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^(n-k) + t*q^k)^(n-k) * v^k * q^(k^2)
then b(n) = Sum_{k=0..n} C(n,k) * (v*q^(n-k) + t*p^k)^(n-k) * u^k * p^(k^2).
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) * (3^(n-k) + 2^k)^(n-k) * 2^(k^2).
a(n) == 1 (mod 5).
a(n) ~ 3^(n^2). - Vaclav Kotesovec, Sep 03 2017
EXAMPLE
E.g.f.: A(x) = 1 + 6*x + 136*x^2/2! + 23526*x^3/3! + 45511576*x^4/4! +...
Illustration of initial terms:
a(1) = (2 + 1) + 3 = 6;
a(2) = (2^2 + 1)^2 + 2*(2 + 3)*3 + 3^4 = 136;
a(3) = (2^3 + 1)^3 + 3*(2^2 + 3)^2*3 + 3*(2 + 3^2)*3^4 + 3^9 = 23526;
a(4) = (2^4 + 1)^4 + 4*(2^3 + 3)^3*3 + 6*(2^2 + 3^2)^2*3^4 + 4*(2 + 3^3)*3^9 + 3^16 = 45511576;
a(5) = (2^5 + 1)^5 + 5*(2^4 + 3)^4*3 + 10*(2^3 + 3^2)^3*3^4 + 10*(2^2 + 3^3)^2*3^9 + 5*(2 + 3^4)*3^16 + 3^25 = 865387222026; ...
and by the binomial identity:
a(1) = (3 + 1) + 2 = 6;
a(2) = (3^2 + 1)^2 + 2*(3 + 2)*2 + 2^4 = 136;
a(3) = (3^3 + 1)^3 + 3*(3^2 + 2)^2*2 + 3*(3 + 2^2)*2^4 + 2^9 = 23526;
a(4) = (3^4 + 1)^4 + 4*(3^3 + 2)^3*2 + 6*(3^2 + 2^2)^2*2^4 + 4*(3 + 2^3)*2^9 + 2^16 = 45511576;
a(5) = (3^5 + 1)^5 + 5*(3^4 + 2)^4*2 + 10*(3^3 + 2^2)^3*2^4 + 10*(3^2 + 2^3)^2*2^9 + 5*(3 + 2^4)*2^16 + 2^25 = 865387222026; ...
PROG
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*(2^(n-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)*(3^(n-k) + 2^k)^(n-k)*2^(k^2))}
for(n=0, 16, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 12 2014
STATUS
approved