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a(n) = Sum_{k=0..floor((n-4)/5)} Stirling2(n,5*k+4).
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%I #8 Sep 08 2023 07:22:46

%S 0,0,0,0,1,10,65,350,1701,7771,34150,146905,633776,2892032,15526876,

%T 109484545,992589171,10223409493,108982611518,1156117871286,

%U 12062817285396,123603289559039,1245986248828926,12391614409960544,121996350285087172

%N a(n) = Sum_{k=0..floor((n-4)/5)} Stirling2(n,5*k+4).

%F Let A(0)=1, B(0)=0, C(0)=0, D(0)=0 and E(0)=0. Let B(n+1) = Sum_{k=0..n} binomial(n,k)*A(k), C(n+1) = Sum_{k=0..n} binomial(n,k)*B(k), D(n+1) = Sum_{k=0..n} binomial(n,k)*C(k), E(n+1) = Sum_{k=0..n} binomial(n,k)*D(k) and A(n+1) = Sum_{k=0..n} binomial(n,k)*E(k). A365528(n) = A(n), A365529(n) = B(n), A365530(n) = C(n), A365531(n) = D(n) and a(n) = E(n).

%F G.f.: Sum_{k>=0} x^(5*k+4) / Product_{j=1..5*k+4} (1-j*x).

%o (PARI) a(n) = sum(k=0, (n-4)\5, stirling(n, 5*k+4, 2));

%Y Cf. A365528, A365529, A365530, A365531.

%K nonn

%O 0,6

%A _Seiichi Manyama_, Sep 08 2023