OFFSET
1,2
COMMENTS
Compare to: [x^n] exp( n^2 * x ) = n * [x^(n-1)] exp( n^2 * x ) for n>=1.
It is conjectured that this sequence consists entirely of integers.
a(n) is divisible by n (conjecture): A300598(n) = a(n)/n for n>=1.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..200
FORMULA
O.g.f. equals the logarithm of the e.g.f. of A300590.
a(n) ~ c * n!^2 * n^2, where c = 0.1354708370957778563796... - Vaclav Kotesovec, Oct 13 2020
EXAMPLE
O.g.f.: A(x) = x + 2*x^2 + 27*x^3 + 736*x^4 + 30525*x^5 + 1715454*x^6 + 123198985*x^7 + 10931897664*x^8 + 1172808994833*x^9 + 149774206572050*x^10 + ...
where
exp(A(x)) = 1 + x + 5*x^2/2! + 175*x^3/3! + 18385*x^4/4! + 3759701*x^5/5! + 1258735981*x^6/6! + 630063839035*x^7/7! + 445962163492385*x^8/8! + ... + A300590(n)*x^n/n! + ...
such that: [x^n] exp( n^2 * A(x) ) = n^2 * [x^(n-1)] exp( n^2 * A(x) ).
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n+1, A=concat(A, 0); V=Vec(Ser(A)^((#A-1)^2)); A[#A] = ((#A-1)^2*V[#A-1] - V[#A])/(#A-1)^2 ); polcoeff( log(Ser(A)), n)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 09 2018
STATUS
approved