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%I #11 Jul 02 2016 07:53:26
%S 1,5,72,6089,3326498,12405917044,336474648380394,69883583587428350874,
%T 115099747754889610404191160,1536533057081060754026861201898620,
%U 168527150638482484315370462123098294514192
%N a(n) = Sum_{k=0..n} C(n, k) * C(2^k*3^(n-k), n).
%C Equals row sums of triangle A136635.
%F G.f.: A(x) = Sum_{n>=0} log(1 + (2^n + 3^n)*x )^n / n!.
%F a(n) ~ 3^(n^2) / n!. - _Vaclav Kotesovec_, Jul 02 2016
%e More generally,
%e if Sum_{n>=0} log(1 + (p^n + r*q^n)*x )^n / n! = Sum_{n>=0} b(n)*x^n,
%e then b(n) = Sum_{k=0..n} C(n,k)*r^(n-k) * C(p^k*q^(n-k), n)
%e (a result due to _Vladeta Jovovic_, Jan 13 2008).
%t Table[Sum[Binomial[n,k]*Binomial[2^k*3^(n-k),n], {k, 0, n}], {n, 0, 15}] (* _Vaclav Kotesovec_, Jul 02 2016 *)
%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)*binomial(2^k*3^(n-k),n))}
%o (PARI) /* Using g.f.: */ {a(n)=polcoeff(sum(i=0,n,log(1+(2^i+3^i)*x)^i/i!),n,x)}
%Y Cf. A136635 (triangle), A014070 (main diagonal), A136393 (column 0), A136636 (column 1), A136638 (antidiagonal sums).
%K nonn
%O 0,2
%A _Vladeta Jovovic_ and _Paul D. Hanna_, Jan 15 2008
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