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; q 2)i (q; q)x+y+i’1 (q 2x+y+2i ; q)i (q x+2y+2i ; q)i
’2i2 (q
= q . (3.36)
(q; q)x+2i (q; q)y+2i (’q x+y+1 ; q)n’1+i
i=0



The reader should observe that this is not a straightforward q-analogue of (3.35) as it
does contain the terms (’q x+y+1 ; q)i+j in the determinant, respectively (’q x+y+1 ; q)n’1+i
in the denominator of the right-hand side product, which can be cleared only if q = 1.
A similar determinant evaluation, with some overlap with (3.36), was found by An-
drews and Stanton [10, Theorem 8] by making use of LU-factorization, in their “´tude” e
on the Andrews and the Mills“Robbins“Rumsey determinant.
Theorem 42. Let x and E be indeterminates and n be a nonnegative integer. Then
there holds
(E/xq i ; q 2)i’j (q/Exq i; q 2)i’j (1/x2 q 2+4i; q 2)i’j
det
(q; q)2i+1’j (1/Exq 2i; q)i’j (E/xq 1+2i; q)i’j
0¤i,j¤n’1
n’1
(x2q 2i+1; q)i (xq 3+i/E; q 2 )i (Exq 2+i; q 2)i
= . (3.37)
(x2q 2i+2; q 2)i (q; q 2)i+1 (Exq 1+i ; q)i (xq 2+i /E; q)i
i=0



The next group of determinants is (with one exception) from [90]. These determi-
nants were needed in the proof of a conjecture by Robbins and Zeilberger about a
generalization of the enumeration of totally symmetric self-complementary plane parti-
tions.
Theorem 43. Let x, y, n be nonnegative integers. Then
(x + y + i + j ’ 1)! (y ’ x + 3j ’ 3i)
det
(x + 2i ’ j + 1)! (y + 2j ’ i + 1)!
0¤i,j¤n’1
n’1
i! (x + y + i ’ 1)! (2x + y + 2i + 1)i (x + 2y + 2i + 1)i
=
(x + 2i + 1)! (y + 2i + 1)!
i=0
n
n
· (’1)k (x)k (y)n’k . (3.38)
k
k=0
ADVANCED DETERMINANT CALCULUS 41



This is Theorem 8 from [90]. A q-analogue, provided in [89, Theorem 2], is the
following theorem.
Theorem 44. Let x, y, n be nonnegative integers. Then there holds
(q; q)x+y+i+j’1 (1 ’ q y+2j’i ’ q y+2j’i+1 + q x+y+i+j+1 )
det
(q; q)x+2i’j+1 (q; q)y+2j’i+1
0¤i,j¤n’1

q ’2ij
·
(’q x+y+2 ; q)i+j
n’1 2
; q 2)i (q; q)x+y+i’1 (q 2x+y+2i+1 ; q)i (q x+2y+2i+1 ; q)i
’2i2 (q
= q
(q; q)x+2i+1 (q; q)y+2i+1 (’q x+y+2 ; q)n’1+i
i=0
n
n
— (’1)k q nk q yk (q x; q)k (q y ; q)n’k . (3.39)
k q
k=0



Once more, Amdeberhan observed that, in principle, Theorem 43 as well as The-
orem 44 could be proved by means of “condensation”. However, as of now, nobody
provided a proof of the double sum identities which would establish (2.16) in these
cases.
We continue with Theorems 2 and Corollary 3 from [90].
¤ n. Under the convention
Theorem 45. Let x, m, n be nonnegative integers with m
that sums are interpreted by
±
 B
 r=A+1 Expr(r) A<B
B
Expr(r) = 0 A=B

’ A
r=A+1
r=B+1 Expr(r) A > B,
there holds
2x + m + i + j
det
r
0¤i,j¤n’1
x+2i’j<r¤x+m+2j’i
n’1
(2x + m + i)! (3x + m + 2i + 2)i (3x + 2m + 2i + 2)i
=
(x + 2i)! (x + m + 2i)!
i=1
n/2 ’1
(2x + m)!
— · (2x + 2 m/2 + 2i + 1) · P1 (x; m, n), (3.40)
(x + m/2 )! (x + m)! i=0

where P1 (x; m, n) is a polynomial in x of degree ¤ m/2 .
In particular, for m = 0 the determinant equals
±
 n/2’1
n’1
 (2x + 2i + 1)
 i! (2x + i)! (3x + 2i + 2)2 i=0
i
n even (3.41)
(n ’ 1)!!
 i=0 (x + 2i)!2



