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Expansion of ( (1 + x)*(1 - 11*x)*(1 + 121*x) )^(1/3).
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%I #13 Feb 25 2024 10:26:40

%S 1,37,-1776,114096,-9165936,810646320,-76152738288,7450371782832,

%T -750608233752432,77319392827405872,-8104270335592602864,

%U 861419406835986019248,-92621128795282877608560,10055062260891607562940720,-1100545944769838408566122480,121306087657061323164937678512

%N Expansion of ( (1 + x)*(1 - 11*x)*(1 + 121*x) )^(1/3).

%C 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).

%H Paul D. Hanna, <a href="/A370149/b370149.txt">Table of n, a(n) for n = 0..300</a>

%F G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

%F (1) A(x)^3 = (1 + x)*(1 - 11*x)*(1 + 121*x) = (1 + 111*x - 1221*x^2 - 1331*x^3).

%F (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).

%F 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

%e 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 + ...

%e where A(x)^3 = (1 + 111*x - 1221*x^2 - 1331*x^3).

%e RELATED SERIES.

%e We have the following infinite product

%e 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) + ...

%o (PARI) {a(n) = polcoeff( ( (1 + x)*(1 - 11*x)*(1 + 121*x) +x*O(x^n))^(1/3), n)}

%o for(n=0, 40, print1(a(n), ", "))

%Y Cf. A370145, A370147, A370334.

%K sign

%O 0,2

%A _Paul D. Hanna_, Feb 25 2024