login
a(n)=sum{k=0..n, C(n,floor(k/2))*(-1)^k*4^(n-k)}.
1

%I #9 Sep 08 2013 13:31:23

%S 1,3,14,57,246,1038,4424,18777,79846,339258,1442004,6128202,26045436,

%T 110691948,470442924,1999378137,8497365126,36113785698,153483619604,

%U 652305322542,2772297736276,11782265148228,50074627320864,212817165231882,904472953925596

%N a(n)=sum{k=0..n, C(n,floor(k/2))*(-1)^k*4^(n-k)}.

%C Hankel transform is 5^n . Second binomial transform is A076036 .

%H Vincenzo Librandi, <a href="/A133444/b133444.txt">Table of n, a(n) for n = 0..200</a>

%F a(n)=Sum{k, 0<=k<=n} A053121(n,k)*A015521(k+1) = (-1)^n*A127363(n) . G.f.: (1/sqrt(1-4x^2))(1-xc(x^2))/(1-4x*c(x^2)), where c(x) is the g.f. of Catalan numbers A000108 .

%F Recurrence: 4*n*a(n) = (17*n-8)*a(n-1) + 2*(8*n+1)*a(n-2) - 68*(n-2)*a(n-3) . - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) ~ 3*17^n/4^(n+1) . - _Vaclav Kotesovec_, Oct 20 2012

%t Table[Sum[Binomial[n,Floor[k/2]]*(-1)^k*4^(n-k),{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 20 2012 *)

%K nonn

%O 0,2

%A _Philippe Deléham_, Nov 26 2007

%E More terms from _Vincenzo Librandi_, May 25 2013