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G.f.: A(x) = ( Sum_{n>=0} 9^n*(2*n+1) * (-x)^(n*(n+1)/2) )^(-1/3).
2

%I #17 Feb 27 2021 21:30:15

%S 1,9,162,3537,81405,1944243,47615121,1186699005,29960950842,

%T 764012506770,19637356382712,507996422180784,13211600995751697,

%U 345145619340179829,9051411187977957135,238160821447956629934,6284647075107225737511,166263704846500625494533

%N G.f.: A(x) = ( Sum_{n>=0} 9^n*(2*n+1) * (-x)^(n*(n+1)/2) )^(-1/3).

%C Compare to the q-series identity:

%C 1/P(x)^3 = Sum_{n>=0} (-1)^n*(2*n+1) * x^(n*(n+1)/2),

%C where P(x) is the partition function (g.f. of A000041).

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="https://arxiv.org/abs/math/0509316">On the Integrality of n-th Roots of Generating Functions</a>, arXiv:math/0509316 [math.NT], 2005-2006.

%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="https://doi.org/10.1016/j.jcta.2006.03.018">On the Integrality of n-th Roots of Generating Functions</a>, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.

%F a(5*n+4) == 0 (mod 5).

%F Self-convolution cube of A202437.

%F Conjectures: a(25*n+24) == 0 (mod 25) (checked up to n = 50) and a(7*n+5) == 0 (mod 7) (checked up to n = 200). - _Peter Bala_, Feb 26 2021

%e G.f.: A(x) = 1 + 9*x + 162*x^2 + 3537*x^3 + 81405*x^4 + 1944243*x^5 +...

%e where

%e 1/A(x)^3 = 1 - 27*x - 405*x^3 + 5103*x^6 + 59049*x^10 - 649539*x^15 - 6908733*x^21 +...+ 9^n*(2*n+1)*(-x)^(n*(n+1)/2) +...

%t nmax = 18;

%t Sum[9^n (2n+1)(-x)^(n(n+1)/2), {n, 0, nmax}]^(-1/3) + O[x]^nmax // CoefficientList[#, x]& (* _Jean-François Alcover_, Sep 09 2018 *)

%o (PARI) {a(n)=polcoeff(sum(m=0,sqrtint(2*n+1),9^m*(2*m+1)*(-x)^(m*(m+1)/2)+x*O(x^n))^(-1/3),n)}

%Y Cf. A202437, A202210, A193236, A193237, A111984.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Dec 19 2011