From zeilberg Thu Jul 18 11:45:12 1996
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Date: Thu, 18 Jul 1996 11:45:12 -0400
From: Doron Zeilberger
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Subject: Propp's problem
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Status: RO
>From propp@math.mit.edu Mon Jul 1 15:48:11 1996
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Date: Mon, 1 Jul 1996 15:47:56 -0400 (EDT)
From: Jim Propp
Message-Id: <199607011947.PAA20350@pfaff.mit.edu>
To: zeilberg@euclid.math.temple.edu
Subject: determinant
Status: RO
Doron,
>Dear Jim, I enjoyed your talk.
Thanks.
>Could you please send me
>any determinants that evaluate nicely that you can't do
>yet?
Well, the ones that arise in the tilings problem are huge
matrices of 1's, -1's, and 0's whose entries I have no
explicit recipe for (though writing down such recipes
wouldn't be hard); I rely on the computer to turn pictures
into adjacency matrices M and then to change the signs of
strategically chosen entries so that the permanent of M
equals the determinant of the new matrix K.
But here is a determinant that did grow out of some work
Greg and I did on tilings back in 1990:
It appears that the determinant of the (n+1)-by-(n+1) matrix
(indexed by i,j running from 0 to n) whose (i,j)th entry is
(i+j)! (2n-i-j)!
-------------------
i! j! (n-i)! (n-j)!
is equal to
n n-1 n-2 1
(2n+1) (2n) (2n-1) ... (n+2)
-------------------------------------- .
1 2 3 n
(n) (n-1) (n-2) ... (1)
This formula works up to n=9. More generally, if one lets M denote
the (m+1)-by-(n+1) matrix whose (i,j)th entry is
(i+j)! (m+n-i-j)!
-------------------
i! j! (m-i)! (n-j)!
then the determinant of M-transpose times M appears to be
2n+2
[(m+n+1)!] H(m-n)
--------------------- ,
H(2n+2) H(m+n+2)
where H(n) = 1!2!...(n-1)!.
Greg has shown that the formula has a nice q-analogue, though he hasn't
nailed down the exact form of the right hand side, as far as I know.
I should mention that we do not need this conjecture for anything. In
fact, these matrices have no combinatorial interpretation that I know
of --- though they resemble certain matrices that did arise from some
applications (which at this point I have forgotten).
Jim