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%I #15 Sep 26 2023 04:23:07
%S 0,1,7,80,1119,17437,290532,5066364,91311055,1687341227,31797227631,
%T 608727899936,11805599569092,231454163924700,4579765707561240,
%U 91340133073920420,1834295500622405295,37059418988408887015,752741444501505866325,15362331852042084534240,314860558967057266779495
%N G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 4 * A(x)).
%C Reversion of g.f. for heptagonal numbers (with signs).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeptagonalNumber.html">Heptagonal Number</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SeriesReversion.html">Series Reversion</a>
%F a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * 4^k for n > 0.
%F a(n) ~ 5 * (81 + 21*sqrt(21))^n / (sqrt((427 + 93*sqrt(21))*Pi) * n^(3/2) * 2^(3*n + 3/2)). - _Vaclav Kotesovec_, Sep 26 2023
%t nmax = 20; A[_] = 0; Do[A[x_] = x (1 + A[x])^3/(1 - 4 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^3, {x, 0, 20}], x], x]
%t Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] 4^k, {k, 0, n - 1}], {n, 1, 20}]]
%Y Cf. A000566, A064088, A113207, A179848, A263843, A365816, A365818.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Sep 25 2023