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A300595
O.g.f. A(x) satisfies: [x^n] exp( n^3 * A(x) ) = n^3 * [x^(n-1)] exp( n^3 * A(x) ) for n>=1.
12
1, 4, 243, 40448, 12519125, 6111917748, 4308276119854, 4151360558858752, 5268077625693186225, 8567999843251994553500, 17491034862909191177473132, 44081002571786307016424642880, 135294991782833277819666508563622, 499980220350805674732349875409752344, 2203045016526580123761644939382016407000, 11476028442989415865296132639050660100915200
OFFSET
1,2
COMMENTS
Compare to: [x^n] exp( n^3 * x ) = n^2 * [x^(n-1)] exp( n^3 * x ) for n>=1.
It is conjectured that this sequence consists entirely of integers.
LINKS
FORMULA
O.g.f. equals the logarithm of the e.g.f. of A300594.
a(n) ~ c * n!^3 * n^3, where c = 0.40774346023... - Vaclav Kotesovec, Oct 14 2020
EXAMPLE
O.g.f.: A(x) = x + 4*x^2 + 243*x^3 + 40448*x^4 + 12519125*x^5 + 6111917748*x^6 + 4308276119854*x^7 + 4151360558858752*x^8 + 5268077625693186225*x^9 + ...
where
exp(A(x)) = 1 + x + 9*x^2/2! + 1483*x^3/3! + 976825*x^4/4! + 1507281021*x^5/5! + 4409747597401*x^6/6! + 21744850191313999*x^7/7! + ... + A300594(n)*x^n/n! + ...
such that: [x^n] exp( n^3 * A(x) ) = n^3 * [x^(n-1)] exp( n^3 * A(x) ).
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n+1, A=concat(A, 0); V=Vec(Ser(A)^((#A-1)^3)); A[#A] = ((#A-1)^3*V[#A-1] - V[#A])/(#A-1)^3 ); polcoeff( log(Ser(A)), n)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 09 2018
STATUS
approved