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a(n) = (-1)^(n+1)/3/(2n+1)*sum(k=0,n,16^k*B(2k)*C(2n+1,2k)) where B(k) denotes the k-th Bernoulli number.
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%I #30 Mar 01 2023 11:00:54

%S 1,1,21,461,16841,900921,66453661,6463837381,801626558481,

%T 123457062745841,23116291464379301,5171511387852362301,

%U 1362357503097707964121,417419880467876621822761,147181297749674368184560941,59173130526513096478888263221

%N a(n) = (-1)^(n+1)/3/(2n+1)*sum(k=0,n,16^k*B(2k)*C(2n+1,2k)) where B(k) denotes the k-th Bernoulli number.

%C Terms are of form 10k+1.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/EulerPolynomial.html">Euler Polynomial</a>.

%F From _Peter Bala_, Jul 26 2013: (Start)

%F It appears that a(n) = 1/3*(A000364(n) - 2*(-1)^n). See A060082.

%F Conjectural e.g.f. with offset 0 (checked up to a(14)): 1/3*(2 - cos(x)^2 + 2*cos(x)^4)/cos(x)^3 = 1 + x^2/2! + 21*x^4/4! + 461*x^6/6! + .... (End)

%F G.f.: 1/(Q(0)*3*x) + 2/(3*x^2*(1+x)) - 2/(3*x^2) + 1/(3*x), where Q(k) = 1 - x*(k+1)^2/Q(k+1) ; (continued fraction). - _Sergei N. Gladkovskii_, Sep 19 2013

%F a(n) = (2n)! * [x^(2n)] 1/3-2*sin(x)/(3*tan(2*x)). - _Vladimir Kruchinin_, Apr 08 2015

%F Conjecture: a(n) = -1/3*(-4)^n*E(2*n,-1/2), where E(n,x) is the n-th Euler polynomial. - _Peter Bala_, Sep 25 2016

%t max = 28; CoefficientList[Series[1/3-2*Sin[x]/(3*Tan[2*x]), {x, 0, max}], x^2] * Range[0, max, 2]! // Rest (* _Jean-François Alcover_, Apr 08 2015, after _Vladimir Kruchinin_ *)

%o (PARI) a(n)=(-1)^(n+1)/3/(2*n+1)*sum(k=0,n,16^k*bernfrac(2*k)*binomial(2*n+1,2*k))

%Y Cf. A000364, A060082.

%K nonn,easy

%O 1,3

%A _Benoit Cloitre_, Oct 19 2002