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G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 4 * A(x)).
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%I #7 Sep 27 2023 04:42:48

%S 0,1,9,127,2165,40914,824859,17383720,378373437,8440227235,

%T 191938302578,4433259845898,103716352560119,2452629475989840,

%U 58529969579982600,1407775987050271920,34092047564798908045,830565580516900384329,20342106952028722530603,500573735323751221019425,12370242700776737398052970

%N G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 4 * A(x)).

%C Reversion of g.f. for 4-dimensional figurate numbers A002418 (with signs).

%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(5*n,n-k-1) * 4^k for n > 0.

%F a(n) ~ sqrt(34*sqrt(6) - 81) * 2^(n - 11/4) * 3^(n - 5/4) * (3/2 - 1/sqrt(6))^(5*n) / (sqrt(Pi) * n^(3/2) * (3*sqrt(6) - 7)^n). - _Vaclav Kotesovec_, Sep 27 2023

%t nmax = 20; A[_] = 0; Do[A[x_] = x (1 + A[x])^5/(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)^5, {x, 0, 20}], x], x]

%t Join[{0}, Table[1/n Sum[Binomial[n + k - 1, k] Binomial[5 n, n - k - 1] 4^k, {k, 0, n - 1}], {n, 1, 20}]]

%Y Cf. A002294, A002418, A064088, A365668, A365755, A365817, A366016, A366035, A366037.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Sep 26 2023