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%I #15 Oct 14 2016 15:04:38
%S 1,7,41,225,1195,6227,32059,163727,831505,4206145,21215481,106782837,
%T 536618341,2693492305,13507578125,67693008145,339066121115,
%U 1697664211795,8497396194275,42522326235175,212749477704695,1064285646397915,5323532330953295,26625895085494075
%N a(n) = 5^n*Sum_{ k=0..n } binomial(2*k,k)/5^k.
%F From _Vaclav Kotesovec_, Jun 12 2013: (Start)
%F G.f.: 1/((1-5*x)*sqrt(1-4*x)).
%F Recurrence: n*a(n) = (9*n-2)*a(n-1) - 10*(2*n-1)*a(n-2).
%F a(n) ~ 5^(n+1/2). (End)
%F a(n) = 5^(n+1/2) - 2^(n+1)*(2*n+1)!!*hypergeom([1,n+3/2], [n+2], 4/5]/(5*(n+1)!). - _Vladimir Reshetnikov_, Oct 14 2016
%t Table[5^n Sum[Binomial[2k,k]/5^k,{k,0,n}],{n,0,30}] (* _Harvey P. Dale_, Aug 08 2011 *)
%t Round@Table[5^(n + 1/2) - 2^(n + 1) (2 n + 1)!! Hypergeometric2F1[1, n + 3/2, n + 2, 4/5]/(5 (n + 1)!), {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - _Vladimir Reshetnikov_, Oct 14 2016 *)
%o (PARI) a(n) = 5^n*sum(k=0, n, binomial(2*k,k)/5^k); \\ _Michel Marcus_, Oct 14 2016
%Y Cf. A006134, A082590, A132310, A002457.
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Jan 21 2009
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