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A366036 G.f. A(x) satisfies: A(x) = x * (1 + A(x))^5 / (1 - 4 * A(x)). 3
0, 1, 9, 127, 2165, 40914, 824859, 17383720, 378373437, 8440227235, 191938302578, 4433259845898, 103716352560119, 2452629475989840, 58529969579982600, 1407775987050271920, 34092047564798908045, 830565580516900384329, 20342106952028722530603, 500573735323751221019425, 12370242700776737398052970 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Reversion of g.f. for 4-dimensional figurate numbers A002418 (with signs).
LINKS
Eric Weisstein's World of Mathematics, Series Reversion
FORMULA
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.
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
MATHEMATICA
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]
CoefficientList[InverseSeries[Series[x (1 - 4 x)/(1 + x)^5, {x, 0, 20}], x], x]
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}]]
CROSSREFS
Sequence in context: A338077 A034301 A092651 * A258294 A362776 A365033
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Sep 26 2023
STATUS
approved

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Last modified June 4 15:08 EDT 2024. Contains 373099 sequences. (Running on oeis4.)