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

0,3

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

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Peter Luschny, Generalized Bernoulli Numbers and Polynomials

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

Example

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A286114 A214073 A335265 * A318142 A176727 A080924

Adjacent sequences: A141456 A141457 A141458 * A141460 A141461 A141462

Keyword

nonn

Author

Paul Curtz, Aug 08 2008

Extensions

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

New name from Peter Luschny, Nov 22 2015

Status

approved