // 1 OD;
FOR m FROM 0 TO n DO
a[ m ] := 1 // ( m + 1 );
FOR j FROM m BY -1 TO 1 DO
a[ j - 1 ] := j * ( a[ j - 1 ] - a[ j ] )
OD
OD;
a[ 0 ]
FI # bernoulli # ;
FOR n FROM 0 TO 60 DO
FRAC bn := bernoulli( n );
IF num OF bn /= 0 THEN
# have a non-0 Bn #
print( ( "B(", whole( n, -2 ), ") ", whole( num OF bn, -50 ), " / ", whole( den OF bn, 0 ), newline ) )
FI
OD
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
Bracmat
<lang bracmat> ( BernoulliList
= B Bs answer indLn indexLen indexPadding
, n numberPadding p solPos solidusPos sp
. ( B
= m A a j b
. -1:?m
& :?A
& whl
' ( 1+!m:~>!arg:?m
& ((!m+1:?j)^-1:?a)
map
$ ( (
= .(-1+!j:?j)*(!arg+-1*!a):?a
)
. !A
)
: ?A
)
& !A:? @?b
& !b
)
& -1:?n
& :?Bs
& whl
' ( 1+!n:~>!arg:?n
& B$!n !Bs:?Bs
)
& @(!arg:? [?indexLen)
& 1+!indexLen:?indexLen
& !Bs:%@(?:? "/" [?solidusPos ?) ?
& 1+!solidusPos:?solidusPos:?p
& :?sp
& whl
' (!p+-1:~<0:?p&" " !sp:?sp)
& :?answer
& whl
' ( !Bs:%?B ?Bs
& ( !B:0
| (!B:/|str$(!B "/1"):?B)
& @(!B:? "/" [?solPos ?)
& @(!arg:? [?indLn)
& !sp
: ? [(-1*!indexLen+!indLn) ?indexPadding
: ? [(-1*!solidusPos+!solPos) ?numberPadding
& "B("
!arg
")="
!indexPadding
!numberPadding
(!B:>0&" "|)
!B
\n
!answer
: ?answer
)
& -1+!arg:?arg
)
& str$!answer
)
& BernoulliList$60;</lang>
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
<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, 61) // the second parameter is the number of range elements, and is not the final item of the range.
.Select(n => new {N = n, BernoulliNumber = CalculateBernoulli(n)})
.Where(b => !b.BernoulliNumber.Numerator.IsZero)
.Select(b => string.Format("B({0, 2}) = {1, 44} / {2}", b.N, b.BernoulliNumber.Numerator, b.BernoulliNumber.Denominator))
.ToList()
.ForEach(WriteLine);
}
}
} </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 System.Numerics
Algo based on the example provided in the header of this RC page (the one from Wikipedia).
Extra feature - one can override the default of 60 by supplying a suitable number on the command line. The column widths are not hard-coded, but will adapt to the widths of the items listed.
<lang csharp>using System;
using System.Numerics;
using System.Collections.Generic;
namespace bern {
class Program
{
struct BerNum { public int index; public BigInteger Numer, Denomin; };
static int w1 = 1, w2 = 1; // widths for formatting output
static int max = 60; // default maximum, can override on command line
// returns nth Bernoulli number
static BerNum CalcBernoulli(int n)
{
BerNum res;
BigInteger f;
BigInteger[] nu = new BigInteger[n + 1],
de = new BigInteger[n + 1];
for (int m = 0; m <= n; m++)
{
nu[m] = 1; de[m] = m + 1;
for (int j = m; j > 0; j--)
if ((f = BigInteger.GreatestCommonDivisor(
nu[j - 1] = j * (de[j] * nu[j - 1] - de[j - 1] * nu[j]),
de[j - 1] *= de[j])) != BigInteger.One)
{ nu[j - 1] /= f; de[j - 1] /= f; }
}
res.index = n; res.Numer = nu[0]; res.Denomin = de[0];
w1 = Math.Max(n.ToString().Length, w1); // ratchet up widths
w2 = Math.Max(res.Numer.ToString().Length, w2);
if (max > 50) Console.Write("."); // progress dots appear for larger values
return res;
}
static void Main(string[] args)
{
List<BerNum> BNumbList = new List<BerNum>();
// defaults to 60 when no (or invalid) command line parameter is present
if (args.Length > 0) {
int.TryParse(args[0], out max);
if (max < 1 || max > Int16.MaxValue) max = 60;
if (args[0] == "0") max = 0;
}
for (int i = 0; i <= max; i++) // fill list with values
{
BerNum BNumb = CalcBernoulli(i);
if (BNumb.Numer != BigInteger.Zero) BNumbList.Add(BNumb);
}
if (max > 50) Console.WriteLine();
string strFmt = "B({0, " + w1.ToString() + "}) = {1, " + w2.ToString() + "} / {2}";
// display formatted list
foreach (BerNum bn in BNumbList)
Console.WriteLine(strFmt , bn.index, bn.Numer, bn.Denomin);
if (System.Diagnostics.Debugger.IsAttached) Console.Read();
}
}
} </lang>
- Output:
Default (nothing entered on command line):
............................................................. B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
Output with "8" entered on command line:
B(0) = 1 / 1 B(1) = 1 / 2 B(2) = 1 / 6 B(4) = -1 / 30 B(6) = 1 / 42 B(8) = -1 / 30
Output with "126" entered on the command line:
............................................................................................................................... B( 0) = 1 / 1 B( 1) = 1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B( 10) = 5 / 66 B( 12) = -691 / 2730 B( 14) = 7 / 6 B( 16) = -3617 / 510 B( 18) = 43867 / 798 B( 20) = -174611 / 330 B( 22) = 854513 / 138 B( 24) = -236364091 / 2730 B( 26) = 8553103 / 6 B( 28) = -23749461029 / 870 B( 30) = 8615841276005 / 14322 B( 32) = -7709321041217 / 510 B( 34) = 2577687858367 / 6 B( 36) = -26315271553053477373 / 1919190 B( 38) = 2929993913841559 / 6 B( 40) = -261082718496449122051 / 13530 B( 42) = 1520097643918070802691 / 1806 B( 44) = -27833269579301024235023 / 690 B( 46) = 596451111593912163277961 / 282 B( 48) = -5609403368997817686249127547 / 46410 B( 50) = 495057205241079648212477525 / 66 B( 52) = -801165718135489957347924991853 / 1590 B( 54) = 29149963634884862421418123812691 / 798 B( 56) = -2479392929313226753685415739663229 / 870 B( 58) = 84483613348880041862046775994036021 / 354 B( 60) = -1215233140483755572040304994079820246041491 / 56786730 B( 62) = 12300585434086858541953039857403386151 / 6 B( 64) = -106783830147866529886385444979142647942017 / 510 B( 66) = 1472600022126335654051619428551932342241899101 / 64722 B( 68) = -78773130858718728141909149208474606244347001 / 30 B( 70) = 1505381347333367003803076567377857208511438160235 / 4686 