A TALE OF TOEPLITZ PLUS HANKEL DETERMINANTS

TEWODROS AMDEBERHAN, CHRISTOPH KOUTSCHAN, AND DORON ZEILBERGER

Abstract. Three versatile tools (Dodgson’s Condensation, the Holonomic Anstaz and Constant Term evaluations) developed by the third-named author are shown to operate in tandem to prove a certain intriguing determinantal formula conjectured by the first-named author.

1. An intriguing determinant

The discussion in this article is motivated by a question on Mathoverflow [3] posed by Johann Cigler wherein he asked for a simple direct proof of

det A_{n,m} := det [\binom{2m}{j − i + m} − \binom{2m}{m − i − j − 1}]_{i,j=0}^{n−1} = ∏_{i≤j}^{1,m−1} \frac{2n + i + j}{i + j} .

The expression on the right-hand side of the identity is connected to a host of interesting combinatorial interpretations (see references [6], [7], [10], [16] and [17]).

Our journey began with a generalization (true to Zeilberger’s tradition: generalize to trivialize). Introduce a new parameter x and a matrix given by

T_{n,m}(x) : = [\binom{x + m}{j − i + m} − \binom{x + m}{m − i − j − 1}]_{i,j=0}^{n−1} .

After enough experimentation, we managed to guess

Conjecture 1. We have

det T_{n,m}(x) = ∏_{i=1}^{n} ∏_{j=1}^{m} \frac{(x + i − j)(x + 2i + j − 2)}{(x + 2i − j)(i + j − 1)}.

None of our initial attempts at proving this claim worked.

Remark 1. In the meantime, we made the following curious observation. If

φ_{n}(y) := \frac{(1 + y^{n} − y^{n+1} + y^{2n})(1 − y^{n})^{2}}{y^{n−1}(1 + y)(1 − y)^{2}} ∈ ℤ[y, y^{−1}]

then

det T_{n,n}(x) = n!^{−n} ∏_{i=0}^{n−1} \binom{n + i}{i}^{−1} · ∏_{s=1−n}^{3n−2}(x + s)^{CT_{y}(y^{−s}φ_{n}(y))};

where CT_{y} F (y) denotes the constant term of a Laurent polynomial F (y).

2. Enter Dodgson

A relatively simple choice for conjecturing and proving explicit determinant evaluations is a method inspired by Dodgson’s Condensation [19], that goes back to Jacobi and Desanot, and popularized by the Reverend Charles L. Dodgson (aka Lewis Caroll). However, det T_{n,m}(x) did not yield to this at all. It was necessary to change the approach. Our first breakthrough was realized in the form of discovering a new matrix

D_{n}(a, b) : = [\binom{2i + 2a}{i − j + a − b} − \binom{2i + 2a}{i − j + a − b − 1}]_{i,j=0}^{n−1}

which, happily, is amenable to the present technique.

Theorem 2.1. We have the determinantal evaluation

det D_{n}(a, b) = ∏_{i=1}^{n} ∏_{j=1}^{a−b} \frac{(a + b + i − j)(a + b + 2i + j − 2)}{(a + b + 2i − j)(i + j − 1)}.

Proof. Denoting the determinant by M_{n}(a, b), Dodgson’s Condensation states that

M_{n}(a, b) = \frac{M_{n−1}(a, b)M_{n−1}(a + 1, b + 1) − M_{n−1}(a + 1, b)M_{n−1}(a, b + 1)}{M_{n−2}(a + 1, b + 1)}.

Therefore, we only need to verify (routinely) that the right-hand side does the same. To conclude, simply check both sides agree, say, when n = 1 and n = 2. □

For reasons that will become clearer soon, let’s reformulate the above result.

Corollary 2.2. We have

det B_{n,m}(x) : = det [\binom{x + m + 2i}{i − j + m} − \binom{x + m + 2i}{i − j + m − 1}]_{i,j=0}^{n−1}

= ∏_{i=1}^{n} ∏_{j=1}^{m} \frac{(x + i − j)(x + 2i + j − 2)}{(x + 2i − j)(i + j − 1)}.

Proof. This follows from Theorem 2.1 with a = \frac{x+m}{2} and b = \frac{x−m}{2} in D_{n}(a, b). □

3. Enter Constant Term evaluations

In this section, we describe the connection between the two matrices B_{n,m}(x) and T_{n,m}(x) from the preceding sections.

