A236407 - OEIS

%I #17 Jun 01 2017 13:26:34

%S 0,3,10,41,196,1007,5342,28821,157192,864155,4780018,26572097,

%T 148321356,830764807,4666890950,26283115053,148348809232,838944980531,

%U 4752575891162,26964373486425,153196621856212,871460014012703,4962895187697070,28292329581548741

%N a(n) = 2*Sum_{k=0..n-1} C(n-1,k)*C(n+k,k) + n.

%H G. C. Greubel, <a href="/A236407/b236407.txt">Table of n, a(n) for n = 0..1000</a>

%H J. Brzozowski, M. Szykula, <a href="http://arxiv.org/abs/1401.0157">Large Aperiodic Semigroups</a>, arXiv:1401.0157 [cs.FL], 2013 (Tables 1, 2).

%F a(n) = A002003(n) + n.

%F Conjecture: n*(n-3)*a(n) -4*(2*n-1)*(n-3)*a(n-1) +2*(7*n^2-28*n+20)*a(n-2) -4*(n-1)*(2*n-7)*a(n-3) +(n-1)*(n-4)*a(n-4)=0. - _R. J. Mathar_, Feb 01 2014

%F Recurrence: (n-2)*n*(2*n^2 - 8*n + 7)*a(n) = (14*n^4 - 88*n^3 + 189*n^2 - 158*n + 39)*a(n-1) - (14*n^4 - 80*n^3 + 153*n^2 - 112*n + 24)*a(n-2) + (n-3)*(n-1)*(2*n^2 - 4*n + 1)*a(n-3). - _Vaclav Kotesovec_, Feb 14 2014

%F a(n) ~ 2^(-3/4) * (3+2*sqrt(2))^n / sqrt(Pi*n). - _Vaclav Kotesovec_, Feb 14 2014

%t Table[2*Sum[Binomial[n-1,k]*Binomial[n+k,k],{k,0,n-1}]+n,{n,0,20}] (* _Vaclav Kotesovec_, Feb 14 2014 *)

%t Flatten[{0,Table[n+2*Hypergeometric2F1[1-n,1+n,1,-1],{n,1,20}]}] (* _Vaclav Kotesovec_, Feb 14 2014 *)

%o (PARI) for(n=0,25, print1(n + 2*sum(k=0,n-1, binomial(n-1,k) * binomial(n+k,k)), ", ")) \\ _G. C. Greubel_, Jun 01 2017

%Y Cf. A002003.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Jan 31 2014