%I #11 Mar 22 2020 13:29:31
%S 1,6,75,2916,417605,234812358,527932234375,4755738419928072,
%T 171280331996409907209,24606864966197875457438730,
%U 14080929986159936046600341796875,32073236633246852578917758577924120588,290760173774986242601808360162358149769707533,10492680499171055486742235424276666079725581443186702,1507792223578968167717594884445653164343553232898773193359375
%N a(n) = (n+1) * (2^n + 1)^n.
%C More generally, the following sums are equal:
%C (1) Sum_{n>=0} binomial(n+k-1, n) * y^n * (F + G^n)^n,
%C (2) Sum_{n>=0} binomial(n+k-1, n) * y^n * G^(n^2) / (1 - y*F*G^n)^(n+k),
%C for any fixed integer k; here, k = 2 and y = x, F = 1, G = 2.
%F O.g.f.: Sum_{n>=0} (n+1) * (2^n + 1)^n * x^n.
%F O.g.f.: Sum_{n>=0} (n+1) * 2^(n^2) * x^n / (1 - 2^n*x)^(n+2).
%F E.g.f.: sum_{n>=0} (n+1 + 2^n*x) * 2^(n^2) * exp(2^n*x) * x^n/n!.
%e O.g.f.: A(x) = 1 + 6*x + 75*x^2 + 2916*x^3 + 417605*x^4 + 234812358*x^5 + 527932234375*x^6 + 4755738419928072*x^7 + ... + (n+1)*(2^n + 1)^n*x^n + ...
%e such that
%e A(x) = 1/(1 - x)^2 + 2*2*x/(1 - 2*x)^3 + 3*2^4*x^2/(1 - 2^2*x)^4 + 4*2^9*x^3/(1 - 2^3*x)^5 + 5*2^16*x^4/(1 - 2^4*x)^6 + 6*2^25*x^5/(1 - 2^5*x)^7 + 7*2^36*x^6/(1 - 2^6*x)^8 + ... + (n+1)*2^(n^2)*x^n/(1 - 2^n*x)^(n+2) + ...
%t Table[(n+1)(2^n+1)^n,{n,0,20}] (* _Harvey P. Dale_, Mar 22 2020 *)
%o (PARI) {a(n) = (n+1) * (2^n + 1)^n}
%o for(n=0,15, print1(a(n),", "))
%o (PARI) /* O.g.f. */
%o {a(n) = my(A = sum(m=0,n, (m+1) * 2^(m^2) * x^m / (1 - 2^m*x +x*O(x^n))^(m+2) )); polcoeff(A,n)}
%o for(n=0,15, print1(a(n),", "))
%o (PARI) /* E.g.f. */
%o {a(n) = my(A = sum(m=0,n, (m+1 + 2^m*x) * 2^(m^2) * exp(2^m*x +x*O(x^n)) * x^m/m! )); n!*polcoeff(A,n)}
%o for(n=0,15, print1(a(n),", "))
%Y Cf. A136516, A326012.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jun 05 2019