B( 72) = -5827954961669944110438277244641067365282488301844260429 / 140100870 B( 74) = 34152417289221168014330073731472635186688307783087 / 6 B( 76) = -24655088825935372707687196040585199904365267828865801 / 30 B( 78) = 414846365575400828295179035549542073492199375372400483487 / 3318 B( 80) = -4603784299479457646935574969019046849794257872751288919656867 / 230010 B( 82) = 1677014149185145836823154509786269900207736027570253414881613 / 498 B( 84) = -2024576195935290360231131160111731009989917391198090877281083932477 / 3404310 B( 86) = 660714619417678653573847847426261496277830686653388931761996983 / 6 B( 88) = -1311426488674017507995511424019311843345750275572028644296919890574047 / 61410 B( 90) = 1179057279021082799884123351249215083775254949669647116231545215727922535 / 272118 B( 92) = -1295585948207537527989427828538576749659341483719435143023316326829946247 / 1410 B( 94) = 1220813806579744469607301679413201203958508415202696621436215105284649447 / 6 B( 96) = -211600449597266513097597728109824233673043954389060234150638733420050668349987259 / 4501770 B( 98) = 67908260672905495624051117546403605607342195728504487509073961249992947058239 / 6 B(100) = -94598037819122125295227433069493721872702841533066936133385696204311395415197247711 / 33330 B(102) = 3204019410860907078243020782116241775491817197152717450679002501086861530836678158791 / 4326 B(104) = -319533631363830011287103352796174274671189606078272738327103470162849568365549721224053 / 1590 B(106) = 36373903172617414408151820151593427169231298640581690038930816378281879873386202346572901 / 642 B(108) = -3469342247847828789552088659323852541399766785760491146870005891371501266319724897592306597338057 / 209191710 B(110) = 7645992940484742892248134246724347500528752413412307906683593870759797606269585779977930217515 / 1518 B(112) = -2650879602155099713352597214685162014443151499192509896451788427680966756514875515366781203552600109 / 1671270 B(114) = 21737832319369163333310761086652991475721156679090831360806110114933605484234593650904188618562649 / 42 B(116) = -309553916571842976912513458033841416869004128064329844245504045721008957524571968271388199595754752259 / 1770 B(118) = 366963119969713111534947151585585006684606361080699204301059440676414485045806461889371776354517095799 / 6 B(120) = -51507486535079109061843996857849983274095170353262675213092869167199297474922985358811329367077682677803282070131 / 2328255930 B(122) = 49633666079262581912532637475990757438722790311060139770309311793150683214100431329033113678098037968564431 / 6 B(124) = -95876775334247128750774903107542444620578830013297336819553512729358593354435944413631943610268472689094609001 / 30 B(126) = 5556330281949274850616324408918951380525567307126747246796782304333594286400508981287241419934529638692081513802696639 / 4357878
C++
<lang cpp>/**
* Configured with: --prefix=/Library/Developer/CommandLineTools/usr --with-gxx-include-dir=/usr/include/c++/4.2.1 * Apple LLVM version 9.1.0 (clang-902.0.39.1) * Target: x86_64-apple-darwin17.5.0 * Thread model: posix
- /
- include <boost/multiprecision/cpp_int.hpp> // 1024bit precision
- include <boost/rational.hpp> // Rationals
- include <iostream> // formatting with std::cout
- include <vector> // Container
typedef boost::rational<boost::multiprecision::int1024_t> rational; // reduce boilerplate
rational bernoulli(size_t n) {
auto out = std::vector<rational>();
for (size_t m = 0; m <= n; m++) {
out.emplace_back(1, (m + 1)); // automatically constructs object
for (size_t j = m; j >= 1; j--) {
out[j - 1] = rational(j) * (out[j - 1] - out[j]);
}
}
return out[0];
}
int main() {
for (size_t n = 0; n <= 60; n += n >= 2 ? 2 : 1) {
auto b = bernoulli(n);
std::cout << "B(" << std::right << std::setw(2) << n << ") = ";
std::cout << std::right << std::setw(44) << b.numerator();
std::cout << " / " << b.denominator() << std::endl;
}
return 0;
}</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
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
Crystal
<lang ruby>require "big"
class Bernoulli
include Iterator(Tuple(Int32, BigRational))
def initialize @a = [] of BigRational @m = 0 end
def next
@a << BigRational.new(1, @m+1)
@m.downto(1) { |j| @a[j-1] = j*(@a[j-1] - @a[j]) }
v = @m.odd? && @m != 1 ? BigRational.new(0, 1) : @a.first
return {@m, v}
ensure
@m += 1
end
end
b = Bernoulli.new bn = b.first(61).to_a
max_width = bn.map { |_, v| v.numerator.to_s.size }.max bn.reject { |i, v| v.zero? }.each do |i, v|
puts "B(%2i) = %*i/%i" % [i, max_width, v.numerator, v.denominator]
end </lang>
Version 1: compute each number separately. <lang ruby>require "big"
def bernoulli(n)
ar = [] of BigRational
(0..n).each do |m|
ar << BigRational.new(1, m+1)
m.downto(1) { |j| ar[j-1] = j * (ar[j-1] - ar[j]) }
end
ar[0] # (which is Bn)
end
b_nums = (0..61).map { |i| bernoulli(i) } width = b_nums.map{ |b| b.numerator.to_s.size }.max b_nums.each_with_index { |b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? } </lang>
Version 2: create faster generator to compute array of numbers once. <lang ruby>require "big"
def bernoulli2(limit)
ar = [] of BigRational
(0..limit).each do |m|
ar << BigRational.new(1, m+1)
m.downto(1) { |j| ar[j-1] = j * (ar[j-1] - ar[j]) }
yield ar[0] # use Bn value in required block
end
end
b_nums = [] of BigRational bernoulli2(61){ |b| b_nums << b } width = b_nums.map{ |b| b.numerator.to_s.size }.max b_nums.each_with_index { |b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? } </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.
Delphi
Thanks Rudy Velthuis for the Velthuis.BigRationals library.
