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%I #11 Jul 02 2016 07:53:11
%S 1,3,38,2955,1666194,6775599252,204212962736426,47025953519744215608,
%T 84798028785462127288681736,1219731316443261012339196962784452,
%U 141916030637329352970764084182705691263552
%N a(n) = Sum_{k=0..[n/2]} C(n-k, k) * C(3^(n-2*k)*2^k, n-k).
%C Equals antidiagonal sums of triangle A136635.
%F G.f.: A(x) = Sum_{n>=0} log(1 + 3^n*x + 2^n*x^2)^n / n!.
%F a(n) ~ 3^(n^2) / n!. - _Vaclav Kotesovec_, Jul 02 2016
%e More generally, if Sum_{n>=0} log(1 + b*p^n*x + d*q^n*x^2)^n/n! = Sum_{n>=0} a(n)*x^n then a(n) = Sum_{k=0..[n/2]} C(n-k,k)*b^(n-2k)*d^k*C(p^(n-2k)*q^k,n-k).
%t Table[Sum[Binomial[n-k,k]*Binomial[2^k*3^(n-2*k),n-k], {k, 0, Floor[n/2]}], {n, 0, 15}] (* _Vaclav Kotesovec_, Jul 02 2016 *)
%o (PARI) {a(n)=sum(k=0,n\2,binomial(n-k,k)*binomial(3^(n-2*k)*2^k,n-k))}
%o (PARI) /* Using g.f.: */ {a(n)=polcoeff(sum(i=0,n,log(1+3^i*x+2^i*x^2)^i/i!),n,x)}
%Y Cf. A136635 (triangle), A014070 (main diagonal), A136393 (column 0), A136636 (column 1), A136637 (row sums).
%K nonn
%O 0,2
%A _Vladeta Jovovic_ and _Paul D. Hanna_, Jan 15 2008
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