%I #47 Sep 27 2023 03:57:09
%S 0,1,7,73,905,12354,179305,2715192,42414021,678476755,11058588574,
%T 182999237590,3066447596459,51926183715280,887204891847960,
%U 15276037569668880,264797324173666845,4617195655522976361,80930337327794271445,1425171253004955494215,25202145191953299213490
%N G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 2 * A(x)).
%C Reversion of g.f. for 4-dimensional figurate numbers A001296 (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) * 2^k for n > 0.
%F a(n) ~ sqrt(32 - 19*sqrt(5/2)) * 3^(4*n - 3/2) * 5^(3*n) / (sqrt(Pi) * n^(3/2) * 2^(2*n + 3/2) * (25 + 34*sqrt(10))^n). - _Vaclav Kotesovec_, Sep 27 2023
%t nmax = 20; A[_] = 0; Do[A[x_] = x (1 + A[x])^5/(1 - 2 A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t CoefficientList[InverseSeries[Series[x (1 - 2 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] 2^k, {k, 0, n - 1}], {n, 1, 20}]]
%Y Cf. A001296, A002294, A064063, A365755, A366014, A366035, A366036, A366037.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Sep 26 2023