|
%I #17 Sep 09 2018 08:24:59
%S 1,3,45,900,19305,437076,10254681,246553875,6035226975,149777902710,
%T 3757716928053,95110270281675,2424907723685985,62204709603345075,
%U 1604054030028748830,41549974064592136020,1080505644116115671622,28195636842752845510215,738014045325584817820275
%N G.f.: A(x) = ( Sum_{n>=0} 9^n*(2*n+1) * (-x)^(n*(n+1)/2) )^(-1/9).
%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+2) == a(5*n+3) == a(5*n+4) == 0 (mod 5).
%F Self-convolution cube yields A202438.
%e G.f.: A(x) = 1 + 3*x + 45*x^2 + 900*x^3 + 19305*x^4 + 437076*x^5 +...
%e where
%e 1/A(x)^9 = 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) +...
%e Note that the residues a(n) (mod 5) begin:
%e [1,3,0,0,0,1,1,0,0,0,3,0,0,0,0,0,2,0,0,0,2,2,0,0,0,1,3,0,0,0,3,3,0,0,0,4,4...].
%t nmax = 19;
%t Sum[9^n (2n+1)(-x)^(n(n+1)/2), {n, 0, nmax}]^(-1/9) + 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))^(-1/9),n)}
%Y Cf. A202210, A202436, A202438, A193236, A193237, A111984.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Dec 19 2011
|
