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a(n) = Sum_{k=0..floor(n/3)} (-1)^k * (n-2*k)!/(n-3*k)!.
4

%I #14 Nov 28 2022 12:05:24

%S 1,1,1,0,-1,-2,-1,2,7,8,-1,-26,-49,-28,103,314,359,-344,-2113,-3682,

%T -161,14684,36791,25762,-100297,-373456,-472241,587846,3877487,

%U 7149988,-1111801,-40808566,-103472249,-56751688,424662623,1490284654,1674543359,-4121143444

%N a(n) = Sum_{k=0..floor(n/3)} (-1)^k * (n-2*k)!/(n-3*k)!.

%H Seiichi Manyama, <a href="/A358604/b358604.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.

%o (PARI) a(n) = sum(k=0, n\3, (-1)^k*(n-2*k)!/(n-3*k)!);

%Y Cf. A358603, A358605, A358606.

%Y Cf. A357532.

%K sign

%O 0,6

%A _Seiichi Manyama_, Nov 23 2022