A141459 a(n) = Product_{p-1 divides n} p, where p is an odd prime.

1, 1, 3, 1, 15, 1, 21, 1, 15, 1, 33, 1, 1365, 1, 3, 1, 255, 1, 399, 1, 165, 1, 69, 1, 1365, 1, 3, 1, 435, 1, 7161, 1, 255, 1, 3, 1, 959595, 1, 3, 1, 6765, 1, 903, 1, 345, 1, 141, 1, 23205, 1, 33, 1, 795, 1, 399, 1, 435, 1, 177, 1, 28393365, 1, 3, 1, 255, 1, 32361, 1, 15, 1, 2343, 1, 70050435

Offset: 0

Author

Paul Curtz, Aug 08 2008

Keywords

• nonn

Comments

Previous name was: A027760(n)/2 for n>=1, a(0) = 1.

Conjecture: a(n) = denominator of integral_{0..1}(log(1-1/x)^n) dx. - Jean-François Alcover, Feb 01 2013

Define the generalized Bernoulli function as B(s,z) = -s*z^s*HurwitzZeta(1-s,1/z) for Re(1/z) > 0 and B(0,z) = 1 for all z; further the generalized Bernoulli polynomials as Bp(m,n,z) = Sum_{j=0..n} B(j,m)*C(n,j)*(z-1)^(n-j) then the a(n) are denominators of Bp(2,n,1), i. e. of the generalized Bernoulli numbers in the case m=2. The numerators of these numbers are A157779(n). - Peter Luschny, May 17 2015

From Peter Luschny, Nov 22 2015: (Start)

a(n) are the denominators of the centralized Bernoulli polynomials 2^n*Bernoulli(n, x/2+1/2) evaluated at x=1. The numerators are A239275(n).

a(n) is the odd part of A141056(n).

a(n) is squarefree, by the von Staudt-Clausen theorem. (End)

Apparently a(n) = denominator(Sum_{k=0..n-1}(-1)^k*E2(n-1, k+1)/binomial(2*n-1, k+1)) where E2(n, k) denotes the second-order Eulerian numbers A340556. - Peter Luschny, Feb 17 2021

Examples

			The denominators of 1, 0, -1/3, 0, 7/15, 0, -31/21, 0, 127/15, 0, -2555/33, 0, 1414477/1365, ...
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Peter Luschny, Generalized Bernoulli Numbers and Polynomials

Crossrefs

Cf. A027760, A141056, A141459, A157779, A157818, A160014, A226157, A239275, A340556.

Programs

Maple

Bfun := (s,z) -> `if`(s=0,1,-s*z^s*Zeta(0,1-s,1/z): # generalized Bernoulli function
Bpoly := (m,n,z) -> add(Bfun(j,m)*binomial(n,j)*(z-1)^(n-j),j=0..n): # generalized Bernoulli polynomials
seq(Bpoly(2,n,1),n=0..50): denom([%]);
# which simplifies to:
a := n -> 0^n+add(Zeta(1-j)*(2^j-2)*j*binomial(n,j), j=1..n):
seq(denom(a(n)), n=0..50); # Peter Luschny, May 17 2015
# Alternatively:
with(numtheory):
ClausenOdd := proc(n) local S, m;
S := map(i -> i + 1, divisors(n));
S := select(isprime, S) minus {2};
mul(m, m = S) end: seq(ClausenOdd(n), n=0..72); # Peter Luschny, Nov 22 2015
# Alternatively:
N:= 1000: # to get a(0) to a(N)
V:= Array(0..N, 1):
for p in select(isprime, [seq(i,i=3..N+1,2)]) do
  R:=[seq(j,j=p-1..N, p-1)]:
  V[R]:= V[R] * p;
od:
convert(V,list); # Robert Israel, Nov 22 2015

Mathematica

a[n_] := If[OddQ[n], 1, Denominator[-2*(2^(n - 1) - 1)*BernoulliB[n]]]; Table[a[n], {n, 0, 72}] (* Jean-François Alcover, Jan 30 2013 *)
Table[Times @@ Select[Divisors@ n + 1, PrimeQ@ # && OddQ@ # &] + Boole[n == 0], {n, 0, 72}] (* Michael De Vlieger, Apr 30 2017 *)

PARI

A141056(n) =
{
    p = 1;
    if (n > 0,
        fordiv(n, d,
            r = d + 1;
            if (isprime(r) & r>2, p = p*r)
        )
    );
    return(p)
}
for(n=0, 72, print1(A141056(n), ", ")); \\ Peter Luschny, Nov 22 2015

Sage

def A141459_list(size):
    f = x / sum(x^(n*2+1)/factorial(n*2+1) for n in (0..2*size))
    t = taylor(f, x, 0, size)
    return [(factorial(n)*s).denominator() for n,s in enumerate (t.list())]
print(A141459_list(72)) # Peter Luschny, Jul 05 2016

Formula

a(2*n+1) = 1. a(2*n)= A001897(n).

a(n) = denominator(0^n + Sum_{j=1..n} zeta(1-j)*(2^j-2)*j*C(n,j)). - Peter Luschny, May 17 2015

Let P(x)= Sum_{n>=0} x^(2*n+1)/(2*n+1)! then a(n) = denominator( n! [x^n] x/P(x) ). - Peter Luschny, Jul 05 2016

a(n) = A157818(n)/4^n. See a comment under A157817, also for other Bernoulli numbers B[4,1] and B[4,3] with this denominator. - Wolfdieter Lang, Apr 28 2017

Extensions

1 prepended and offset set to 0 by Peter Luschny, May 17 2015

New name from Peter Luschny, Nov 22 2015