The game of Quod

[Mitch Harris]

 > From: Joshua Zucker [mailto:joshua.zucker at gmail.com]
 >
 > In Ian Stewart's latest book, he mentions the game of Quod. (Ch 7 of
 > _How To Cut A Cake: and other mathematical conundrums_)
 >
 > One sequence of relevance to this game is the number of squares in an
 > nxn board with the four corners deleted.  He says that Denis Borris
 > found that there are (n^4 - n^2 - 48n + 84)/12 of them, which is not
 > in OEIS.
 >
 > He also says that Ken Duisenberg found the formula for the mxn case
 > (which would make a nice triangle for OEIS) but he doesn't give the
 > formula, nor a reference.  Anyone have a suggestion?

Call me kooky (or slow) but a quick check to me says that the 3x3 board
with the 4 corners removed leaves 3^2 - 4 = 5 (unit) squares (and unit
squares are the only ones that fit).

And isn't the general idea just that the only 'bad' squares of any size
k are those that include one of the corners (and for any k sided squares
there are exactly 4 of these 'bad' squares)?

Or formally
   Sum[(n-k)^2 - 4, {k, 0, n-2}] =   Sum[k^2 - 4, {k, 2, n}]
  =
   (2n^3 + 3n^2 - 23n + 18)/6
  =
   0, 5, 17, 38, 70, 115, 175...

(which is also not in the OEIS)

and the formula for nxm rectangles (and squares in them) has formula
(for n<m)

   Sum[(n - k)(m - k) - 4 , {k, 0, n - 2}]
  =
   (3m n^2 - n^3 + 3m n - 5n - 18m + 18)/6

Please tell me where I've gone wrong.

(Unfortunately I have no idea about the original problem.)

Mitch