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%I #13 Jul 06 2019 08:18:08
%S 1,4,18,112,976,11424,169936,3101032,67876608,1746757504,52034505376,
%T 1771434644544,68180144988928,2939951026982272,140920461751138176,
%U 7457658363325181824,433145750643704774656,27464893679743640343552,1892311278990953945563648,141074242336048184406390784,11336870115013701213795557376
%N E.g.f.: Sum_{n>=0} ((1+x)^n + 3)^n * x^n/n!.
%C More generally, the following sums are equal:
%C (1) Sum_{n>=0} (q^n + p)^n * x^n/n!,
%C (2) Sum_{n>=0} q^(n^2) * exp(p*q^n*x) * x^n/n!;
%C here, q = (1+x) and p = 3.
%C In general, let F(x) be a formal power series in x such that F(0)=1, then
%C Sum_{n>=0} m^n * F(q^n*r)^p * log( F(q^n*r) )^n / n! =
%C Sum_{n>=0} r^n * [y^n] F(y)^(m*q^n + p);
%C here, F(x) = exp(x), q = 1+x, p = 3, r = x, m = 1.
%H Paul D. Hanna, <a href="/A326093/b326093.txt">Table of n, a(n) for n = 0..300</a>
%F E.g.f.: Sum_{n>=0} ((1+x)^n + 3)^n * x^n/n!,
%F E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(3*x*(1+x)^n) * x^n/n!.
%F a(n) = 0 (mod 4) for n > 2.
%e E.g.f.: A(x) = 1 + 4*x + 18*x^2/2! + 112*x^3/3! + 976*x^4/4! + 11424*x^5/5! + 169936*x^6/6! + 3101032*x^7/7! + 67876608*x^8/8! + 1746757504*x^9/9! + 52034505376*x^10/10! + ...
%e such that
%e A(x) = 1 + ((1+x) + 3)*x + ((1+x)^2 + 3)^2*x^2/2! + ((1+x)^3 + 3)^3*x^3/3! + ((1+x)^4 + 3)^4*x^4/4! + ((1+x)^5 + 3)^5*x^5/5! + ((1+x)^6 + 3)^6*x^6/6! + ((1+x)^7 + 3)^7*x^7/7! + ...
%e also
%e A(x) = 1 + (1+x)*exp(3*x*(1+x))*x + (1+x)^4*exp(3*x*(1+x)^2)*x^2/2! + (1+x)^9*exp(3*x*(1+x)^3)*x^3/3! + (1+x)^16*exp(3*x*(1+x)^4)*x^4/4! + (1+x)^25*exp(3*x*(1+x)^5)*x^5/5! + (1+x)^36*exp(3*x*(1+x)^6)*x^6/6! + ...
%o (PARI) /* E.g.f.: Sum_{n>=0} ((1+x)^n + 3)^n * x^n/n! */
%o {a(n) = my(A = sum(m=0,n, ((1+x)^m + 3 +x*O(x^n))^m * x^m/m! )); n!*polcoeff(A,n)}
%o for(n=0,25, print1(a(n),", "))
%o (PARI) /* E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(3*x*(1+x)^n) * x^n/n! */
%o {a(n) = my(A = sum(m=0,n, (1+x +x*O(x^n))^(m^2) * exp(3*x*(1+x)^m +x*O(x^n)) * x^m/m! )); n!*polcoeff(A,n)}
%o for(n=0,25, print1(a(n),", "))
%Y Cf. A326096, A326092, A326094.
%Y Cf. A326273.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jun 21 2019
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