OFFSET
0,2
COMMENTS
The cube root of F(x) = (1 + x)*(1 - 11*x)*(1 + 121*x) = (1 + 111*x - 1221*x^2 - 1331*x^3) has integer coefficients because F(x) == (1+x)^3 (mod 9).
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..300
FORMULA
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x)^3 = (1 + x)*(1 - 11*x)*(1 + 121*x) = (1 + 111*x - 1221*x^2 - 1331*x^3).
(2) Product_{n>=1} A( 11^(n-1)*x^n )^3 = Sum_{n>=0} 11^(n*(n-1)/2) * (1 + 11^(2*n+1))/12 * x^(n*(n+1)/2).
a(n) ~ (-1)^(n+1) * 2^(5/3) * 5^(1/3) * 11^(2*n-1) / (3^(1/3) * Gamma(2/3) * n^(4/3)). - Vaclav Kotesovec, Feb 25 2024
EXAMPLE
G.f.: A(x) = 1 + 37*x - 1776*x^2 + 114096*x^3 - 9165936*x^4 + 810646320*x^5 - 76152738288*x^6 + 7450371782832*x^7 - 750608233752432*x^8 + ...
where A(x)^3 = (1 + 111*x - 1221*x^2 - 1331*x^3).
RELATED SERIES.
We have the following infinite product
A(x)^3 * A(11*x^2)^3 * A(11^2*x^3)^3 * A(11^3*x^4)^3 * ... = 1 + 111*x + 147631*x^3 + 2161452161*x^6 + 348104014265601*x^10 + 616687495357008127151*x^15 + ... + 11^(n*(n-1)/2) * (1 + 11^(2*n+1))/12 * x^(n*(n+1)/2) + ...
PROG
(PARI) {a(n) = polcoeff( ( (1 + x)*(1 - 11*x)*(1 + 121*x) +x*O(x^n))^(1/3), n)}
for(n=0, 40, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Feb 25 2024
STATUS
approved