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
From Seiichi Manyama, Apr 12 2026: (Start)
G.f. A(x) satisfies [x^n] exp(n^2*A(x)) * (1 - n^3*x) = 0, for n >= 1.
a(n) = n^(3*n-3) - (1/n) * Sum_{k=1..n-1} k * c_n(k) * e_n(n-k),
where c_n(k) = a(k) - n^(3*k-2)/k for 1 <= k <= n-1,
and e_n(0) = 1, e_n(k) = (n^2/k) * Sum_{j=1..k} j * c_n(j) * e_n(k-j) for 1 <= k <= n-1. (End)
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), ", "))
(Ruby)
def A300593(n)
a = [0]
(1..n).each{|i|
c = [0] + (1..i - 1).map{|k| a[k] - i ** (3 * k - 2) / k.to_r}
e = [1]
(1..i - 1).each{|k| e << i ** 2 / k.to_r * (1..k).inject(0){|s, j| s + j * c[j] * e[k - j]}}
a << i ** (3 * i - 3) - (1..i - 1).inject(0){|s, k| s + k * c[k] * e[i - k]}.to_i / i
}
a[1..-1]
end
p A300593(20) # Seiichi Manyama, Apr 12 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 09 2018
STATUS
approved