A076552 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.

1, 1, 21, 461, 16841, 900921, 66453661, 6463837381, 801626558481, 123457062745841, 23116291464379301, 5171511387852362301, 1362357503097707964121, 417419880467876621822761, 147181297749674368184560941, 59173130526513096478888263221

Offset: 1

Author

Benoit Cloitre, Oct 19 2002

Keywords

• nonn
• easy

Comments

Terms are of form 10k+1.

Links

• Eric Weisstein's World of Mathematics, Euler Polynomial.

Crossrefs

Cf. A000364, A060082.

Programs

Mathematica

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 *)

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))

Formula

From Peter Bala, Jul 26 2013: (Start)

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

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)

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

a(n) = (2n)! * [x^(2n)] 1/3-2*sin(x)/(3*tan(2*x)). - Vladimir Kruchinin, Apr 08 2015

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