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%I #10 Mar 23 2022 09:41:26
%S 1,1,2,6,24,120,721,5054,40488,364896,3654000,40249441,483659508,
%T 6296246424,88269037584,1325861901000,21243052172161,361630022931666,
%U 6518319228715302,124018898163736536,2483799332459535000,52231733840672804881,1150683180739820615582,26502219276887376327696
%N a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/5)} binomial(n,5*k+1) * a(n-5*k-1).
%H Seiichi Manyama, <a href="/A352430/b352430.txt">Table of n, a(n) for n = 0..449</a>
%F E.g.f.: 1 / (1 - Sum_{k>=0} x^(5*k+1) / (5*k+1)!).
%t a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, 5 k + 1] a[n - 5 k - 1], {k, 0, Floor[(n - 1)/5]}]; Table[a[n], {n, 0, 23}]
%t nmax = 23; CoefficientList[Series[1/(1 - Sum[x^(5 k + 1)/(5 k + 1)!, {k, 0, nmax}]), {x, 0, nmax}], x] Range[0, nmax]!
%o (PARI) my(N=40, x='x+O('x^N)); Vec(serlaplace(1/(1-sum(k=0, N\5, x^(5*k+1)/(5*k+1)!)))) \\ _Seiichi Manyama_, Mar 23 2022
%Y Cf. A000670, A006154, A243666, A333882, A352428, A352429.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Mar 16 2022