A177735 a(0)=1, a(n)=A002445(n)/6 for n>=1.

1, 1, 5, 7, 5, 11, 455, 1, 85, 133, 55, 23, 455, 1, 145, 2387, 85, 1, 319865, 1, 2255, 301, 115, 47, 7735, 11, 265, 133, 145, 59, 9464455, 1, 85, 10787, 5, 781, 23350145, 1, 5, 553, 38335, 83, 567385, 1, 10235, 45353, 235, 1, 750295, 1, 5555, 721, 265, 107

Offset: 0

Author

Paul Curtz, May 12 2010

Keywords

• nonn

Comments

For n>=1: denominators of the Bernoulli numbers (A002445) divided by 6.

All entries are odd.

a(n)= A002445(n) / A020793(n).

5 divides a(2*n) for n>=1.

These numbers also equal to the lengths of the repeating patterns for the excluded integer values of c/6, when both p^n + c and p^n - c are prime, for an infinite number of primes p>2, and a given integer n>0, arising from the union of one or more prime-based modulo cycles, determined by the divisors of n. See A005097 for details and connection to the von Staudt-Clausen Theorem below. - Richard R. Forberg, Jul 19 2016

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• C. M. Bender and K. A. Milton, Continued fraction as a discrete nonlinear transform, arXiv:hep-th/9304062, 1993.
• Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem

Crossrefs

Cf. A002445, A165949.

Programs

Maple

A002445 := proc(n) bernoulli(2*n) ; denom(%) ; end proc:
A177735 := proc(n) if n = 0 then 1; else A002445(n)/6 ; end if; end proc:
seq(A177735(n),n=0..60) ; # R. J. Mathar, Aug 15 2010

Mathematica

Join[{1},Denominator[BernoulliB[Range[2,120,2]]]/6] (* Harvey P. Dale, Oct 19 2012 *)
result = {}; Do[prod = 1; Do[If[PrimeQ[2*Divisors[n][[i]] + 1], prod *= (2*Divisors[n][[i]] + 1)], {i, 2, Length[Divisors[n]]}];
AppendTo[result, prod] , {n, 1, 100}]  ; result (* Richard R. Forberg, Jul 19 2016 *)

PARI

a(n)=
{
    my(bd=1);
    forprime (p=5, 2*n+1, if( (2*n)%(p-1)==0, bd*=p ) );
    bd;
}
/* Joerg Arndt, May 06 2012 */

PARI

a(n)=if(n<2, return(1)); my(s=1); fordiv(n,d, if(isprime(2*d+1) && d>1, s *= 2*d+1)); s \\ Charles R Greathouse IV, Jul 20 2016

Sage

def A177735(n):
    if n == 0: return 1
    M = map(lambda i: i+1, divisors(2*n))
    return mul(filter(lambda s: is_prime(s), M))//6
print([A177735(n) for n in (0..53)]) # Peter Luschny, Feb 20 2016

Formula

a(n) = denominator(BernoulliB(2*n, 1/2))/(3*2^(2*n)). - Jean-François Alcover, Apr 16 2013

A simple direct calculation of the denominators, for n>=1, is based on the von Staudt-Clausen Theorem: Product{d|n}(2d+1), for d>1 and 2d+1 prime. See in the Mathematica section below. - Richard R. Forberg, Jul 19 2016