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%I #15 Jul 06 2019 09:30:33
%S 1,5,27,185,1693,20565,316375,5948465,133579065,3517749125,
%T 107024710675,3714813650025,145570443534805,6383184292589525,
%U 310815510350462415,16694390352153656225,983323269272332915825,63186890982241624232325,4409134435821084657726475,332714992062735780407411225
%N E.g.f.: Sum_{n>=0} ((1+x)^n + 4)^n * x^n/n!.
%C More generally, the following sums are equal:
%C (1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
%C (2) Sum_{n>=0} q^(n^2) * exp(p*q^n*x) * r^n/n!;
%C here, q = (1+x) and p = 4, r = x.
%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 = 4, r = x, m = 1.
%H Paul D. Hanna, <a href="/A326094/b326094.txt">Table of n, a(n) for n = 0..300</a>
%F E.g.f.: Sum_{n>=0} ((1+x)^n + 4)^n * x^n/n!,
%F E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(4*x*(1+x)^n) * x^n/n!.
%F a(n) = 0 (mod 5) for n > 4.
%e E.g.f.: A(x) = 1 + 5*x + 27*x^2/2! + 185*x^3/3! + 1693*x^4/4! + 20565*x^5/5! + 316375*x^6/6! + 5948465*x^7/7! + 133579065*x^8/8! + 3517749125*x^9/9! + 107024710675*x^10/10! + ...
%e such that
%e A(x) = 1 + ((1+x) + 4)*x + ((1+x)^2 + 4)^2*x^2/2! + ((1+x)^3 + 4)^3*x^3/3! + ((1+x)^4 + 4)^4*x^4/4! + ((1+x)^5 + 4)^5*x^5/5! + ((1+x)^6 + 4)^6*x^6/6! + ((1+x)^7 + 4)^7*x^7/7! + ...
%e also
%e A(x) = 1 + (1+x)*exp(4*x*(1+x))*x + (1+x)^4*exp(4*x*(1+x)^2)*x^2/2! + (1+x)^9*exp(4*x*(1+x)^3)*x^3/3! + (1+x)^16*exp(4*x*(1+x)^4)*x^4/4! + (1+x)^25*exp(4*x*(1+x)^5)*x^5/5! + (1+x)^36*exp(4*x*(1+x)^6)*x^6/6! + ...
%o (PARI) /* E.g.f.: Sum_{n>=0} ((1+x)^n + 4)^n * x^n/n! */
%o {a(n) = my(A = sum(m=0,n, ((1+x)^m + 4 +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(4*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(4*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, A326093.
%Y Cf. A326274.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jun 21 2019
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