A136637 a(n) = Sum_{k=0..n} C(n, k) * C(2^k*3^(n-k), n).
1, 5, 72, 6089, 3326498, 12405917044, 336474648380394, 69883583587428350874, 115099747754889610404191160, 1536533057081060754026861201898620, 168527150638482484315370462123098294514192
Offset: 0
Keywords
Examples
More generally, if Sum_{n>=0} log(1 + (p^n + r*q^n)*x )^n / n! = Sum_{n>=0} b(n)*x^n, then b(n) = Sum_{k=0..n} C(n,k)*r^(n-k) * C(p^k*q^(n-k), n) (a result due to _Vladeta Jovovic_, Jan 13 2008).
Crossrefs
Programs
-
Mathematica
Table[Sum[Binomial[n,k]*Binomial[2^k*3^(n-k),n], {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Jul 02 2016 *)
-
PARI
{a(n)=sum(k=0,n,binomial(n,k)*binomial(2^k*3^(n-k),n))}
-
PARI
/* Using g.f.: */ {a(n)=polcoeff(sum(i=0,n,log(1+(2^i+3^i)*x)^i/i!),n,x)}
Formula
G.f.: A(x) = Sum_{n>=0} log(1 + (2^n + 3^n)*x )^n / n!.
a(n) ~ 3^(n^2) / n!. - Vaclav Kotesovec, Jul 02 2016
Comments