|
%I #17 Mar 08 2015 18:15:05
%S -2,10,-170,6370,-415826,41649850,-5922729722,1134081384850,
%T -281284596509858,87722769712529770,-33597252908389628234,
%U 15502327024398065811010,-8481855507605264686660850,5429636257086663655134162970
%N a(n) = Sum_{i=0..2n} B(i)*C(2n+1,i)*6^i where B(i) are the Bernoulli numbers, C(2n,i) the binomial coefficients.
%C 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 Wikipedia, <a href="http://en.wikipedia.org/wiki/Bernoulli_polynomials">Bernoulli Polynomials</a>
%F From _Peter Bala_, Mar 02 2015: (Start)
%F a(n) = 6^(2*n - 1)*B(2*n - 1,1/6), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A009843, A069852.
%F Conjecturally, a(n) = 2 * the unsigned numerator of B(2*n - 1,1/6). If true then this sequence is a bisection of 2*A158073.
%F G.f.: -3*t*sinh(2*t)/sinh(3*t) = -2*t + 10*t^3/3! - 170*t^5/5! + ....
%F G.f.: Sum_{n >= 0} { 2/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (6*k + 1)*x)*(1 - (6*k + 5)*x) ) } = -2 + 10*x^2 - 170*x^4 + 6370*x^6 - ....
%F (End)
%p seq(6^(2*n-1)*bernoulli(2*n-1,1/6),n=1..14); # (after Peter Bala) _Peter Luschny_, Mar 08 2015
%o (PARI) for(n=1,25,print1(sum(i=0,2*n,binomial(2*n+1,i)*bernfrac(i)*6^i),","))
%Y Cf. A002111, A009843, A069852, A158073.
%K easy,sign
%O 1,1
%A _Benoit Cloitre_, May 01 2002
|
