OFFSET
0,2
COMMENTS
Here we set p=2, q=3, t=u=v=1, in the binomial identity:
if b(n) = Sum_{k=0..n} C(n,k) * (u*p^k + t*q^k)^(n-k) * v^k * q^(k^2)
then b(n) = Sum_{k=0..n} C(n,k) * (t + v*p^(n-k)*q^k)^k * u^(n-k).
This is a special case of the more general binomial identity:
if b(n) = Sum_{k=0..n} C(n,k) * (t*p^(n-k)*r^k + u*q^(n-k)*s^k)^(n-k) * (v*p^(n-k)*r^k + w*q^(n-k)*s^k)^k
then b(n) = Sum_{k=0..n} C(n,k) * (t*p^(n-k)*q^k + v*r^(n-k)*s^k)^(n-k) * (u*p^(n-k)*q^k + w*r^(n-k)*s^k)^k.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..45
FORMULA
a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 2^(n-k)*3^k)^k.
a(n) ~ 3^(n^2). - Vaclav Kotesovec, Sep 03 2017
EXAMPLE
E.g.f.: A(x) = 1 + 5*x + 115*x^2/2! + 23075*x^3/3! + 45885991*x^4/4! +...
Illustration of initial terms:
a(1) = (1 + 1) + 3 = 5;
a(2) = (1 + 1)^2 + 2*(2 + 3)*3 + 3^4 = 115;
a(3) = (1 + 1)^3 + 3*(2 + 3)^2*3 + 3*(2^2 + 3^2)*3^4 + 3^9 = 23075;
a(4) = (1 + 1)^4 + 4*(2 + 3)^3*3 + 6*(2^2 + 3^2)^2*3^4 + 4*(2^3 + 3^3)*3^9 + 3^16 = 45885991;
a(5) = (1 + 1)^5 + 5*(2 + 3)^4*3 + 10*(2^2 + 3^2)^3*3^4 + 10*(2^3 + 3^3)^2*3^9 + 5*(2^4 + 3^4)*3^16 + 3^25 = 868409174855; ...
and by the binomial identity:
a(1) = 1 + (1 + 3) = 5;
a(2) = 1 + 2*(1 + 2*3) + (1 + 3^2)^2 = 115;
a(3) = 1 + 3*(1 + 2^2*3) + 3*(1 + 2*3^2)^2 + (1 + 3^3)^3 = 23075;
a(4) = 1 + 4*(1 + 2^3*3) + 6*(1 + 2^2*3^2)^2 + 4*(1 + 2*3^3)^3 + (1 + 3^4)^4 = 45885991;
a(5) = 1 + 5*(1 + 2^4*3) + 10*(1 + 2^3*3^2)^2 + 10*(1 + 2^2*3^3)^3 + 5*(1 + 2*3^4)^4 + (1 + 3^5)^5 = 868409174855; ...
MATHEMATICA
Table[Sum[Binomial[n, k](2^k+3^k)^(n-k) 3^(k^2), {k, 0, n}], {n, 0, 10}] (* Harvey P. Dale, Mar 14 2020 *)
PROG
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*(2^k + 3^k)^(n-k)*3^(k^2))}
for(n=0, 16, print1(a(n), ", "))
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*(1 + 2^(n-k)*3^k)^k)}
for(n=0, 16, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 12 2014
STATUS
approved