OFFSET
0,2
COMMENTS
This sequence illustrates the identity:
Sum_{n>=0} [(n+1)(n+2)/2]*q^(n^2)*G(q^n*x)^n*x^n = Sum_{n>=0} c(n)*x^n
where c(n) = [x^n] 1/(1 - q^n*x*G(x))^3.
FORMULA
a(n) = [x^n] 1/(1 - 2^n*x*(1+x))^3.
a(n) = Sum_{k=0..[n/2]} [(n-k+1)(n-k+1)/2]*C(n-k,k)*2^(n(n-k)).
a(n) ~ n^2 * 2^(n^2 - 1). - Vaclav Kotesovec, Oct 07 2020
EXAMPLE
G.f.: A(x) = 1 + 6*x + 108*x^2 + 5888*x^3 + 1107456*x^4 +...
MATHEMATICA
Table[Sum[(n - k + 1)*(n - k + 2)/2 * Binomial[n - k, k]*2^(n*(n - k)), {k, 0, n/2}], {n, 0, 15}] (* Vaclav Kotesovec, Oct 07 2020 *)
PROG
(PARI) {a(n)=polcoeff(sum(m=0, n, (m+1)*(m+2)/2*(1+2^m*x)^m*2^(m^2)*x^m)+x*O(x^n), n)}
(PARI) {a(n)=polcoeff(1/(1-2^n*x*(1+x)+x*O(x^n))^3, n)}
(PARI) {a(n)=sum(k=0, n\2, (n-k+1)*(n-k+2)/2*binomial(n-k, k)*2^(n*(n-k)))}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Nov 26 2009
STATUS
approved