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A002445
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Denominators of Bernoulli numbers B_{2n}.
(Formerly M4189 N1746)
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145
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1, 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, 1806, 690, 282, 46410, 66, 1590, 798, 870, 354, 56786730, 6, 510, 64722, 30, 4686, 140100870, 6, 30, 3318, 230010, 498, 3404310, 6, 61410, 272118, 1410, 6, 4501770, 6, 33330, 4326, 1590, 642, 209191710, 1518, 1671270, 42
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OFFSET
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0,2
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COMMENTS
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From the von Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Equals row products of even rows in triangle A143343. In triangle A080092, row products = denominators of B1, B2, B4, B6, ... . - Gary W. Adamson, Aug 09 2008
Julius Worpitzky's 1883 algorithm for generating Bernoulli numbers is shown in A028246. - Gary W. Adamson, Aug 09 2008
There is a relation between the Euler numbers E_n and the Bernoulli numbers B_{2*n}, for n>0, namely, B_{2n} = A000367(n)/a(n) = ((-1)^n/(2*(1-2^{2*n})) * Sum_{k = 0..n-1} (-1)^k*2^{2*k}*C(2*n,2*k)*A000364(n-k)*A000367(k)/a(k). (See Bucur, et al.) - L. Edson Jeffery, Sep 17 2012
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REFERENCES
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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 136.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
See A000367 for further references and links (there are a lot).
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LINKS
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FORMULA
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E.g.f: x/(exp(x) - 1); take denominators of even powers.
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k=1..inf} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/ (2*Pi)^(2*n).
a(n) = denominator(-I*(2*n)!/(Pi*(1-2*n))*integral(log(1-1/t)^(1-2*n) dt, t=0..1)). - Gerry Martens, May 17 2011
a(n) = 2*denominator((2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 28 2012
a(n) = gcd(2!S(2n+1,2),...,(2n+1)!S(2n+1,2n+1)). Here S(n,k) is the Stirling number of the second kind. See the paper of Komatsu et al. - Istvan Mezo, May 12 2016
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EXAMPLE
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B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].
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MAPLE
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A002445 := n -> mul(i, i=select(isprime, map(i->i+1, numtheory[divisors] (2*n)))): seq(A002445(n), n=0..40); # Peter Luschny, Aug 09 2011
# Alternative
N:= 1000: # to get a(0) to a(N)
A:= Vector(N, 2):
for p in select(isprime, [seq(2*i+1, i=1..N)]) do
r:= (p-1)/2;
for n from r to N by r do
A[n]:= A[n]*p
od
od:
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MATHEMATICA
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Take[Denominator[BernoulliB[Range[0, 100]]], {1, -1, 2}] (* Harvey P. Dale, Oct 17 2011 *)
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PROG
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(PARI) a(n)=prod(p=2, 2*n+1, if(isprime(p), if((2*n)%(p-1), 1, p), 1)) \\ Benoit Cloitre
(PARI) a(n) = denominator(bernfrac(2*n)); \\ Michel Marcus, Jul 16 2021
(Magma) [Denominator(Bernoulli(2*n)): n in [0..60]]; // Vincenzo Librandi, Nov 16 2014
(Sage)
if n == 0:
return 1
M = (i + 1 for i in divisors(2 * n))
return prod(s for s in M if is_prime(s))
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CROSSREFS
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Cf. A090801 (distinct numbers appearing as denominators of Bernoulli numbers)
See A000367 for numerators. Cf. A027762, A027641, A027642, A002882, A003245, A127187, A127188, A138239, A028246, A143343, A080092, A001897, A277087.
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KEYWORD
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nonn,frac,nice
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AUTHOR
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STATUS
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approved
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