Theorem 3.1. It holds true that det T_{n,m}(x) = det B_{n,m}(x), so does Conjecture 1.

Proof. Denote the constant term of the Laurent series expansion of F (t) by CT_{t}F. In view of this, \binom{x+m}{j−i+m} − \binom{x+m}{m−i−j−1} is the constant term (in the variable t):

CT_{t} \{(1 + \frac{1}{t})^{x+m} t^{m}(t^{j−i} − t^{−i−j−1})\} .

The entries in the i^{th} row of the matrix T_{n,m}(x) are (we need a different variable for each row, let’s call it t_{i})

CT_{t_{i}} (1 + \frac{1}{t_{i}} )^{x+m} t_{i}^{m}(t_{i}^{j−i} − t_{i}^{−i−j−1}).

Let t⃗ = (t_{0}, t_{1}, …, t_{n−1}). We apply multi-linearity of determinants to obtain

det T_{n,m}(x) = det [CT_{t_{i}} (1 + \frac{1}{t_{i}} )^{x+m} t_{i}^{m−i} (t_{i}^{j} − t_{i}^{−j−1})]

= CTt⃗ [∏_{i=0}^{n−1} (1 + t_{i})^{x+m} · t_{i}^{−x−i− \frac{1}{2}} · det (t_{i}^{j+ \frac{1}{2}} − t_{i}^{−j− \frac{1}{2}} )]

= CTt⃗ [∏_{i=0}^{n−1} \frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+n+i}} · ∏_{i<j}^{0,n−1}(t_{i} − t_{j})(1 − t_{i}t_{j})];

where we have utilized ([11], Lemma 2). Let’s symmetrize the right-hand side (by averaging over the symmetric group 𝔖_{n}) and note that only the product of two factors ∏_{i=0}^{n−1} t_{i}^{−i} · ∏_{i<j}(t_{i} − t_{j}) will be affected to introduce a new Vandermonde:

det T_{n,m}(x)

= \frac{1}{n!}CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+n}}] · ∏_{i<j}^{0,n−1}(t_{i} − t_{j})(t_{j}^{−1} − t_{i}^{−1})(1 − t_{i}t_{j})\}

= \frac{1}{n!}CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+2n−1}}] · ∏_{i<j}^{0,n−1}(t_{i} − t_{j})^{2}(1 − t_{i}t_{j})\} .

Now, repeat the constant term extraction on the determinant from the previous section. The outcome of this procedure is:

det B_{n,m}(x)

= det [CT_{t_{i}} (1 + \frac{1}{t_{i}} )^{x+m+2i} t_{i}^{m}(t_{i}^{i−j} − t_{i}^{i−j−1})]

= CTt⃗ \{∏_{i=0}^{n−1} [(1 + \frac{1}{t_{i}} )^{x+m+2i} t_{i}^{m+i−1}(t_{i} − 1)] ∏_{i<j}^{0,n−1}(t_{j}^{−1} − t_{i}^{−1})\}

= CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+n}}] (2 + t_{i} + t_{i}^{−1})^{i} ∏_{i<j}^{0,n−1}(t_{i} − t_{j})\}

= \frac{1}{n!}CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+n}}] ∏_{i<j}^{0,n−1}(t_{i} − t_{j}) (t_{j} + t_{j}^{−1} − t_{i} − t_{i}^{−1})\}

= \frac{1}{n!} CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+2n−1}}] ∏_{i<j}^{0,n−1}(t_{i} − t_{j})^{2}(1 − t_{i}t_{j})\} .

Based on our analysis shown above, both determinants det T_{n,m}(x) and det B_{n,m}(x) are equal to the same constant term

\frac{1}{n!} CTt⃗ \{∏_{i=0}^{n−1} [(1 + t_{i})^{x+m}(t_{i} − 1)t_{i}^{−x−2n+1}] ∏_{i<j}^{0,n−1}(t_{i} − t_{j})^{2}(1 − t_{i}t_{j})\} .

Hence det T_{n,m}(x) and det B_{n,m}(x) must be equal to each other, as claimed. □

In particular, we thus recover the determinant of Cigler as stated in Section 1:

Corollary 3.2. (The case x = m) We have

det [\binom{2m}{j − i + m} − \binom{2m}{m − i − j − 1}]_{i,j=0}^{n−1} = ∏_{1≤i≤j≤m−1} \frac{2n + i + j}{i + j} .

