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a(n) = Sum_{k=0..n/2} binomial(n+3,k)*binomial(n+1-k,k+1).
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%I #25 Jun 06 2017 04:21:11

%S 1,2,8,22,68,198,586,1718,5047,14808,43470,127636,374957,1102078,

%T 3241082,9537070,28079357,82718212,243809138,718994032,2121378272,

%U 6262089436,18493519148,54639914652,161503493023,477558890378,1412658185320

%N a(n) = Sum_{k=0..n/2} binomial(n+3,k)*binomial(n+1-k,k+1).

%H G. C. Greubel, <a href="/A262720/b262720.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: B'(x)*(1-x/B(x))^2/x^4, where B(x)/x is g.f. of A005043.

%F Recurrence: n*(n+4)*a(n) = (n^2 + 3*n + 6)*a(n-1) + (n+2)*(5*n + 6)*a(n-2) + 3*(n+1)*(n+2)*a(n-3). - _Vaclav Kotesovec_, Sep 29 2015

%F a(n) ~ 3^(n+7/2)/(8*sqrt(Pi*n)). - _Vaclav Kotesovec_, Sep 29 2015

%t Table[Sum[Binomial[n+3,k]*Binomial[n+1-k,k+1], {k,0,n/2}], {n,0,25}] (* _Vaclav Kotesovec_, Sep 29 2015 *)

%o (Maxima)

%o B(x):=(1+x-sqrt(1-2*x-3*x^2))/(2*(1+x));

%o taylor(diff(B(x),x,1)*(1-x/B(x))^2/x^4,x,0,30);

%o (PARI) a(n) = sum(k=0, n/2, (binomial(n+3,k)*binomial(n+1-k,k+1))) ;

%o vector(30, n, a(n-1)) \\ _Altug Alkan_, Sep 28 2015

%Y Cf. A005043.

%K nonn

%O 0,2

%A _Vladimir Kruchinin_, Sep 28 2015