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

%I #15 Nov 28 2022 12:05:21

%S 1,1,1,1,0,-1,-2,-3,-2,1,6,13,16,9,-14,-59,-108,-119,-26,261,736,1177,

%T 1026,-731,-4964,-11079,-14978,-6299,30024,102841,189466,190917,

%U -97004,-921191,-2301354,-3396539,-1674368,7265241,27311794,53600101,56943756,-31760903,-310594514,-809146971

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

%H Seiichi Manyama, <a href="/A358605/b358605.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = (3 * a(n-1) - n * a(n-4) + 1)/4 for n > 3.

%o (PARI) a(n) = sum(k=0, n\4, (-1)^k*(n-3*k)!/(n-4*k)!);

%Y Cf. A358603, A358604, A358606.

%Y Cf. A357533.

%K sign

%O 0,7

%A _Seiichi Manyama_, Nov 23 2022