Proof. What we need here is the specialization x = m in det T_{n,m}(x) resulting in det T_{n,m}(m) = ∏_{1≤i≤j≤m−1} \frac{2n+i+j}{i+j} after some simplification. The outcome is exactly det A_{n,m}. □

In addition, we believe to have uncovered a new constant term evaluation which appears not to fit any of the existing formulas in the literature.

Corollary 3.3. We have

CTt⃗ \{∏_{i=0}^{n−1} [\frac{(1 + t_{i})^{x+m}(t_{i} − 1)}{t_{i}^{x+2n−1}}] ∏_{i<j}^{0,n−1}(t_{i} − t_{j})^{2}(1 − t_{i}t_{j})\}

= n! · ∏_{i=1}^{n} ∏_{j=1}^{m} \frac{(x + i − j)(x + 2i + j − 2)}{(x + 2i − j)(i + j − 1)}.

Proof. This is a consequence of Corollary 2.2 and the proof of Theorem 3.1. □

4. Enter the Holonomic Ansatz

In this section, we use an automated method called the Holonomic Ansatz due to the third author [21], and extended by the second author to discover and prove many new determinant identities, which will provide yet another proof of Conjecture 1. Note: We also reindex 1 ≤ i, j ≤ n instead of 0 ≤ i, j ≤ n − 1. Let us recall the matrix A_{n} := A_{n}(m, x) := (a_{i,j})_{1≤i,j≤n} whose entries a_{i,j} are given as follows:

a_{i,j} := a_{i,j}(m, x) := \binom{m + x}{m − i + j} − \binom{m + x}{m − i − j + 1}.

Denote by A_{n}^{(i,j)} the matrix A_{n} with the i^{th}-row and the j^{th}-column being removed. Then we define

(1)  c_{n,j} := c_{n,j}(m, x) := (−1)^{n+j} \frac{det A_{n}^{(n,j)}}{det A_{n−1}} .

Employing an ansatz with undetermined coefficients (“guessing”), we find plausible recurrence relations that are conjecturally satisfied by the sequence c_{n,j}, which suggest that it is holonomic with rank 2:

j(j − n − 1)(m + n)(2n + x − 1)(2n + x)c_{n+1,j}+
(2j − 1)n(j − m)(2n + x)(j − n − x + 1)c_{n,j+1}+
n(j + n + x − 1)(4j^{2}n − (m^{2} + 4n^{2} − m)j
− 2mn + (2j^{2} − 4jn − m)x − jx^{2})c_{n,j} = 0, and
j(j − m + 1)(j + n + 1)(j − n − x + 2)c_{n,j+2}
+ (2j^{4} + 4j^{3} − j^{2}m^{2} − 2j^{2}n^{2} − 2j^{2}nx + 2j^{2}n − j^{2}x^{2} + 2j^{2}x + 2j^{2} − jm^{2}
− 2jn^{2} − 2jnx + 2jn − jx^{2} + 2jx − mn^{2} − mnx + mn)c_{n,j+1}

+ (j + 1)(j + m)(j − n)(j + n + x − 1)c_{n,j} = 0.

For the purpose of guessing, one has to evaluate c_{n,j} for concrete values of n and j; in this particular example it suffices to consider 1 ≤ j ≤ n ≤ 15. For a concrete integer n, the vector (c_{n,j})_{1≤j≤n} can be determined by computing the kernel of the matrix (a_{i,j})_{1≤i<n,1≤j≤n}. Note that this kernel computation appears to be challenging, because the matrix entries are not rational functions in the parameters m and x. To overcome this problem, and also for efficiency reasons, we employ the evaluation-interpolation technique: when m and x are substituted by concrete integers, then the matrix entries turn into integers, and the recurrences for c_{n,j} can be easily guessed. Performing the same computation for different choices of m and x allows one to reconstruct recurrences with symbolic m and x, as given above. The shape of the above recurrences, i.e., their support and their leading coefficients, implies that the following three initial conditions are sufficient to define a unique bivariate sequence: c_{1,1} = 1, c_{1,2} = 0, c_{2,2} = 1. From now on, let c_{n,j} denote this unique bivariate sequence that is defined by unrolling the recurrences, starting from the given initial conditions. We want to prove that this conjectured definition of c_{n,j} agrees with its original definition (1). For this purpose it suffices to establish the two identities:

