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%I #14 May 13 2016 05:21:36
%S 1,1,6,37,235,1539,10392,72267,516474,3783115,28317562,215969271,
%T 1673702191,13148444197,104494340880,838670818365,6788255966595,
%U 55346471893395,454123503938490,3746885525588175,31066887028255065
%N Sequence with g.f. 1+(x/(1-5*x))*c(x/(1-5*x)), c(x) the g.f. of A000108.
%C Hankel transform is F(4n+1) (A033889).
%C a(n+1) is the 5th binomial transform of A000108.
%H Vincenzo Librandi, <a href="/A154623/b154623.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (1/2)*(3-sqrt((1-9*x)/(1-5*x))).
%F a(n)=(4/5)*0^n+sum{k=0..n-1, C(n-1,k)*A000108(k)*5^(n-k-1)}.
%F Conjecture: n*a(n) +2*(8-7n)*a(n-1) +45*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
%F a(n) ~ 3^(2*n+1)/(8*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) = (-1)^(n-1)*(GegenbauerC(n-1,-n+1,7/2) + GegenbauerC(n-2,-n+1,7/2)) for n>=1. - _Peter Luschny_, May 13 2016
%t CoefficientList[Series[1/2*(3-Sqrt[(1-9*x)/(1-5*x)]), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 20 2012 *)
%K easy,nonn
%O 0,3
%A _Paul Barry_, Jan 13 2009
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