<lang Delphi> program Bernoulli_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils, Velthuis.BigRationals;
function b(n: Integer): BigRational; begin
var a: TArray<BigRational>;
SetLength(a, n + 1);
for var m := 0 to High(a) do
begin
a[m] := BigRational.Create(1, m + 1);
for var j := m downto 1 do
begin
a[j - 1] := (a[j - 1] - a[j]) * j;
end;
end;
Result := a[0];
end;
begin
for var n := 0 to 60 do
begin
var bb := b(n);
if bb.Numerator.BitLength > 0 then
writeln(format('B(%2d) =%45s/%s', [n, bb.Numerator.ToString, bb.Denominator.ToString]));
end;
readln;
end.</lang>
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>
Fermat
<lang fermat>Func Bern(m) = Sigma<k=0,m>[Sigma<v=0,k>[(-1)^v*Bin(k,v)*(v+1)^m/(k+1)]].; for i=0, 60 do b:=Bern(i); if b<>0 then !!(i,b) fi od;</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 / 35460 -1215233140483755572040304994079820246041491 / 56786730
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
Frink
<lang frink>BernoulliNumber[n] := {
a = new array
for m = 0 to n
{
a@m = 1/(m+1)
for j = m to 1 step -1
a@(j-1) = j * (a@(j-1) - a@j)
}
return a@0
}
result = new array for n=0 to 60 {
b = BernoulliNumber[n]
if b != 0
{
[num,den] = numeratorDenominator[b]
result.pushn, num, "/", den
}
}
println[formatTable[result, "right"]]</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
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
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
GAP
<lang gap>for a in Filtered(List([0 .. 60], n -> [n, Bernoulli(n)]), x -> x[2] <> 0) do
Print(a, "\n");
od;
[ 0, 1 ] [ 1, -1/2 ] [ 2, 1/6 ] [ 4, -1/30 ] [ 6, 1/42 ] [ 8, -1/30 ] [ 10, 5/66 ] [ 12, -691/2730 ] [ 14, 7/6 ] [ 16, -3617/510 ] [ 18, 43867/798 ] [ 20, -174611/330 ] [ 22, 854513/138 ] [ 24, -236364091/2730 ] [ 26, 8553103/6 ] [ 28, -23749461029/870 ] [ 30, 8615841276005/14322 ] [ 32, -7709321041217/510 ] [ 34, 2577687858367/6 ] [ 36, -26315271553053477373/1919190 ] [ 38, 2929993913841559/6 ] [ 40, -261082718496449122051/13530 ] [ 42, 1520097643918070802691/1806 ] [ 44, -27833269579301024235023/690 ] [ 46, 596451111593912163277961/282 ] [ 48, -5609403368997817686249127547/46410 ] [ 50, 495057205241079648212477525/66 ] [ 52, -801165718135489957347924991853/1590 ] [ 54, 29149963634884862421418123812691/798 ] [ 56, -2479392929313226753685415739663229/870 ] [ 58, 84483613348880041862046775994036021/354 ] [ 60, -1215233140483755572040304994079820246041491/56786730 ]</lang>
Go
<lang go>package main
import ( "fmt" "math/big" )
func b(n int) *big.Rat { var f big.Rat a := make([]big.Rat, n+1) for m := range a { a[m].SetFrac64(1, int64(m+1)) for j := m; j >= 1; j-- { d := &a[j-1] d.Mul(f.SetInt64(int64(j)), d.Sub(d, &a[j])) } } return f.Set(&a[0]) }
func main() { for n := 0; n <= 60; n++ { if b := b(n); b.Num().BitLen() > 0 { fmt.Printf("B(%2d) =%45s/%s\n", n, b.Num(), b.Denom()) } } }</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.Bool (bool) import Data.Ratio (Ratio, denominator, numerator, (%))
BERNOULLI NUMBERS -------------------
bernouillis :: Integer -> [Rational] bernouillis =
fmap head . tail . scanl faulhaber [] . enumFromTo 0
faulhaber :: [Ratio Integer] -> Integer -> [Ratio Integer] faulhaber rs n =
(:) =<< (-) 1 . sum $ zipWith ((*) . (n %)) [2 ..] rs
TEST -------------------------
main :: IO () main = do
let xs = bernouillis 60
w = length $ show (numerator (last xs))
putStrLn $
fTable
"Bernouillis from Faulhaber triangle:\n"
(show . fst)
(showRatio w . snd)
id
(filter ((0 /=) . snd) $ zip [0 ..] xs)
FORMATTING ----------------------
fTable ::
String -> (a -> String) -> (b -> String) -> (a -> b) -> [a] -> String
fTable s xShow fxShow f xs =
let w = maximum (length . xShow <$> xs)
in unlines $
s :
fmap
( ((<>) . rjust w ' ' . xShow)
<*> ((" -> " <>) . fxShow . f)
)
xs
showRatio :: Int -> Rational -> String showRatio w r =
let d = denominator r
in rjust w ' ' $ show (numerator r)
<> bool [] (" / " <> show d) (1 /= d)
rjust :: Int -> a -> [a] -> [a] rjust n c = drop . length <*> (replicate n c <>)</lang>
- Output:
Bernouillis from Faulhaber triangle: 0 -> 1 1 -> 1 / 2 2 -> 1 / 6 4 -> -1 / 30 6 -> 1 / 42 8 -> -1 / 30 10 -> 5 / 66 12 -> -691 / 2730 14 -> 7 / 6 16 -> -3617 / 510 18 -> 43867 / 798 20 -> -174611 / 330 22 -> 854513 / 138 24 -> -236364091 / 2730 26 -> 8553103 / 6 28 -> -23749461029 / 870 30 -> 8615841276005 / 14322 32 -> -7709321041217 / 510 34 -> 2577687858367 / 6 36 -> -26315271553053477373 / 1919190 38 -> 2929993913841559 / 6 40 -> -261082718496449122051 / 13530 42 -> 1520097643918070802691 / 1806 44 -> -27833269579301024235023 / 690 46 -> 596451111593912163277961 / 282 48 -> -5609403368997817686249127547 / 46410 50 -> 495057205241079648212477525 / 66 52 -> -801165718135489957347924991853 / 1590 54 -> 29149963634884862421418123812691 / 798 56 -> -2479392929313226753685415739663229 / 870 58 -> 84483613348880041862046775994036021 / 354 60 -> -1215233140483755572040304994079820246041491 / 56786730
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:
See Bernoulli Numbers Essay on the J wiki. <lang j>B=: {.&1 %. (i. ! ])@>:@i.@x:</lang>
Task:
<lang j> 'B' ,. rplc&'r/_-'"1": (#~ 0 ~: {:"1)(i. ,. B) 61 B 0 1 B 1 -1/2 B 2 1/6 B 4 -1/30 B 6 1/42 B 8 -1/30 B10 5/66 B12 -691/2730 B14 7/6 B16 -3617/510 B18 43867/798 B20 -174611/330 B22 854513/138 B24 -236364091/2730 B26 8553103/6 B28 -23749461029/870 B30 8615841276005/14322 B32 -7709321041217/510 B34 2577687858367/6 B36 -26315271553053477373/1919190 B38 2929993913841559/6 B40 -261082718496449122051/13530 B42 1520097643918070802691/1806 B44 -27833269579301024235023/690 B46 596451111593912163277961/282 B48 -5609403368997817686249127547/46410 B50 495057205241079648212477525/66 B52 -801165718135489957347924991853/1590 B54 29149963634884862421418123812691/798 B56 -2479392929313226753685415739663229/870 B58 84483613348880041862046775994036021/354 B60 -1215233140483755572040304994079820246041491/56786730</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}}(undef, n + 1)
for m = 0 : n
A[m + 1] = 1 // (m + 1)
for j = m : -1 : 1
A[j] = j * (A[j] - A[j + 1])
end
end
return A[1]
end
function display(n)
B = map(bernoulli, 0 : n)
pad = mapreduce(x -> ndigits(numerator(x)) + Int(x < 0), max, B)
argdigits = ndigits(n)
for i = 0 : n
if numerator(B[i + 1]) & 1 == 1
println(
"B(", lpad(i, argdigits), ") = ",
lpad(numerator(B[i + 1]), pad), " / ", denominator(B[i + 1])
)
end
end
end
display(60)
- Alternative: Following the comment in the Perl section it is much more efficient
- to compute the list of numbers instead of one number after the other.
