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%I #14 Nov 12 2023 12:07:14
%S 1,3,13,69,411,2633,17739,124029,892327,6567285,49235715,374841195,
%T 2890994445,22545855855,177524073021,1409591810133,11275693221519,
%U 90792020672429,735367765159347,5987665336600683,48987680485918149
%N a(n) = Sum_{k=0..n} A000984(k)*A001263(n+1,k+1), where A000984 is the central binomial coefficients and A001263 is the Narayana triangle.
%H Vincenzo Librandi, <a href="/A128079/b128079.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = Sum_{k=0..n} C(2k,k)*C(n,k)*C(n+1,k)/(k+1).
%F Recurrence: (n+1)*(n+2)*a(n) = (7*n^2+11*n+6)*a(n-1) + 3*(7*n^2-19*n+6)*a(n-2) - 27*(n-2)*(n-1)*a(n-3) . - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) ~ 3^(2*n+7/2)/(8*Pi*n^2) . - _Vaclav Kotesovec_, Oct 20 2012
%F a(n) = ((n+3)^2*A005802(n+1)-(n-3)*(n+1)*A005802(n))/12. - _Mark van Hoeij_, Nov 12 2023
%e Illustrate a(n) = Sum_{k=0..n} A000984(k)*A001263(n+1,k+1) by:
%e a(2) = 1*(1) + 2*(3) + 6*(1) = 13;
%e a(3) = 1*(1) + 2*(6) + 6*(6) + 20*(1) = 69;
%e a(4) = 1*(1) + 2*(10)+ 6*(20)+ 20*(10)+ 70*(1) = 411.
%e The Narayana triangle A001263(n+1,k+1) = C(n,k)*C(n+1,k)/(k+1) begins:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 6, 6, 1;
%e 1, 10, 20, 10, 1;
%e 1, 15, 50, 50, 15, 1; ...
%t Table[Sum[Binomial[2*k,k]*Binomial[n,k]*Binomial[n+1,k]/(k+1),{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 20 2012 *)
%o (PARI) {a(n)=sum(k=0,n,binomial(2*k,k)*binomial(n,k)*binomial(n+1,k)/(k+1))}
%Y Cf. A000984, A001263.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Feb 23 2007
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