OFFSET
0,2
COMMENTS
From the von Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Row products of A138239. - Mats Granvik, Mar 08 2008
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
REFERENCES
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).
LINKS
T. D. Noe, Table of n, a(n) for n = 0..10000
A. Bucur, J. Lopez-Bonilla and J. Robles-Garcia, A note on the Namias identity for Bernoulli numbers, Journal of Scientific Research (Banaras Hindu University, Varanasi), Vol. 56 (2012), 117-120.
G. Everest, A. J. van der Poorten, Y. Puri and T. Ward, Integer Sequences and Periodic Points, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.3
S. Kaji, T. Maeno, K. Nuida and Y. Numata, Polynomial Expressions of Carries in p-ary Arithmetics, arXiv preprint arXiv:1506.02742 [math.CO], 2015.
T. Komatsu, F. Luca and C. de J. Pita Ruiz V., A note on the denominators of Bernoulli numbers, Proc. Japan Acad., 90, Ser. A (2014), p. 71-72.
Guo-Dong Liu, H. M. Srivastava and Hai-Quing Wang, Some Formulas for a Family of Numbers Analogous to the Higher-Order Bernoulli Numbers, J. Int. Seq. 17 (2014) # 14.4.6
H.-M. Liu, S-H. Qi and S.-Y. Ding, Some Recurrence Relations for Cauchy Numbers of the First Kind, JIS 13 (2010) # 10.3.8.
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
Niels Nielsen, Traite Elementaire des Nombres de Bernoulli, Gauthier-Villars, 1923, pp. 398.
N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, Berlin, 1924 [Annotated scanned copy of pages 144-151 and 456-463]
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
Simon Plouffe, The First 498 Bernoulli numbers [Project Gutenberg Etext]
FORMULA
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).
If n>=3 is prime,then a((n+1)/2)==(-1)^((n-1)/2)*12*|A000367((n+1)/2)|(mod n). - Vladimir Shevelev, Sep 04 2010
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
EXAMPLE
B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].
MAPLE
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:
1, seq(A[n], n=1..N); # Robert Israel, Nov 16 2014
MATHEMATICA
Take[Denominator[BernoulliB[Range[0, 100]]], {1, -1, 2}] (* Harvey P. Dale, Oct 17 2011 *)
PROG
(PARI) a(n)=prod(p=2, 2*n+1, if(isprime(p), if((2*n)%(p-1), 1, p), 1)) \\ Benoit Cloitre
(PARI) A002445(n, P=1)=forprime(p=2, 1+n*=2, n%(p-1)||P*=p); P \\ M. F. Hasler, Jan 05 2016
(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)
def A002445(n):
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))
[A002445(n) for n in (0..57)] # Peter Luschny, Feb 20 2016
CROSSREFS
KEYWORD
nonn,frac,nice
AUTHOR
STATUS
approved