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%I #14 Sep 22 2015 18:10:44
%S 5,37,275,2071,15781,121395,940915,7337560,57507892,452598884,
%T 3574599205,28316957579,224901946395,1790287826789,14279629073403,
%U 114097695427295,913103420246956,7317725618907700,58719917176448820,471733089071984376
%N a(n) = (n+1)*Sum_{k=1..n} binomial(n-1,k-1)*binomial(n+2*k+2,k+1)/(n+k+2).
%F G.f.: B'(x)/B(x)-x*B'(x)-B(x), where B(x) is g.f. of A108447.
%F Recurrence: 2*n*(n+1)*(2*n + 1)*(74*n^3 - 183*n^2 + 129*n - 18)*a(n) = 2*n*(1184*n^5 - 1744*n^4 - 242*n^3 + 857*n^2 - 201*n + 26)*a(n-1) + 2*(n-2)*(296*n^5 + 8*n^4 - 410*n^3 + 35*n^2 + 29*n - 18)*a(n-2) - 5*(n-3)*(n-2)*(n+2)*(74*n^3 + 39*n^2 - 15*n + 2)*a(n-3). - _Vaclav Kotesovec_, Sep 22 2015
%t Table[(n + 1) Sum[Binomial[n - 1, k - 1] Binomial[n + 2 k + 2, k + 1]/(n + k + 2), {k, n}], {n, 20}] (* _Michael De Vlieger_, Sep 22 2015 *)
%o (Maxima)
%o b(n):=sum((binomial(n-1,n-k)*binomial(2*k+n,k))/(n+k+1),k,0,n);
%o B(x):=sum(b(n)*x^n,n,0,30);
%o taylor(diff(B(x),x,1)/B(x)-x*(diff(B(x),x,1))-B(x),x,0,10);
%o (PARI) a(n) = (n+1)*sum(k=1,n,((binomial(n-1,k-1) *binomial(n+2*k+2,k+1))/(n+k+2))) \\ _Anders Hellström_, Sep 22 2015
%Y Cf. A108447.
%K nonn
%O 1,1
%A _Vladimir Kruchinin_, Sep 22 2015
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