OFFSET
0,3
COMMENTS
It appears that for n > 6, a(n) (mod 6) equals [1, 4, 3, 4, 5, 4] repeating.
In general, 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*r) * r^n/n!;
here, q = A(x) with p = x, r = x.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..200
FORMULA
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = Sum_{n>=0} (A(x)^n + x)^n * x^n / n!.
(2) A(x) = Sum_{n>=0} A(x)^(n^2) * exp(x^2*A(x)^n) * x^n / n!.
EXAMPLE
E.g.f.: A(x) = 1 + x + 5*x^2/2! + 34*x^3/3! + 437*x^4/4! + 7996*x^5/5! + 191497*x^6/6! + 5679178*x^7/7! + 200959929*x^8/8! + 8269303384*x^9/9! + ...
where
A(x) = 1 + (A(x) + x)*x + (A(x)^2 + x)^2*x^2/2! + (A(x)^3 + x)^3*x^3/3! + (A(x)^4 + x)^4*x^2/4! + (A(x)^5 + x)^5*x^5/5! + ...
Also,
A(x) = exp(x^2) + A(x)*exp(x^2*A(x))*x + A(x)^4*exp(x^2*A(x)^2)*x^2/2! + A(x)^9*exp(x^2*A(x)^3)*x^3/3! + A(x)^16*exp(x^2*A(x)^4)*x^4/4! + ...
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = polcoeff( sum(m=0, #A, (Ser(A)^m + x)^m * x^m/m! ), #A-1) ); H=A; n!*A[n+1]}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 08 2025
STATUS
approved
