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%I #40 Feb 24 2020 15:35:39
%S 1,1,5,7,5,11,455,1,85,133,55,23,455,1,145,2387,85,1,319865,1,2255,
%T 301,115,47,7735,11,265,133,145,59,9464455,1,85,10787,5,781,23350145,
%U 1,5,553,38335,83,567385,1,10235,45353,235,1,750295,1,5555,721,265,107
%N a(0)=1, a(n)=A002445(n)/6 for n>=1.
%C For n>=1: denominators of the Bernoulli numbers (A002445) divided by 6.
%C All entries are odd.
%C a(n)= A002445(n) / A020793(n).
%C 5 divides a(2*n) for n>=1.
%C 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
%H Harvey P. Dale, <a href="/A177735/b177735.txt">Table of n, a(n) for n = 0..1000</a>
%H C. M. Bender and K. A. Milton, <a href="http://arxiv.org/abs/hep-th/9304062">Continued fraction as a discrete nonlinear transform</a>, arXiv:hep-th/9304062, 1993.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html">von Staudt-Clausen Theorem</a>
%F a(n) = denominator(BernoulliB(2*n, 1/2))/(3*2^(2*n)). - _Jean-François Alcover_, Apr 16 2013
%F 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
%p A002445 := proc(n) bernoulli(2*n) ; denom(%) ; end proc:
%p A177735 := proc(n) if n = 0 then 1; else A002445(n)/6 ; end if; end proc:
%p seq(A177735(n),n=0..60) ; # _R. J. Mathar_, Aug 15 2010
%t Join[{1},Denominator[BernoulliB[Range[2,120,2]]]/6] (* _Harvey P. Dale_, Oct 19 2012 *)
%t result = {}; Do[prod = 1; Do[If[PrimeQ[2*Divisors[n][[i]] + 1], prod *= (2*Divisors[n][[i]] + 1)], {i, 2, Length[Divisors[n]]}];
%t AppendTo[result, prod] , {n, 1, 100}] ; result (* _Richard R. Forberg_, Jul 19 2016 *)
%o (PARI)
%o a(n)=
%o {
%o my(bd=1);
%o forprime (p=5, 2*n+1, if( (2*n)%(p-1)==0, bd*=p ) );
%o bd;
%o }
%o /* _Joerg Arndt_, May 06 2012 */
%o (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
%o (Sage)
%o def A177735(n):
%o if n == 0: return 1
%o M = map(lambda i: i+1, divisors(2*n))
%o return mul(filter(lambda s: is_prime(s), M))//6
%o print([A177735(n) for n in (0..53)]) # _Peter Luschny_, Feb 20 2016
%Y Cf. A002445, A165949.
%K nonn
%O 0,3
%A _Paul Curtz_, May 12 2010
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