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a(0)=1; thereafter a(n) = -2*Sum_{k=1..n} binomial(2n,2k)*a(n-k).
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%I #79 Sep 03 2022 12:13:07

%S 1,-2,22,-602,30742,-2523002,303692662,-50402079002,11030684333782,

%T -3077986048956602,1066578948824962102,-449342758735568563802,

%U 226182806795367665865622,-134065091768709178087428602,92423044260377387363207812342,-73323347841467639992211297199002

%N a(0)=1; thereafter a(n) = -2*Sum_{k=1..n} binomial(2n,2k)*a(n-k).

%C The version without signs has an interpretation as a sum over marked Schröder paths. See the Josuat-Verges and Kim reference.

%C Consider the sequence defined by a(0)=1; thereafter a(n) = c*Sum_{k=1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.

%C Apparently a(n) = 2*(-1)^n*A002114(n). - _R. J. Mathar_, Mar 01 2015

%H Seiichi Manyama, <a href="/A210657/b210657.txt">Table of n, a(n) for n = 0..200</a>

%H Matthieu Josuat-Vergès and Jang Soo Kim, <a href="http://arxiv.org/abs/1101.5608">Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity</a>, arXiv:1101.5608 [math.CO], 2011.

%H Zhi-Hong Sun, <a href="http://arxiv.org/abs/1203.5977">On the further properties of U_n</a>, arXiv:1203.5977 [math.NT], 2012.

%F O.g.f.: Sum_{n>=0} (2*n)! * (-x)^n / Product_{k=1..n} (1 - k^2*x). - _Paul D. Hanna_, Sep 17 2012

%F E.g.f.: 1/(2*cosh(x) - 1) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. - _Paul D. Hanna_, Oct 30 2014

%F E.g.f.: cos(z/2)/cos(3z/2) = Sum_{n>=0} abs(a(n))*x^(2*n)/(2*n)!. - _Olivier Gérard_, Feb 12 2014

%F From _Peter Bala_, Mar 09 2015: (Start)

%F a(n) = 3^(2*n)*E(2*n,1/3), where E(n,x) is the n-th Euler polynomial.

%F O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - x*(3*k + 1)^2).

%F O.g.f. as a continued fraction: 1/(1 + (3^2 - 1^2)*x/(4 + 12^2*x/(4 + (18^2 - 2^2)*x/(4 + 24^2*x/(4 + (30^2 - 2^2)*x/(4 + 36^2*x/(4 + ... ))))))) = 1 - 2*x + 22*x^2 - 602*x^3 + 30742*x^4 - .... See Josuat-Vergès and Kim, p. 23.

%F The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255882. (End)

%F a(n) = 2*36^n*(zeta(-n*2,1/6)-zeta(-n*2,2/3)), where zeta(a,z) is the generalized Riemann zeta function. - _Peter Luschny_, Mar 11 2015

%F a(n) ~ 2 * (-1)^n * (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - _Vaclav Kotesovec_, Mar 14 2015

%F a(n) = Sum_{k=0..n} A241171(n, k)*(-2)^k. - _Peter Luschny_, Sep 03 2022

%p A210657:=proc(n) option remember;

%p if n=0 then 1

%p else -2*add(binomial(2*n,2*k)*procname(n-k),k=1..floor(n)); fi;

%p end;

%p [seq(f(n),n=0..20)];

%p # Second program:

%p a := (n) -> 2*36^n*(Zeta(0,-n*2,1/6)-Zeta(0,-n*2,2/3)):

%p seq(a(n), n=0..15); # _Peter Luschny_, Mar 11 2015

%t nmax=20; Table[(CoefficientList[Series[1/(2*Cosh[x]-1), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[2*n+1]], {n,0,nmax}] (* _Vaclav Kotesovec_, Mar 14 2015 *)

%t Table[9^n EulerE[2 n, 1/3], {n, 0, 20}] (* _Vladimir Reshetnikov_, Jun 05 2016 *)

%o (PARI) a(n)=polcoeff(sum(m=0, n, (2*m)!*(-x)^m/prod(k=1, m, 1-k^2*x +x*O(x^n)) ), n)

%o for(n=0,20,print1(a(n),", ")) \\ _Paul D. Hanna_, Sep 17 2012

%Y Cf. A002114, A028296, A094088, A210657, A210672, A210674, A210676, A249939, A255882, A241171.

%K sign

%O 0,2

%A _N. J. A. Sloane_, Mar 28 2012