function BernoulliList(len)
A = Vector{Rational{BigInt}}(undef, len + 1)
B = similar(A)
for n in 0 : len
A[n + 1] = 1 // (n + 1)
for j = n : -1 : 1
A[j] = j * (A[j] - A[j + 1])
end
B[n + 1] = A[1]
end
return B
end
for (n, b) in enumerate(BernoulliList(60))
isodd(numerator(b)) && println("B($(n-1)) = $b")
end </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.
Lua
LuaJIT version with FFI and GMP library
<lang lua>#!/usr/bin/env luajit local gmp = require 'gmp' ('libgmp') local ffi = require'ffi' local mpz, mpq = gmp.types.z, gmp.types.q local function mpq_for(buf, op, n) for i=0,n-1 do op(buf[i]) end end local function bernoulli(rop, n) local a=ffi.new("mpq_t[?]", n+1) mpq_for(a, gmp.q_init, n+1)
for m=0,n do gmp.q_set_ui(a[m],1, m+1) for j=m,1,-1 do gmp.q_sub(a[j-1], a[j], a[j-1]) gmp.q_set_ui(rop, j, 1) gmp.q_mul(a[j-1], a[j-1], rop) end end gmp.q_set(rop,a[0]) mpq_for(a, gmp.q_clear, n+1) end do --MAIN local rop=mpq() local n,d=mpz(),mpz() gmp.q_init(rop) gmp.z_inits(n, d) local to=arg[1] and tonumber(arg[1]) or 60 local from=arg[2] and tonumber(arg[2]) or 0 if from~=0 then to,from=from,to end
for i=from,to do
bernoulli(rop, i)
if gmp.q_cmp_ui(rop, 0, 1)~=0 then
gmp.q_get_num(n, rop)
gmp.q_get_den(d, rop)
gmp.printf("B(%-2g) = %44Zd / %Zd\n", i, n, d)
end
end
gmp.z_clears(n,d)
gmp.q_clear(rop)
end</lang>
- Output:
> time ./bernoulli_gmp.lua B(0 ) = 1 / 1 B(1 ) = -1 / 2 B(2 ) = 1 / 6 B(4 ) = -1 / 30 B(6 ) = 1 / 42 B(8 ) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730 ./bernoulli_gmp.lua 0,02s user 0,00s system 97% cpu 0,022 total
Time compare: Python 0.591 sec, C 0.023 sec, Lua 0.022-0.025
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)
Nim
<lang Nim>import bignum import strformat
const Lim = 60
- ---------------------------------------------------------------------------------------------------
proc bernoulli(n: Natural): Rat =
## Compute a Bernoulli number using Akiyama–Tanigawa algorithm.
var a = newSeq[Rat](n + 1)
for m in 0..n:
a[m] = newRat(1, m + 1)
for j in countdown(m, 1):
a[j-1] = j * (a[j] - a[j-1])
result = a[0]
- ———————————————————————————————————————————————————————————————————————————————————————————————————
type Info = tuple
n: int # Number index in Bernoulli sequence. val: Rat # Bernoulli number.
var values: seq[Info] # List of values as Info tuples. var maxLen = -1 # Maximum length.
- First step: compute the values and prepare for display.
for n in 0..Lim:
# Compute value. if n != 1 and (n and 1) == 1: continue # Ignore odd "n" except 1. let b = bernoulli(n) # Check numerator length. let len = ($b.num).len if len > maxLen: maxLen = len # Store information for next step. values.add((n, b))
- Second step: display the values with '/' aligned.
for (n, b) in values:
let s = fmt"{($b.num).alignString(maxLen, '>')} / {b.denom}"
echo fmt"{n:2}: {s}"</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
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
Tested with fpc 3.0.4 <lang Pascal> (* Taken from the 'Ada 99' project, https://marquisdegeek.com/code_ada99 *)
program BernoulliForAda99;
uses BigDecimalMath; {library for arbitary high precision BCD numbers}
type
Fraction = object private numerator, denominator: BigDecimal; public procedure assign(n, d: Int64); procedure subtract(rhs: Fraction); procedure multiply(value: Int64); procedure reduce(); procedure writeOutput();
end;
function gcd(a, b: BigDecimal):BigDecimal;
begin
if (b = 0) then begin gcd := a; end else begin gcd := gcd(b, a mod b); end;
end;
procedure Fraction.writeOutput();
var sign : char;
begin
sign := ' ';
if (numerator<0) then sign := '-';
if (denominator<0) then sign := '-';
write(sign + BigDecimalToStr(abs(numerator)):45);
write(' / ');
write(BigDecimalToStr(abs(denominator)));
end;
procedure Fraction.assign(n, d: Int64);
begin
numerator := n; denominator := d;
end;
procedure Fraction.subtract(rhs: Fraction);
begin
numerator := numerator * rhs.denominator; numerator := numerator - (rhs.numerator * denominator); denominator := denominator * rhs.denominator;
end;
procedure Fraction.multiply(value: Int64);
var
temp :BigDecimal;
begin
temp := value; numerator := numerator * temp;
end;
procedure Fraction.reduce();
var gcdResult: BigDecimal;
begin
gcdResult := gcd(numerator, denominator); begin numerator := numerator div gcdResult; (* div is Int64 division *) denominator := denominator div gcdResult; (* could also use round(d/r) *) end;
end;
function calculateBernoulli(n: Int64) : Fraction;
var
m, j: Int64;
results: array of Fraction;
begin
setlength(results, 60) ; {largest value 60}
for m:= 0 to n do
begin
results[m].assign(1, m+1);
for j:= m downto 1 do
begin
results[j-1].subtract(results[j]);
results[j-1].multiply(j);
results[j-1].reduce();
end;
end;
calculateBernoulli := results[0];
end;
(* Main program starts here *)
var
b: Int64; result: Fraction;
begin
writeln('Calculating Bernoulli numbers...');
writeln('B( 0) : 1 / 1');
for b:= 1 to 60 do
begin
if (b<3) or ((b mod 2) = 0) then begin
result := calculateBernoulli(b);
write('B(',b:2,')');
write(' : ');
result.writeOutput();
writeln;
end;
end;
end. </lang>
- Output:
Calculating Bernoulli numbers... B( 0) : 1 / 1 B( 1) : 1 / 2 B( 2) : 1 / 6 B( 4) : -1 / 30 B( 6) : 1 / 42 B( 8) : -1 / 30 B(10) : 5 / 66 B(12) : -691 / 2730 B(14) : -7 / 6 B(16) : -3617 / 510 B(18) : 43867 / 798 B(20) : -174611 / 330 B(22) : 854513 / 138 B(24) : -236364091 / 2730 B(26) : 8553103 / 6 B(28) : -23749461029 / 870 B(30) : 8615841276005 / 14322 B(32) : -7709321041217 / 510 B(34) : 2577687858367 / 6 B(36) : -26315271553053477373 / 1919190 B(38) : 2929993913841559 / 6 B(40) : -261082718496449122051 / 13530 B(42) : 1520097643918070802691 / 1806 B(44) : -27833269579301024235023 / 690 B(46) : -596451111593912163277961 / 282 B(48) : -5609403368997817686249127547 / 46410 B(50) : 495057205241079648212477525 / 66 B(52) : -801165718135489957347924991853 / 1590 B(54) : 29149963634884862421418123812691 / 798 B(56) : -2479392929313226753685415739663229 / 870 B(58) : 84483613348880041862046775994036021 / 354 B(60) : -1215233140483755572040304994079820246041491 / 56786730
Perl
The only thing in the suggested algorithm which depends on N is the number of times through the inner block. This means that all but the last iteration through the loop produce the exact same values of A.
