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A365818 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 5 * A(x)).

%I A365818 #20 Feb 16 2025 08:34:06
%S A365818 0,1,8,107,1760,32298,634128,13034247,276943568,6033834950,
%T A365818 134069957840,3026476515790,69213144181888,1600157697995092,
%U A365818 37337615574348960,878166685063548639,20797051344280763184,495509950454603339310,11869278747340342255440,285669061791469915886250,6904850429493240677872320
%N A365818 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - 5 * A(x)).
%C A365818 Reversion of g.f. for octagonal numbers (with signs).
%H A365818 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/OctagonalNumber.html">Octagonal Number</a>
%H A365818 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SeriesReversion.html">Series Reversion</a>
%F A365818 a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) * 5^k for n > 0.
%F A365818 a(n) ~ 3^(3/2) * 2^(n - 1/2) * (154 + 31*sqrt(31))^n / (sqrt((2821 + 506*sqrt(31))*Pi) * n^(3/2) * 5^(2*n)). - _Vaclav Kotesovec_, Sep 26 2023
%t A365818 nmax = 20; A[_] = 0; Do[A[x_] = x (1 + A[x])^3/(1 - 5 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t A365818 CoefficientList[InverseSeries[Series[x (1 - 5 x)/(1 + x)^3, {x, 0, 20}], x], x]	
%t A365818 Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[3 n, n - k - 1] 5^k, {k, 0, n - 1}], {n, 1, 20}]]
%Y A365818 Cf. A000567, A064089, A113207, A179848, A263843, A365816, A365817.
%K A365818 nonn
%O A365818 0,3
%A A365818 _Ilya Gutkovskiy_, Sep 25 2023