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%I #16 Sep 02 2019 07:36:48
%S 1,1,1,0,1,0,0,0,1,1,0,0,1,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,1,0,0,
%T 0,0,0,0,0,1,1,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,
%U 0,0,0,1,0,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,0,0,0,1,0
%N (D(p)-6)/(12p) where D(p) denotes the denominator of the 2p-th Bernoulli number and p runs through the primes.
%C If p is a Sophie Germain prime (A005384) then denominator(B(2p))= 6*(2p+1).
%F a(A053176(n))=0; a(A005384(n))=1.
%F a(n) = pi(2*prime(n) + 1) - pi(2*prime(n)), where pi(n) = A000720(n) and prime(n) = A000040(n). - _Ridouane Oudra_, Sep 02 2019
%t dbn[n_]:=Module[{d=Denominator[BernoulliB[2n]]},(d-6)/(12n)]; dbn/@ Prime[ Range[100]] (* _Harvey P. Dale_, May 19 2012 *)
%o (PARI) a(n) = my(p=prime(n)); (denominator(bernfrac(2*p)) - 6)/(12*p); \\ _Michel Marcus_, Sep 02 2019
%Y Cf. A002445, A053176, A005384, A000040, A000720.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Feb 09 2003