A019318 formula

[David Wilson]

Per A019318:

I used Burnside's Lemma, as suggested, to compute a formula for
A019318.  It was fairly easy once the application of the Lemma to
the problem in question was explained.  The formula is still
somewhat ugly, and I used some cheats, but here goes.

Let
 C(n, k) = choose
 [n]  = floor
 n mod k = remainder

Define

A(n) = C(n^2, n)
B(n) = C((n^2-(n mod 2))/2, n/2);
C(n) = C((n^2-(n mod 2))/4, n/4);
D(n) = SUM(p = 0 to [n/2], C((n^2-n)/2, p)*C(n, n-2p))

Then

a(n) =
 (A(n) + 3B(n) + 2C(n) + 2D(n))/8 if n == 0 (mod 4)
 (A(n) + B(n) + 2C(n) + 4D(n))/8 if n == 1 (mod 4)
 (A(n) + 3B(n) + 2D(n))/8 if n == 2 (mod 4)
 (A(n) + B(n) + 4D(n))/8 if n == 3 (mod 4)

I used this formula to confirm and extend A019318 to full length.
I have submitted the terms, but not the formula, due to its length.
Maybe someone could submit the formula in an acceptable form,
it it is deemed worthy.