0 n odd,
42 C. KRATTENTHALER

for m = 1, n ≥ 1, it equals
n/2 ’1
(2x + 2i + 3)
n’1
i! (2x + i + 1)! (3x + 2i + 3)i (3x + 2i + 4)i i=0
, (3.42)
(2 n/2 ’ 1)!!
(x + 2i)! (x + 2i + 1)!
i=0
for m = 2, n ≥ 2, it equals
n/2 ’1
(2x + 2i + 3)
n’1
i! (2x + i + 2)! (3x + 2i + 4)i (3x + 2i + 6)i i=0
(2 n/2 ’ 1)!!
(x + 2i)! (x + 2i + 2)!
i=0

1 (x + n + 1) n even
— · (3.43)
(x + 1) (2x + n + 2) n odd,
for m = 3, n ≥ 3, it equals
n/2 ’1
(2x + 2i + 5)
n’1
i! (2x + i + 3)! (3x + 2i + 5)i (3x + 2i + 8)i i=0
(2 n/2 ’ 1)!!
(x + 2i)! (x + 2i + 3)!
i=0

1 (x + 2n + 1) n even
— · (3.44)
(x + 1) (3x + 2n + 5) n odd,
and for m = 4, n ≥ 4, it equals
n/2 ’1
(2x + 2i + 5)
n’1
i! (2x + i + 4)! (3x + 2i + 6)i (3x + 2i + 10)i i=0
(2 n/2 ’ 1)!!
(x + 2i)! (x + 2i + 4)!
i=0

(x2 + (4n + 3)x + 2(n2 + 4n + 1)) n even
1
— · (3.45)
(x + 1)(x + 2) (2x + n + 4)(2x + 2n + 4) n odd.



One of the most embarrassing failures of “identi¬cation of factors,” respectively of
LU-factorization, is the problem of q-enumeration of totally symmetric plane partitions,
as stated for example in [164, p. 289] or [165, p. 106]. It is now known for quite a
while that also this problem can be reduced to the evaluation of a certain determinant,
by means of Okada™s result [123, Theorem 4] about the sum of all minors of a given
matrix, that was already mentioned in Section 2.8. In fact, in [123, Theorem 5], Okada
succeeded to transform the resulting determinant into a reasonably simple one, so that
the problem of q-enumerating totally symmetric plane partitions reduces to resolving
the following conjecture.
Conjecture 46. For any nonnegative integer n there holds
2
1 ’ q i+j+k’1
(1) (2)
det Tn + Tn = , (3.46)
1 ’ q i+j+k’2
1¤i,j¤n
1¤i¤j¤k¤n
ADVANCED DETERMINANT CALCULUS 43

where
i+j’2 i+j’1
(1)
q i+j’1
Tn = +q
i’1 i
q q 1¤i,j¤n
and « 
1+q
¬ ’1 1 + q 2 ·
0
¬ ·
¬ ·
’1 1 + q 3
Tn = ¬ ·.
(2)
¬ ’1 1 + q ·
4
¬ ·
 
.. ..
. .
0
’1 1 + q n
While the problem of (plain) enumeration of totally symmetric plane partitions was
solved a few years ago by Stembridge [170] (by some ingenious transformations of the
determinant which results directly from Okada™s result on the sum of all minors of a
matrix), the problem of q-enumeration is still wide open. “Identi¬cation of factors”
cannot even get started because so far nobody came up with a way of introducing a
parameter in (3.46) or any equivalent determinant (as it turns out, the parameter q
cannot serve as a parameter in the sense of Section 2.4), and, apparently, guessing the
LU-factorization is too di¬cult.
Let us proceed by giving a few more determinants which arise in the enumeration of
rhombus tilings.
Our next determinant evaluation is the evaluation of a determinant which, on dis-
regarding the second binomial coe¬cient, would be just a special case of (3.13), and
which, on the other hand, resembles very much the q = 1 case of (3.18). (It is the
determinant that was shown as (1.4) in the Introduction.) However, neither Lemma 3
nor Lemma 5 su¬ce to give a proof. The proof in [30] by means of “identi¬cation of
factors” is unexpectedly di¬cult.
Theorem 47. Let n be a positive integer, and let x and y be nonnegative integers.
Then the following determinant evaluation holds:
x+y+j x+y+j

det
x ’ i + 2j x + i + 2j
1¤i,j¤n
n
(j ’ 1)! (x + y + 2j)! (x ’ y + 2j + 1)j (x + 2y + 3j + 1)n’j
= . (3.47)
(x + n + 2j)! (y + n ’ j)!
j=1



