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A366015 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 3 * A(x)).

%I A366015 #7 Feb 16 2025 08:34:06
%S A366015 0,1,7,76,995,14433,223300,3611016,60305787,1032115315,18007816255,
%T A366015 319110233104,5727667197044,103913426353324,1902498385538520,
%U A366015 35106179258551632,652236828560562987,12190651925663309175,229059610932456616501,4324334144117016053500,81983637468108446363755
%N A366015 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^4 / (1 - 3 * A(x)).
%C A366015 Reversion of g.f. for hexagonal pyramidal numbers (with signs).
%H A366015 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/HexagonalPyramidalNumber.html">Hexagonal Pyramidal Number</a>
%H A366015 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SeriesReversion.html">Series Reversion</a>
%F A366015 a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(4*n,n-k-1) * 3^k for n > 0.
%t A366015 nmax = 20; A[_] = 0; Do[A[x_] = x (1 + A[x])^4/(1 - 3 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t A366015 CoefficientList[InverseSeries[Series[x (1 - 3 x)/(1 + x)^4, {x, 0, 20}], x], x]	
%t A366015 Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[4 n, n - k - 1] 3^k, {k, 0, n - 1}], {n, 1, 20}]]
%Y A366015 Cf. A002293, A002412, A064087, A365754, A365816, A366014, A366016, A366017.
%K A366015 nonn
%O A366015 0,3
%A A366015 _Ilya Gutkovskiy_, Sep 26 2023