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A370149 Expansion of ( (1 + x)*(1 - 11*x)*(1 + 121*x) )^(1/3).

%I A370149 #13 Feb 25 2024 10:26:40
%S A370149 1,37,-1776,114096,-9165936,810646320,-76152738288,7450371782832,
%T A370149 -750608233752432,77319392827405872,-8104270335592602864,
%U A370149 861419406835986019248,-92621128795282877608560,10055062260891607562940720,-1100545944769838408566122480,121306087657061323164937678512
%N A370149 Expansion of ( (1 + x)*(1 - 11*x)*(1 + 121*x) )^(1/3).
%C A370149 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 A370149 Paul D. Hanna, <a href="/A370149/b370149.txt">Table of n, a(n) for n = 0..300</a>
%F A370149 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
%F A370149 (1) A(x)^3 = (1 + x)*(1 - 11*x)*(1 + 121*x) = (1 + 111*x - 1221*x^2 - 1331*x^3).
%F A370149 (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 A370149 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 A370149 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 A370149 where A(x)^3 = (1 + 111*x - 1221*x^2 - 1331*x^3).
%e A370149 RELATED SERIES.
%e A370149 We have the following infinite product
%e A370149 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 A370149 (PARI) {a(n) = polcoeff( ( (1 + x)*(1 - 11*x)*(1 + 121*x) +x*O(x^n))^(1/3), n)}
%o A370149 for(n=0, 40, print1(a(n), ", "))
%Y A370149 Cf. A370145, A370147, A370334.
%K A370149 sign
%O A370149 0,2
%A A370149 _Paul D. Hanna_, Feb 25 2024