Instead of doing the same calculations over and over again, I retain the A array until the final Bernoulli number is produced.
<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.
Phix
with javascript_semantics include builtins/mpfr.e procedure bernoulli(mpq rop, integer n) sequence a = mpq_inits(n+1) for m=1 to n+1 do mpq_set_si(a[m], 1, m) for j=m-1 to 1 by -1 do mpq_sub(a[j], a[j+1], a[j]) mpq_set_si(rop, j, 1) mpq_mul(a[j], a[j], rop) end for end for mpq_set(rop, a[1]) a = mpq_free(a) end procedure mpq rop = mpq_init() mpz n = mpz_init(), d = mpz_init() for i=0 to 60 do bernoulli(rop, i) if mpq_cmp_si(rop, 0, 1) then mpq_get_num(n, rop) mpq_get_den(d, rop) string ns = mpz_get_str(n), ds = mpz_get_str(d) printf(1,"B(%2d) = %44s / %s\n", {i,ns,ds}) end if end for {n,d} = mpz_free({n,d}) rop = mpq_free(rop)
- Output:
B( 0) = 1 / 1 B( 1) = -1 / 2 B( 2) = 1 / 6 B( 4) = -1 / 30 B( 6) = 1 / 42 B( 8) = -1 / 30 B(10) = 5 / 66 B(12) = -691 / 2730 B(14) = 7 / 6 B(16) = -3617 / 510 B(18) = 43867 / 798 B(20) = -174611 / 330 B(22) = 854513 / 138 B(24) = -236364091 / 2730 B(26) = 8553103 / 6 B(28) = -23749461029 / 870 B(30) = 8615841276005 / 14322 B(32) = -7709321041217 / 510 B(34) = 2577687858367 / 6 B(36) = -26315271553053477373 / 1919190 B(38) = 2929993913841559 / 6 B(40) = -261082718496449122051 / 13530 B(42) = 1520097643918070802691 / 1806 B(44) = -27833269579301024235023 / 690 B(46) = 596451111593912163277961 / 282 B(48) = -5609403368997817686249127547 / 46410 B(50) = 495057205241079648212477525 / 66 B(52) = -801165718135489957347924991853 / 1590 B(54) = 29149963634884862421418123812691 / 798 B(56) = -2479392929313226753685415739663229 / 870 B(58) = 84483613348880041862046775994036021 / 354 B(60) = -1215233140483755572040304994079820246041491 / 56786730
PicoLisp
Brute force and method by Srinivasa Ramanujan. <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.
Quackery
<lang Quackery> $ "bigrat.qky" loadfile
[ 1+
' [ [] ] over of swap
times
[ i^ 1+ n->v 1/v
join swap i^ poke
i^ times
[ dup i 1+ peek do
dip over swap i peek do
v- i 1+ n->v v*
join swap i poke ] ]
1 split drop do ] is bernoulli ( n --> n/d )
61 times
[ i^ bernoulli
2dup v0= iff
2drop
else
[ i^ 10 < if sp
i^ echo sp
vulgar$
char / over find
44 swap - times sp
echo$ cr ] ]</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
R
<lang rsplus>
library(pracma)
for (idx in c(1,2*0:30)) {
b <- bernoulli(idx)
d <- as.character(denominator(b))
n <- as.character(numerator(b))
cat("B(",idx,") = ",n,"/",d,"\n", sep = "")
} </lang>
- Output:
B(1) = 1/2 B(0) = 1/1 B(2) = 1/6 B(4) = -1/30 B(6) = 1/42 B(8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Racket
This implements, firstly, the algorithm specified with the task... then the better performing bernoulli.3, which uses the "double sum formula" listed under REXX. The number generators all (there is also a bernoulli.2) use the same emmitter... it's just a matter of how long to wait for the emission.
<lang>#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
Raku
(formerly 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 Raku by not defining it as a function at all, but by defining it as an infinite sequence that we can read however many values we like from (52, in this case, to get up to B(100)). In this solution we've also avoided subscripting operations; rather we use a sequence operator (...) iterated over the list of the previous solution to find the next solution. We reverse the array in this case to make reference to the previous value in the list more natural, which means we take the last value of the list rather than the first value, and do so conditionally to avoid 0 values.
<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 ... *)
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>
Same output as memoization example
REXX
The double sum formula used is number (33) from the entry Bernoulli number on Wolfram MathWorldTM.
- where is a binomial coefficient.
- where is a binomial coefficient.
