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A069852 a(n) = Sum_{i=0..2n} B(i)*C(2n+1,i)*5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.

%I A069852 #27 Jun 03 2019 12:25:28
%S A069852 6,-74,1946,-88434,6154786,-607884394,80834386026,-13923204233954,
%T A069852 3015393801263666,-801997872697905114,256982712667627683706,
%U A069852 -97641716941862894337874,43406301788286350509870146,-22319737637152541506923644234
%N A069852 a(n) = Sum_{i=0..2n} B(i)*C(2n+1,i)*5^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial numbers.
%C A069852 Related to those formulas derived from Bernoulli polynomials: Sum_{k>0} sin(k*x)/k^(2n+1) = (-1)^(n+1)/2*x^(2n+1)/(2n+1)!*Sum_{i=0..2n}(2Pi/x)^i*B(i)*C(2n+1,i).
%H A069852 Robert Israel, <a href="/A069852/b069852.txt">Table of n, a(n) for n = 1..232</a>
%H A069852 Wikipedia, <a href="http://en.wikipedia.org/wiki/Bernoulli_polynomials">Bernoulli Polynomials</a>
%F A069852 From _Peter Bala_, Mar 02 2015: (Start)
%F A069852 a(n) = 5^(2*n + 1)*B(2*n + 1,1/5), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A009843 and A069994.
%F A069852 Conjecturally, a(n) = the signed numerator of B(2*n + 1,1/5).
%F A069852 G.f.: t/2*( 3 - 5*sinh(3*t/2)/sinh(5*t/2) ) = 6*t^3/3! - 74*t^5/5! + 1946*t^7/7! - ....
%F A069852 G.f. for signed version of sequence: 3/2 + 3/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (5*k + 1)*x)*(1 - (5*k + 4)*x) ) } = 6*x^2 - 74*x^4 + 1946*x^6 + .... (End)
%p A069852 seq(5^(2*n+1)*bernoulli(2*n+1,1/5),n=1..14); # (after _Peter Bala_) _Peter Luschny_, Mar 08 2015
%t A069852 Table[5^(2n+1) BernoulliB[2n+1, 1/5], {n, 1, 14}] (* _Jean-François Alcover_, Jun 03 2019, from Maple *)
%o A069852 (PARI) for(n=1,25,print1(sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*5^i),","))
%Y A069852 Cf. A002111, A009843, A069994.
%K A069852 easy,sign
%O A069852 1,1
%A A069852 _Benoit Cloitre_, May 01 2002