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COMMENTS
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When p=-1, q=3, this sequence illustrates the following identity.
Given e.g.f.: Sum_{n>=0} (p^n+q^n)^n*exp((p^n+q^n)*x)*x^n/n! = Sum_{n>=0} a(n)*x^n/n!,
then a(n) = Sum_{k=0..n} C(n,k)*(p^k + q^k)^n = Sum_{k=0..n} C(n,k)*(1 + p^(n-k)*q^k)^n;
which is a special case of the more general binomial identity:
Sum_{k=0..n} C(n,k)*(s*p^k + t*q^k)^(n-k) * (u*p^k + v*q^k)^k = Sum_{k=0..n} C(n,k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k.
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EXAMPLE
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E.g.f.: A(x) = 1 + 4*x + 112*x^2/2! + 20608*x^3/3! + 47100160*x^4/4! +...
where
_ A(x) = exp((1+1)*x) + (3-1)*exp((3-1)*x)*x + (3^2+1)^2*exp((3^2+1)*x)*x^2/2! + (3^3-1)^3*exp((3^3-1)*x)*x^3/3! +...
or, equivalently,
_ A(x) = exp(2*x) + 2*exp(2*x)*x + 10^2*exp(10*x)*x^2/2! + 26^3*exp(26*x)*x^3/3! + 82^4*exp(82*x)*x^4/4! + 242^5*exp(242*x)*x^5/5! +...
Illustrate the formula for the terms:
a(1) = (1+1) + (3-1) = 4 ;
a(2) = (1+1)^2 + 2*(3-1)^2 + (3^2+1)^2 = 2^2 + 2*2^2 + 10^2 = 112 ;
a(3) = (1+1)^3 + 3*(3-1)^3 + 3*(3^2+1)^3 + (3^3-1)^3 = 2^3 + 3*2^3 + 3*10^3 + 26^3 = 20608 ;
a(4) = (1+1)^4 + 4*(3-1)^4 + 6*(3^2+1)^4 + 4*(3^3-1)^4 + (3^4+1)^4 = 2^4 + 4*2^4 + 6*10^4 + 4*26^4 + 82^4 = 47100160.
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PROG
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(PARI) {a(n)=n!*polcoeff(sum(m=0, n, exp((3^m+(-1)^m+x*O(x^n))*x)*(3^m+(-1)^m)^m*x^m/m!), n)}
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*(3^k + (-1)^k)^n)}
(PARI) {a(n)=sum(k=0, n, binomial(n, k)*(1 + (-1)^(n-k)*3^k)^n)}
(PARI) {a(n)=local(p=-1, q=3); n!*polcoeff(sum(m=0, n, (p^m+q^m)^m*exp((p^m+q^m+x*O(x^n))*x)*x^m/m!), n)}
(PARI) {a(n)=local(p=-1, q=3, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s*p^k + t*q^k)^(n-k)*(u*p^k + v*q^k)^k)}
(PARI) /* right side of the general binomial identity: */
{a(n)=local(p=-1, q=3, s=1, t=1, u=1, v=1);
sum(k=0, n, binomial(n, k)*(s + u*p^(n-k)*q^k)^(n-k) * (t + v*p^(n-k)*q^k)^k)}
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