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.
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 A300592.
a(n) ~ c * n!^3, where c = 3.10566781078993955626127892108166... - Vaclav Kotesovec, Oct 14 2020
EXAMPLE
O.g.f.: A(x) = x + 6*x^2 + 216*x^3 + 18016*x^4 + 2718575*x^5 + 667151244*x^6 + 249904389518*x^7 + 136335045655680*x^8 + 104258627494173747*x^9 ...
where
exp(A(x)) = 1 + x + 13*x^2/2! + 1333*x^3/3! + 438073*x^4/4! + 328561681*x^5/5! + 482408372341*x^6/6! + 1262989939509733*x^7/7! + ... + A300592(n)*x^n/n! + ...
such that: [x^n] exp( n^2 * A(x) ) = n^3 * [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)^3*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