A349335 G.f. A(x) satisfies A(x) = 1 + x * A(x)^8 / (1 - x).
1, 1, 9, 109, 1541, 23823, 390135, 6651051, 116798643, 2098313686, 38382509118, 712447023590, 13385500614902, 254065657922154, 4864482597112186, 93840443376075810, 1822169236520766546, 35586928273002974487, 698572561837366684479, 13775697096997873764647
Offset: 0
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..756
Crossrefs
Programs
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Maple
a:= n-> coeff(series(RootOf(1+x*A^8/(1-x)-A, A), x, n+1), x, n): seq(a(n), n=0..19); # Alois P. Heinz, Nov 15 2021
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Mathematica
nmax = 19; A[] = 0; Do[A[x] = 1 + x A[x]^8/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] Table[Sum[Binomial[n - 1, k - 1] Binomial[8 k, k]/(7 k + 1), {k, 0, n}], {n, 0, 19}]
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PARI
{a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[#A] = 1 + sum(k=1, m-1, (polcoeff(Ser(A)^8, k)) )); A[n+1]} for(n=0, 30, print1(a(n), ", ")) \\ Vaclav Kotesovec, Nov 23 2024, after Paul D. Hanna
Formula
a(n) = Sum_{k=0..n} binomial(n-1,k-1) * binomial(8*k,k) / (7*k+1).
a(n) ~ 17600759^(n + 1/2) / (2048 * sqrt(Pi) * n^(3/2) * 7^(7*n + 3/2)). - Vaclav Kotesovec, Nov 15 2021
Comments