This determinant evaluation is used in [30] to enumerate rhombus tilings of a certain
pentagonal subregion of a hexagon.
To see an example of di¬erent nature, I present a determinant evaluation from [50,
Lemma 2.2], which can be considered as a determinant of a mixture of two matrices,
out of one we take all rows except the l-th, while out of the other we take just the l-th
row. The determinants of both of these matrices could be straightforwardly evaluated
by means of Lemma 3. (They are in fact equivalent to special cases of (3.13).) However,
to evaluate this mixture is much more challenging. In particular, the mixture does not
44 C. KRATTENTHALER

anymore factor completely into “nice” factors, i.e., the result is of the form (2.21), so
that for a proof one has to resort to the extension of “identi¬cation of factors” described
there.
Theorem 48. Let n, m, l be positive integers such that 1 ¤ l ¤ n. Then there holds
«± 
 n+m’i (m+ 2 ) if i = l
n’j+1


det  m+i’j (n+j’2i+1) 
 n+m’i
1¤i,j¤n
if i = l
m+i’j
n/2
n
(n + m ’ i)! 1
= (m + i)n’2i+1 (m + i + )n’2i
(m + i ’ 1)! (2n ’ 2i + 1)! 2
i=1 i=1

’ 1)!
n l’1
(n ’ 2e) ( 1 )e
(m)n+1 j=1 (2j n
(n’1)(n’2)
—2 e 2
(’1) . (3.48)
2
e (m + e) (m + n ’ e) ( 1 ’ n)e
n/2
n! (2i)2n’4i+1 2
e=0
i=1



In [50], this result was applied to enumerate all rhombus tilings of a symmetric
hexagon that contain a ¬xed rhombus. In Section 4 of [50] there can be found several
conjectures about the enumeration of rhombus tilings with more than just one ¬xed
rhombus, all of which amount to evaluating other mixtures of the above-mentioned two
determinants.
As last binomial determinants, I cannot resist to show the, so far, weirdest deter-
minant evaluations that I am aware of. They arose in an attempt [16] by Bombieri,
Hunt and van der Poorten to improve on Thue™s method of approximating an algebraic
number. In their paper, they conjectured the following determinant evaluation, which,
thanks to van der Poorten [132], has recently become a theorem (see the subsequent
paragraphs and, in particular, Theorem 51 and the remark following it).
Theorem 49. Let N and l be positive integers. Let M be the matrix with rows labelled
by pairs (i1, i2) with 0 ¤ i1 ¤ 2l(N ’ i2) ’ 1 (the intuition is that the points (i1, i2 ) are
the lattice points in a right-angled triangle), with columns labelled by pairs (j1, j2 ) with
0 ¤ j2 ¤ N and 2l(N ’ j2 ) ¤ j1 ¤ l(3N ’ 2j2 ) ’ 1 (the intuition is that the points
(j1, j2 ) are the lattice points in a lozenge), and entry in row (i1, i2) and column (j1, j2 )
equal to
j1 j2
.
i1 i2
Then the determinant of M is given by
(N3 )
+2
l’1 3l’1
k=0 k! k=2l k!
± .
2l’1
k=l k!

This determinant evaluation is just one in a whole series of conjectured determinant
evaluations and greatest common divisors of minors of a certain matrix, many of them
reported in [16]. These conjectures being settled, the authors of [16] expect important
implications in the approximation of algebraic numbers.
The case N = 1 of Theorem 49 is a special case of (3.11), and, thus, on a shallow
level. On the other hand, the next case, N = 2, is already on a considerably deeper
ADVANCED DETERMINANT CALCULUS 45

level. It was ¬rst proved in [94], by establishing, in fact, a much more general result,
given in the next theorem. It reduces to the N = 2 case of Theorem 49 for x = 0,
b = 4l, and c = 2l. (In fact, the x = 0 case of Theorem 50 had already been conjectured
in [16].)
Theorem 50. Let b, c be nonnegative integers, c ¤ b, and let ∆(x; b, c) be the determi-
nant of the (b + c) — (b + c) matrix

« 0¤j<c c ¤ j <b b ¤ j <b+c
. .
. .
. . 2x + j
¬ ·
. .
x+j x+j
. .
0¤i<c ¬ ·
. .
. .
¬............................................................ ·
. .
. .
i i i
. .
¬ ·
. .
. .
¬ ·
. . 2x + j
. .
¬ ·.
x+j
. . (3.49)
c¤i<b 0 . .
¬ ·
. .
. .
¬............................................................ ·
. .
i i
. .
¬ ·
. .
. .
 
. .
. .
x+j . x+j .
b¤i <b+c 0
. .
2 . .
i’b . .
i’b
. .
. .
Then
(i) ∆(x; b, c) = 0 if b is even and c is odd;
(ii) if any of these conditions does not hold, then

i+ 1 ’
b’c b
2 2
cc c
∆(x; b, c) = (’1) 2
(i)c
i=1
+ c+i b’c+i
c x x+
b’c+ i/2 ’ (c+i)/2 (b+i)/2 ’ (b’c+i)/2
2 2
— .
’ ’
1 b
+ c+i 1 b b’c+i
+
b’c+ i/2 ’ (c+i)/2 (b+i)/2 ’ (b’c+i)/2
2 2 2 2 2 2
i=1

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