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a:= proc(n) option remember; `if`(n<4, [1$3, 7][n+1], ((2*n-3)
*a(n-1)+(n-1)*(2*n^2-8*n+7)*a(n-2) + (n-2)*(n-1)*(2*n-5)
*a(n-3)-(n-4)*(n-3)*(n-2)^2*(n-1)*a(n-4))/(n-2))
end:
seq(a(n), n=0..25); # Alois P. Heinz, Dec 01 2021
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a(n) = (n!*sum(m=floor((n+1)/2)..n, (binomial(n-1,2*m-n-1))/(2*m-n)!)). [ _- _Vladimir Kruchinin_, Mar 10 2013]
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Vincenzo Librandi, <a href="/A096965/b096965_1.txt">Table of n, a(n) for n = 0..200</a>