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a(n) = Sum_{k=0..floor(n/3)} (n-2*k)!/(n-3*k)!.
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%I #30 Nov 25 2022 06:33:49

%S 1,1,1,2,3,4,7,12,19,34,63,112,211,414,799,1588,3267,6706,13999,30024,

%T 64723,141142,314271,705724,1599619,3685338,8573167,20112016,47804499,

%U 114743614,277615903,679057092,1676636611,4171532674,10477002159,26545428568,67755344467,174386589606

%N a(n) = Sum_{k=0..floor(n/3)} (n-2*k)!/(n-3*k)!.

%H Seiichi Manyama, <a href="/A357532/b357532.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = (2 * a(n-1) + n * a(n-3) + 1)/3 for n > 2.

%F a(n) ~ c * n^(n/3 + 1/2) / (3^(n/3) * exp(n/3 - n^(2/3)/3^(2/3) - 2*n^(1/3) / 3^(7/3))) * (1 + 1235/(729 * 3^(2/3) * n^(1/3)) + 9452027/(15943230 * 3^(1/3) * n^(2/3)) + 16015315669/(41841412812*n)), where c = 0.50682110703119..., conjecture: c = exp(4/81) * sqrt(2*Pi) / 3^(3/2). - _Vaclav Kotesovec_, Nov 25 2022

%o (PARI) a(n) = sum(k=0, n\3, (n-2*k)!/(n-3*k)!);

%Y Cf. A072374, A122852, A357533, A357570.

%K nonn,easy

%O 0,4

%A _Seiichi Manyama_, Nov 19 2022