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a(n) = 1 + Sum_{k=0..n-5} a(k) * a(n-k-5).
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%I #13 Jan 22 2023 09:15:11

%S 1,1,1,1,1,2,3,4,5,6,9,14,21,30,41,59,89,136,205,301,443,664,1011,

%T 1545,2341,3530,5341,8143,12487,19148,29299,44817,68721,105742,163025,

%U 251392,387595,597988,924047,1430167,2215595,3433788,5323915,8260652,12829849

%N a(n) = 1 + Sum_{k=0..n-5} a(k) * a(n-k-5).

%H Seiichi Manyama, <a href="/A346074/b346074.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x^5 * A(x)^2.

%F Conjecture D-finite with recurrence (n+5)*a(n) +2*(-n-4)*a(n-1) +(n+3)*a(n-2) +2*(-2*n+5)*a(n-5) +4*(n-3)*a(n-6)=0. - _R. J. Mathar_, Feb 17 2022

%F a(n) = Sum_{k=0..floor(n/5)} binomial(n-4*k,k) * Catalan(k). - _Seiichi Manyama_, Jan 22 2023

%t a[n_] := a[n] = 1 + Sum[a[k] a[n - k - 5], {k, 0, n - 5}]; Table[a[n], {n, 0, 44}]

%t nmax = 44; A[_] = 0; Do[A[x_] = 1/(1 - x) + x^5 A[x]^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%o (PARI) a(n) = sum(k=0, n\5, binomial(n-4*k, k)*binomial(2*k, k)/(k+1)); \\ _Seiichi Manyama_, Jan 22 2023

%Y Cf. A007317, A090344, A216604, A307972, A346073.

%K nonn

%O 0,6

%A _Ilya Gutkovskiy_, Jul 04 2021