A050932 Denominator of (n+1)*Bernoulli(n).

1, 1, 2, 1, 6, 1, 6, 1, 10, 1, 6, 1, 210, 1, 2, 1, 30, 1, 42, 1, 110, 1, 6, 1, 546, 1, 2, 1, 30, 1, 462, 1, 170, 1, 6, 1, 51870, 1, 2, 1, 330, 1, 42, 1, 46, 1, 6, 1, 6630, 1, 22, 1, 30, 1, 798, 1, 290, 1, 6, 1, 930930, 1, 2, 1, 102, 1, 966, 1, 10, 1, 66, 1, 1919190

Offset

0,3

Comments

Apparently a(n) = denominator(Sum_{k=0..n-1} (-1)^(n-k+1)*E1(n, k+1)/binomial(n, k+1)), where E1(n, k) denotes the first-order Eulerian numbers A123125. - Peter Luschny, Feb 17 2021

Links

Antti Karttunen, Table of n, a(n) for n = 0..20000 (terms 0..200 from N. J. A. Sloane)

M. Kaneko, A recurrence formula for the Bernoulli numbers, Proc. Japan Acad., 71 A (1995), 192-193.

S. C. Woon, A tree for generating Bernoulli numbers, Math. Mag., 70 (1997), 51-56.

Index entries for sequences related to Bernoulli numbers.

Mathematica

Denominator/@Table[(n+1)BernoulliB[n], {n, 0, 80}] (* Harvey P. Dale, May 19 2011 *)

Prog

(Haskell)
a050932 n = a050932_list !! n
a050932_list = 1 : map (denominator . sum) (zipWith (zipWith (%))
   (zipWith (map . (*)) (drop 2 a000142_list) a242179_tabf) a106831_tabf)
-- Reinhard Zumkeller, Jul 04 2014
(PARI) a(n)=denominator(bernfrac(n)*(n+1)) \\ Charles R Greathouse IV, Feb 07 2017
(Python)
from sympy import bernoulli, gcd
def A050932(n):
    q = bernoulli(n).q
    return q//gcd(q, n+1) # Chai Wah Wu, Apr 02 2021

Crossrefs

Cf. A050925, A000367/A002445, A027641/A027642, A002882, A003245, A127187, A127188, A242179, A106831, A000142, A123125.

Sequence in context: A328580 A362784 A088123 * A366573 A366570 A286515

Adjacent sequences: A050929 A050930 A050931 * A050933 A050934 A050935

Keyword

nonn,frac,nice,easy

Author

N. J. A. Sloane, Dec 30 1999

Status

approved