<lang rexx>/*REXX program calculates N number of Bernoulli numbers expressed as vulgar fractions.*/ parse arg N .; if N== | N=="," then N= 60 /*Not specified? Then use the default.*/ numeric digits max(9, n*2) /*increase the decimal digits if needed*/ w= max(length(N), 4); Nw= N + w + N % 4 /*used for aligning (output) fractions.*/ say 'B(n)' center("Bernoulli numbers expressed as vulgar fractions", max(78-w, Nw) ) say copies('─',w) copies("─", max(78-w,Nw+2*w)) /*display 2nd line of title, separators*/ !.= .; do #=0 to N /*process the numbers from 0 ──► N. */
b= bern(#); if b==0 then iterate /*calculate Bernoulli number, skip if 0*/
indent= max(0, nW - pos('/', b) ) /*calculate the alignment (indentation)*/
say right(#, w) left(, indent) b /*display the indented Bernoulli number*/
end /*#*/ /* [↑] align the Bernoulli fractions. */
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ bern: parse arg x; if x==0 then return '1/1' /*handle the special case of zero. */
if x==1 then return '-1/2' /* " " " " " one. */
if x//2 then return 0 /* " " " " " odds > 1.*/
do j=2 to x by 2; jp= j+1 /*process the positive integers up to X*/
sn= 1 - j /*define the numerator. */
sd= 2 /* " " denominator. */
do k=2 to j-1 by 2 /*calculate a SN/SD sequence. */
parse var @.k bn '/' ad /*get a previously calculated fraction.*/
an= comb(jp, k) * bn /*use COMBination for the next term. */
$lcm= LCM(sd, ad) /*use Least Common Denominator function*/
sn= $lcm % sd * sn; sd= $lcm /*calculate the current numerator. */
an= $lcm % ad * an /* " " next " */
sn= sn + an /* " " current " */
end /*k*/ /* [↑] calculate the SN/SD sequence.*/
sn= -sn /*flip the sign for the numerator. */
sd= sd * jp /*calculate the denominator. */
if sn\==1 then do; _= GCD(sn, sd) /*get the Greatest Common Denominator.*/
sn= sn%_; sd= sd%_ /*reduce the numerator and denominator.*/
end /* [↑] done with the reduction(s). */
@.j= sn'/'sd /*save the result for the next round. */
end /*j*/ /* [↑] done calculating Bernoulli #'s.*/
return sn'/'sd
/*──────────────────────────────────────────────────────────────────────────────────────*/ comb: procedure expose !.; parse arg x,y; if x==y then return 1
if !.C.x.y\==. then return !.C.x.y /*combination computed before?*/
if x-y < y then y= x-y /*x-y < y? Then use a new Y.*/
z= perm(x, y); do j=2 for y-1; z= z % j
end /*j*/
!.C.x.y= z; return z /*assign memoization & return.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ GCD: procedure; parse arg x,y; x= abs(x)
do until y==0; parse value x//y y with y x; end; return x
/*──────────────────────────────────────────────────────────────────────────────────────*/ LCM: procedure; parse arg x,y /*X=ABS(X); Y=ABS(Y) not needed for Bernoulli #s.*/
/*IF Y==0 THEN RETURN 0 " " " " " */
$= x * y /*calculate part of the LCM here. */
do until y==0; parse value x//y y with y x
end /*until*/ /* [↑] this is a short & fast GCD*/
return $ % x /*divide the pre─calculated value.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ perm: procedure expose !.; parse arg x,y; if !.P.x.y\==. then return !.P.x.y
z= 1; do j=x-y+1 to x; z= z*j; end; !.P.x.y= z; return z</lang>
- output when using the default input:
B(n) Bernoulli numbers expressed as vulgar fractions ──── ─────────────────────────────────────────────────────────────────────────────────────── 0 1/1 1 -1/2 2 1/6 4 -1/30 6 1/42 8 -1/30 10 5/66 12 -691/2730 14 7/6 16 -3617/510 18 43867/798 20 -174611/330 22 854513/138 24 -236364091/2730 26 8553103/6 28 -23749461029/870 30 8615841276005/14322 32 -7709321041217/510 34 2577687858367/6 36 -26315271553053477373/1919190 38 2929993913841559/6 40 -261082718496449122051/13530 42 1520097643918070802691/1806 44 -27833269579301024235023/690 46 596451111593912163277961/282 48 -5609403368997817686249127547/46410 50 495057205241079648212477525/66 52 -801165718135489957347924991853/1590 54 29149963634884862421418123812691/798 56 -2479392929313226753685415739663229/870 58 84483613348880041862046775994036021/354 60 -1215233140483755572040304994079820246041491/56786730
Output notes: This version of REXX can compute and display all values up to B110 in sub─second.
Ruby
<lang ruby>bernoulli = Enumerator.new do |y|
ar = []
0.step do |m|
ar << Rational(1, m+1)
m.downto(1){|j| ar[j-1] = j*(ar[j-1] - ar[j]) }
y << ar.first # yield
end
end
b_nums = bernoulli.take(61) width = b_nums.map{|b| b.numerator.to_s.size}.max b_nums.each_with_index {|b,i| puts "B(%2i) = %*i/%i" % [i, width, b.numerator, b.denominator] unless b.zero? }
</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.value.numer() != ToBigInt::to_bigint(&0).unwrap()) {
println!("B({:>2}) = {:>2$} / {denom}", r.index, r.value.numer(), width,
denom = r.value.denom());
}
}
// Implementation with no reused calculations. fn _bernoulli_naive(n: usize, c: &mut Context) -> BigRational {
for m in 0..n + 1 {
c.a.push(BigRational::new(c.bigone_const.clone(), (m + 1).to_bigint().unwrap()));
for j in (1..m + 1).rev() {
c.a[j - 1] = (c.a[j - 1].clone().sub(c.a[j].clone())).mul(
BigRational::new(j.to_bigint().unwrap(), c.bigone_const.clone())
);
}
}
c.a[0].reduced()
}
// Implementation with reused calculations (does not require sequential calls). fn bernoulli(n: usize, c: &mut Context) -> BigRational {
for i in 0..n + 1 {
if i >= c.a.len() {
c.a.push(BigRational::new(c.bigone_const.clone(), (i + 1).to_bigint().unwrap()));
for j in (1..i + 1).rev() {
c.a[j - 1] = (c.a[j - 1].clone().sub(c.a[j].clone())).mul(
BigRational::new(j.to_bigint().unwrap(), c.bigone_const.clone())
);
}
}
}
c.a[0].reduced()
}
- [cfg(test)]
mod tests {
use super::{Bn, Context, bernoulli, _bernoulli_naive};
use num::rational::{BigRational};
use std::str::FromStr;
use test::Bencher;
// [tests elided]
#[bench]
fn bench_bernoulli_naive(b: &mut Bencher) {
let mut context = Context::new();
b.iter(|| {
let mut res: Vec<Bn> = vec![];
for n in 0..30 + 1 {
let b = _bernoulli_naive(n, &mut context);
res.push(Bn { value:b.clone(), index: n as i32});
}
});
}
#[bench]
fn bench_bernoulli(b: &mut Bencher) {
let mut context = Context::new();
b.iter(|| {
let mut res: Vec<Bn> = vec![];
for n in 0..30 + 1 {
let b = bernoulli(n, &mut context);
res.push(Bn { value:b.clone(), index: n as i32});
}
});
}
#[bench]
fn bench_bernoulli_iter(b: &mut Bencher) {
b.iter(|| {
let context = Context::new();
let _res = context.take(30 + 1).collect::<Vec<_>>();
});
}
} </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
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
Scheme
<lang scheme>; Return the n'th Bernoulli number.
