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a(n) = [n/2]!*[(n+1)/2]!*C([n/2],4)*C([(n+1)/2],4).
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%I #15 Jul 12 2024 15:15:58

%S 0,0,0,0,0,0,0,576,14400,360000,6480000,116640000,1905120000,

%T 31116960000,497871360000,7965941760000,129048256512000,

%U 2090581755494400,34843029258240000,580717154304000000,10038110810112000000,173515915431936000000,3123286477774848000000

%N a(n) = [n/2]!*[(n+1)/2]!*C([n/2],4)*C([(n+1)/2],4).

%C Number of permutations of n elements with 8 odd displacements.

%C Column 5 of A226288.

%C Column k=8 of A145893, a diagonal of A145894. - _Alois P. Heinz_, May 29 2014

%H R. H. Hardin, <a href="/A226285/b226285.txt">Table of n, a(n) for n = 1..210</a>

%t a[n_]:= Floor[n/2]!*Floor[(n+1)/2]!*Binomial[Floor[n/2],4]*Binomial[Floor[(n+1)/2],4]; Array[a,21] (* _Stefano Spezia_, Jul 12 2024 *)

%Y Cf. A145893, A145894, A226288.

%K nonn

%O 1,8

%A _R. H. Hardin_, connection of formula with combinatoric problem via _N. J. A. Sloane_ in the Sequence Fans Mailing List, Jun 02 2013