OFFSET
0,3
COMMENTS
Compare: the factorial series F(x) = Sum_{n>=0} n!*x^n satisfies
(1) [x^n] exp( n^2*x*F(x) ) * (n + 1 - n*F(x)) = 0 for n>=1,
(2) [x^n] exp( n^3*x*F(x) ) * (n^2 + 1 - n^2*F(x)) = 0 for n>=1.
It is remarkable that this sequence should consist entirely of integers.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..300
FORMULA
a(n) ~ sqrt(1-c) * 3^(3*n) * n^(2*n - 3/2) / (sqrt(2*Pi) * c^n * (3-c)^(2*n) * exp(2*n)), where c = -LambertW(-3*exp(-3)) = 0.1785606278779211065968086697... = -A226750. - Vaclav Kotesovec, Aug 06 2018
EXAMPLE
O.g.f.: A(x) = 1 + x + 12*x^2 + 729*x^3 + 111440*x^4 + 31377625*x^5 + 14001201036*x^6 + 9064452341847*x^7 + 8027821828474816*x^8 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp( n^3*x*A(x) ) * (n+1 - n*A(x)) begins:
n=1: [1, 0, -23, -4376, -2674995, -3765464504, ...];
n=2: [1, 6, 0, -8908, -5494464, -7640806512, ...];
n=3: [1, 24, 549, 0, -8632395, -12056269968, ...];
n=4: [1, 60, 3616, 204712, 0, -17114998496, -45010750350080, ...];
n=5: [1, 120, 14505, 1750880, 197597325, 0, -60559334101475, ...];
n=6: [1, 210, 44352, 9406044, 1987128000, 391935493296, 0, ...]; ...
in which the coefficients of x^n in row n form a diagonal of zeros.
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp( (m-1)^3*x*(Ser(A)) ) * (m - (m-1)*Ser(A)) )[m]/(m-1) ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 01 2018
STATUS
approved