Dear Dave and Richard,
Richard just told me that you are still not convinced about
the proof of (Chip). I apologize for being a bit impatient,
and of telling you quick (and wrong) ways to "get out of it",
just to "shut you up". On Richard's urging, I looked
at it carefully, and I do agree that
(1) My (Chip) does not follow directly from Morris's identity,
since I claim it for negative a, b, c in Morris's, and Morris
only proved it for positive a,b,c.
(2) There seem to be some problems with adapting Aomoto's
or Selberg's original proof, since they use symmetry and/or
anti-symmetry, which is wrong for an arbitrary c.
(3) Making c an "indeterminate" in (Chip), the way it stands,
does not make sense.
BUT, two of my remedies described in my message
to Dave from 4/15/99 (reproduced below),
are perfectly valid.
1) Adapting the proof of Selberg's identity given in my paper
with Wilf: Inv. Math. 108 (1992), 575-633,
(subsubsection 6.5.1) p. 627-628.
Here is a sketch of a WZ-proof that (Chip) is valid for
(positive) INTEGERS a,b,c (i.e. -a, -b, -c are negative)
(i) Let F_n (a,b,c) be the left side of (Chip)
Then, by taking the inner-most CT (w.r.t x_1)
F_n(0,0,c)=F_{n-1} (0,c,c).
So the "base case" a=b=0 (for n) follows by induction from the
(n-1)-case
(ii) With the certifacate R at the bottom of p. 627 (with the
obvious shadowing) (WZ-Selberg) still holds.
(iii) Now apply the functional CT_{x_n} .... CT_{x_1} to both
sides of (WZ-Selberg) and get the desired recurrence.
2) Another way is to refer to (q-Morris),
due, independently to Laurent Habsieger and
Kevin Kadell, and given an elementary
proof in my paper Discrete Math 79 (1989) 313-322.
(i) Combine (x_i)_b (q/x_i)_c into
(q^{-c} x_i)_{b+c}/(some power of x_i * some power of q).
This will make it look more like (Chip)
(ii) Convert (1.1) into a more general-looking form
by repeatedly using
(Z)_a=(Z)_{\infinity}/(q^a Z)_{\infinity}
and then replacing q^a by A, q^b by B, and q^c by C
(iii) Do the same for (1.3)
Now Theorem' (eq. (1.4)) is an identity involving general
variables A,B,C
(iv) Note that it is valid in general (if nothing else as
formal power series in q) each and every coeff. of q,
of either side of (1.4) is
a rational function of (A,B,C), and proving it for infinitely
many values like is done in my paper (or in
Habsieger's and Kadell's), proves it in general.
(v) Specialize A,B,C to be q^{-a},q^{-b}, q^{-c) respectively.
(vi) Take q->1
It would be too boring to give any more details. However, Dave,
if you are willing to pay $3000 consultant's fees, it will
make this task much less daunting, and I'll be glad to furnish
excrutiating details. To make it even more interesting (to you, Dave),
I'll gladly pay YOU $3000 if the detailed write-up will not
satisfy a mutually-agreed-on arbiter.
Best wishes,
Doron
Appendix:
-----message from DZ to Dave Robbins 4/15/99-------
Date: Thu, 15 Apr 1999 17:42:09 -0400 (EDT)
From: Doron Zeilberger
Message-Id: <199904152142.RAA23800@euclid.math.temple.edu>
To: robbins@idaccr.org
Subject: Re: CRY proof
Content-Length: 585
Dear Dave,
OK, let's make c an integer again, for starters, then
Aomoto, or WZ or my proof of q-Morris realy should go verbatim
(the same recurrences), and similar for initial condition
(if you look at my proof of q-Morris (`A Stembridge-Stanton...')
then it is much easier to juggle a, b and c to negative using
standard q-conversion). Then let q->1. Then
use the properties of Gamma to extend to all c
(see a paper by Remmert a few years ago in the Monthly).
Best wishes
Doron
P.S. Perhaps Frank Garvan has more patience
for this kind of details, ask him!
------end message