(define bernoulli
(lambda (n)
(let ((a (make-vector (1+ n))))
(do ((m 0 (1+ m)))
((> m n))
(vector-set! a m (/ 1 (1+ m)))
(do ((j m (1- j)))
((< j 1))
(vector-set! a (1- j) (* j (- (vector-ref a (1- j)) (vector-ref a j))))))
(vector-ref a 0))))
- Convert a rational to a string. If an integer, ends with "/1".
(define rational->string
(lambda (rational) (format "~a/~a" (numerator rational) (denominator rational))))
- Returns the string length of the numerator of a rational.
(define rational-numerator-length
(lambda (rational) (string-length (format "~a" (numerator rational)))))
- Formats a rational with left-padding such that total length to the slash is as given.
(define rational-padded
(lambda (rational total-length-to-slash)
(let* ((length-padding (- total-length-to-slash (rational-numerator-length rational)))
(padding-string (make-string length-padding #\ )))
(string-append padding-string (rational->string rational)))))
- Return the Bernoulli numbers 0 through n in a list.
(define make-bernoulli-list
(lambda (n)
(if (= n 0)
(list (bernoulli n))
(append (make-bernoulli-list (1- n)) (list (bernoulli n))))))
- Print the non-zero Bernoulli numbers 0 through 60 aligning the slashes.
(let* ((bernoullis-list (make-bernoulli-list 60))
(numerator-lengths (map rational-numerator-length bernoullis-list))
(max-numerator-length (apply max numerator-lengths)))
(let print-bernoulli ((index 0) (numbers bernoullis-list))
(cond
((null? numbers))
((= 0 (car numbers))
(print-bernoulli (1+ index) (cdr numbers)))
(else
(printf "B(~2@a) = ~a~%" index (rational-padded (car numbers) max-numerator-length))
(print-bernoulli (1+ index) (cdr numbers))))))</lang>
- Output:
$ scheme --script bernoulli.scm B( 0) = 1/1 B( 1) = 1/2 B( 2) = 1/6 B( 4) = -1/30 B( 6) = 1/42 B( 8) = -1/30 B(10) = 5/66 B(12) = -691/2730 B(14) = 7/6 B(16) = -3617/510 B(18) = 43867/798 B(20) = -174611/330 B(22) = 854513/138 B(24) = -236364091/2730 B(26) = 8553103/6 B(28) = -23749461029/870 B(30) = 8615841276005/14322 B(32) = -7709321041217/510 B(34) = 2577687858367/6 B(36) = -26315271553053477373/1919190 B(38) = 2929993913841559/6 B(40) = -261082718496449122051/13530 B(42) = 1520097643918070802691/1806 B(44) = -27833269579301024235023/690 B(46) = 596451111593912163277961/282 B(48) = -5609403368997817686249127547/46410 B(50) = 495057205241079648212477525/66 B(52) = -801165718135489957347924991853/1590 B(54) = 29149963634884862421418123812691/798 B(56) = -2479392929313226753685415739663229/870 B(58) = 84483613348880041862046775994036021/354 B(60) = -1215233140483755572040304994079820246041491/56786730
Seed7
The program below uses bigRational numbers. The Bernoulli numbers are written as fraction and as decimal number, with possible repeating decimals. The conversion of a bigRational number to string is done with the function str. This function automatically writes repeating decimals in parentheses, when necessary.
<lang seed7>$ include "seed7_05.s7i";
include "bigrat.s7i";
const func bigRational: bernoulli (in integer: n) is func
result
var bigRational: bernoulli is bigRational.value;
local
var integer: m is 0;
var integer: j is 0;
var array bigRational: a is 0 times bigRational.value;
begin
a := [0 .. n] times bigRational.value;
for m range 0 to n do
a[m] := 1_ / bigInteger(succ(m));
for j range m downto 1 do
a[pred(j)] := bigRational(j) * (a[j] - a[pred(j)]);
end for;
end for;
bernoulli := a[0];
end func;
const proc: main is func
local
var bigRational: bernoulli is bigRational.value;
var integer: i is 0;
begin
for i range 0 to 60 do
bernoulli := bernoulli(i);
if bernoulli <> bigRational.value then
writeln("B(" <& i lpad 2 <& ") = " <& bernoulli.numerator lpad 44 <&
" / " <& bernoulli.denominator rpad 8 <& " " <& bernoulli);
end if;
end for;
end func;</lang>
- Output:
B( 0) = 1 / 1 1.0 B( 1) = -1 / 2 -0.5 B( 2) = 1 / 6 0.1(6) B( 4) = -1 / 30 -0.0(3) B( 6) = 1 / 42 0.0(238095) B( 8) = -1 / 30 -0.0(3) B(10) = 5 / 66 0.0(75) B(12) = -691 / 2730 -0.2(531135) B(14) = 7 / 6 1.1(6) B(16) = -3617 / 510 -7.0(9215686274509803) B(18) = 43867 / 798 54.9(711779448621553884) B(20) = -174611 / 330 -529.1(24) B(22) = 854513 / 138 6192.1(2318840579710144927536) B(24) = -236364091 / 2730 -86580.2(531135) B(26) = 8553103 / 6 1425517.1(6) B(28) = -23749461029 / 870 -27298231.0(6781609195402298850574712643) B(30) = 8615841276005 / 14322 601580873.9(006423683843038681748359167714) B(32) = -7709321041217 / 510 -15116315767.0(9215686274509803) B(34) = 2577687858367 / 6 429614643061.1(6) B(36) = -26315271553053477373 / 1919190 -13711655205088.3(327721590879485616) B(38) = 2929993913841559 / 6 488332318973593.1(6) B(40) = -261082718496449122051 / 13530 -19296579341940068.1(4863266814) B(42) = 1520097643918070802691 / 1806 841693047573682615.0(005537098560354374307862679955703211517165) B(44) = -27833269579301024235023 / 690 -40338071854059455413.0(7681159420289855072463) B(46) = 596451111593912163277961 / 282 2115074863808199160560.1(4539007092198581560283687943262411347517730496) B(48) = -5609403368997817686249127547 / 46410 -120866265222965259346027.3(119370825253178194354664942900237017884076707606) B(50) = 495057205241079648212477525 / 66 7500866746076964366855720.0(75) B(52) = -801165718135489957347924991853 / 1590 -503877810148106891413789303.0(5220125786163) B(54) = 29149963634884862421418123812691 / 798 36528776484818123335110430842.9(711779448621553884) B(56) = -2479392929313226753685415739663229 / 870 -2849876930245088222626914643291.0(6781609195402298850574712643) B(58) = 84483613348880041862046775994036021 / 354 238654274996836276446459819192192.1(4971751412429378531073446327683615819209039548022598870056) B(60) = -1215233140483755572040304994079820246041491 / 56786730 -21399949257225333665810744765191097.3(926741511617238745742183076926598872659158222352299560126106)
Sidef
Built-in: <lang ruby>say bernoulli(42).as_frac #=> 1520097643918070802691/1806</lang>
Recursive solution (with auto-memoization): <lang ruby>func bernoulli_number(n) is cached {
n.is_one && return 1/2 n.is_odd && return 0
1 - sum(^n, {|k|
binomial(n,k) * __FUNC__(k) / (n - k + 1)
})
}
for n in (0..60) {
var Bn = bernoulli_number(n) || next
printf("B(%2d) = %44s / %s\n", n, Bn.nude)
}</lang>
Using Ramanujan's congruences (pretty fast): <lang ruby>func ramanujan_bernoulli_number(n) is cached {
return 1/2 if n.is_one return 0 if n.is_odd
((n%6 == 4 ? -1/2 : 1) * (n+3)/3 - sum(1 .. (n - n%6)/6, {|k|
binomial(n+3, n - 6*k) * __FUNC__(n - 6*k)
})) / binomial(n+3, n)
}</lang>
Using Euler's product formula for the Riemann zeta function and the Von Staudt–Clausen theorem (very fast): <lang ruby>func bernoulli_number_from_zeta(n) {
n.is_zero && return 1 n.is_one && return 1/2 n.is_odd && return 0
var log2B = (log(4*Num.tau*n)/2 + n*log(n) - n*log(Num.tau) - n)/log(2) local Num!PREC = *(int(n + log2B) + (n <= 90 ? 18 : 0))
var K = 2*(n! / Num.tau**n)
var d = n.divisors.grep {|k| is_prime(k+1) }.prod {|k| k+1 }
var z = ceil((K*d).root(n-1)).primes.prod {|p| 1 - p.float**(-n) }
(-1)**(n/2 + 1) * int(ceil(d*K / z)) / d
}</lang>
The Akiyama–Tanigawa algorithm: <lang ruby>func bernoulli_print {
var a = []
for m in (0..60) {
a << 1/(m+1)
for j in (1..m -> flip) {
(a[j-1] -= a[j]) *= j
}
a[0] || next
printf("B(%2d) = %44s / %s\n", m, a[0].nude)
}
} bernoulli_print()</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
Swift
Uses the Frac type defined in the Rational task.
<lang Swift>import BigInt
public func bernoulli<T: BinaryInteger & SignedNumeric>(n: Int) -> Frac<T> {
guard n != 0 else {
return 1
}
var arr = [Frac<T>]()
for m in 0...n {
arr.append(Frac(numerator: 1, denominator: T(m) + 1))
for j in stride(from: m, through: 1, by: -1) {
arr[j-1] = (arr[j-1] - arr[j]) * Frac(numerator: T(j), denominator: 1)
}
}
return arr[0]
}
for n in 0...60 {
let b = bernoulli(n: n) as Frac<BigInt>
guard b != 0 else {
continue
}
print("B(\(n)) = \(b)")
}</lang>
- Output:
B(0) = Frac(1 / 1) B(1) = Frac(1 / 2) B(2) = Frac(1 / 6) B(4) = Frac(-1 / 30) B(6) = Frac(1 / 42) B(8) = Frac(-1 / 30) B(10) = Frac(5 / 66) B(12) = Frac(-691 / 2730) B(14) = Frac(7 / 6) B(16) = Frac(-3617 / 510) B(18) = Frac(43867 / 798) B(20) = Frac(-174611 / 330) B(22) = Frac(854513 / 138) B(24) = Frac(-236364091 / 2730) B(26) = Frac(8553103 / 6) B(28) = Frac(-23749461029 / 870) B(30) = Frac(8615841276005 / 14322) B(32) = Frac(-7709321041217 / 510) B(34) = Frac(2577687858367 / 6) B(36) = Frac(-26315271553053477373 / 1919190) B(38) = Frac(2929993913841559 / 6) B(40) = Frac(-261082718496449122051 / 13530) B(42) = Frac(1520097643918070802691 / 1806) B(44) = Frac(-27833269579301024235023 / 690) B(46) = Frac(596451111593912163277961 / 282) B(48) = Frac(-5609403368997817686249127547 / 46410) B(50) = Frac(495057205241079648212477525 / 66) B(52) = Frac(-801165718135489957347924991853 / 1590) B(54) = Frac(29149963634884862421418123812691 / 798) B(56) = Frac(-2479392929313226753685415739663229 / 870) B(58) = Frac(84483613348880041862046775994036021 / 354) B(60) = Frac(-1215233140483755572040304994079820246041491 / 56786730)
Tcl
<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 'BigInteger
Module Bernoulli_numbers
Function gcd_BigInt(ByVal x As BigInteger, ByVal y As BigInteger) As BigInteger
Dim y2 As BigInteger
x = BigInteger.Abs(x)
Do
y2 = BigInteger.Remainder(x, y)
x = y
y = y2
Loop Until y = 0
Return x
End Function 'gcd_BigInt
Sub bernoul_BigInt(n As Integer, ByRef bnum As BigInteger, ByRef bden As BigInteger)
Dim j, m As Integer
Dim f As BigInteger
Dim anum(), aden() As BigInteger
ReDim anum(n + 1), aden(n + 1)
For m = 0 To n
anum(m + 1) = 1
aden(m + 1) = m + 1
For j = m To 1 Step -1
anum(j) = j * (aden(j + 1) * anum(j) - aden(j) * anum(j + 1))
aden(j) = aden(j) * aden(j + 1)
f = gcd_BigInt(BigInteger.Abs(anum(j)), BigInteger.Abs(aden(j)))
If f <> 1 Then
anum(j) = anum(j) / f
aden(j) = aden(j) / f
End If
Next
Next
bnum = anum(1) : bden = aden(1)
End Sub 'bernoul_BigInt
Sub bernoulli_BigInt()
Dim i As Integer
Dim bnum, bden As BigInteger
bnum = 0 : bden = 0
For i = 0 To 60
bernoul_BigInt(i, bnum, bden)
If bnum <> 0 Then
Console.WriteLine("B(" & i & ")=" & bnum.ToString("D") & "/" & bden.ToString("D"))
End If
Next i
End Sub 'bernoulli_BigInt
End Module 'Bernoulli_numbers</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
Wren
<lang ecmascript>import "/fmt" for Fmt import "/big" for BigRat
var bernoulli = Fn.new { |n|
if (n < 0) Fiber.abort("Argument must be non-negative")
var a = List.filled(n+1, null)
for (m in 0..n) {
a[m] = BigRat.new(1, m+1)
var j = m
while (j >= 1) {
a[j-1] = (a[j-1] - a[j]) * BigRat.new(j, 1)
j = j - 1
}
}
return (n != 1) ? a[0] : -a[0] // 'first' Bernoulli number
}
for (n in 0..60) {
var b = bernoulli.call(n)
if (b != BigRat.zero) Fmt.print("B($2d) = $44i / $i", n, b.num, b.den)
}</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
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