(2)  c_{n,n} = 1  (n ≥ 1),
(3)  ∑_{j=1}^{n} a_{i,j}c_{n,j} = 0  (1 ≤ i < n),

which can be achieved by holonomic closure properties and creative telescoping, respectively. We use the Mathematica package HolonomicFunctions [9]. The closure property “integer-linear substitution” implies that c_{n,n} satisfies a univariate recurrence of order at most 2. We compute this recurrence and show that it admits a constant solution. Then the two initial values c_{1,1} = c_{2,2} = 1 imply that c_{n,n} = 1 for all n ≥ 1. In order to prove (3), we split the sum into two parts:

s_{i,n} := ∑_{j=1}^{n} a_{i,j}c_{n,j} = ∑_{j=1}^{n} \binom{m + x}{m − i + j}c_{n,j} − ∑_{j=1}^{n} \binom{m + x}{m − i − j + 1}c_{n,j}.

Now creative telescoping can be employed to derive a set of recurrences for each of the sums on the right-hand side, which can be combined to a set of recurrences that is satisfied by the whole expression s_{i,n}. These computations are far from trivial (it took about two hours of CPU time) and the recurrences are too big to be displayed here. It can easily be checked that s_{i,n} = 0 for a few concrete small integers i and n, and the computed recurrences then imply that s_{i,n} = 0 for all 1 ≤ i < n. Finally, we investigate the sum

t_{n} := t_{n}(m, x) := ∑_{j=1}^{n} a_{n,j}c_{n,j} = ∑_{j=1}^{n} \binom{m + x}{m − n + j}c_{n,j} − ∑_{j=1}^{n} \binom{m + x}{m − n − j + 1}c_{n,j}.

Another two applications of creative telescoping exhibit that both sums on the right-hand side satisfy the same recurrence, which therefore is also valid for their difference t_{n}:

(m + n)(2n + x − 1)(2n + x + 1)(2n + x)^{2}(m − n − x)t_{n+1}

+ n(n + x)(m − 2n − x − 1)(m − 2n − x)(m + 2n + x − 1)(m + 2n + x)t_{n} = 0.

Since this is a first-order recurrence, a closed form for t_{n} is immediately obtained in the form of a hypergeometric expression:

t_{n} = \frac{Γ(n) Γ(n + x) Γ(2n + x − m) Γ(2n + x + m − 1)}{Γ(n + m) Γ(n + x − m) Γ(2n + x) Γ(2n + x − 1)}

From the definition (1) of c_{n,j} it follows that

t_{n} = \frac{det A_{n}}{det A_{n−1}} ,

which yields the desired closed form of our determinant:

det A_{n} = ∏_{i=1}^{n} t_{i} = ∏_{i=1}^{n} ∏_{j=1}^{m} \frac{t_{i}(j, x)}{t_{i}(j − 1, x)} = ∏_{i=1}^{n} ∏_{j=1}^{m} \frac{(x + i − j)(x + 2i + j − 2)}{(x + 2i − j)(i + j − 1)}. □

The special case x = m: we point out something very pleasant happening here. The entries a_{i,j} of the matrix T_{n,m}(m) are given by

a_{i,j} = \binom{2m}{m − i + j} − \binom{2m}{m − i − j + 1},

and its corresponding determinant

b_{n} := det T_{n,m}(m) = ∏_{1≤i≤j≤m−1} \frac{2n + i + j}{i + j} .

The key difference: this time, we are able to construct an (explicit) double-indexed sequence c by c_{n,n} = 1 and for 1 ≤ j < n,

c_{n,j} = \frac{(−1)^{n−j}\binom{2n−1}{n−j}\binom{n+m−j−1}{n−j}}{\binom{2n+m−2}{n−j}}.

Then, Zeilberger’s holonomic ansatz rules supreme once the following system of equations are proven to hold true:

\begin{cases}c_{n,n} = 1  n ≥ 1 \\ ∑_{j=1}^{n} a_{i,j} c_{n,j} = 0  1 ≤ i ≤ n − 1 \\ ∑_{j=1}^{n} a_{n,j} c_{n,j} = \frac{b_{n}}{b_{n−1}}  n ≥ 1. \end{cases}

But, these are directly justified with the help of Zeilberger’s algorithm [22]. We will not pursue this matter because we have already proven Conjecture 1 (see above), in its full generality.

5. T_{n,m}(x) finds its twin

Suppressing the parameters a and b, and defining u_{k} := \binom{a+b}{a−k}, the matrix T_{c,a}(b) is nothing but the difference of a Toeplitz and a Hankel matrix:

T_{c,a}(b) = [u_{i−j} − u_{i+j+1}].

By analogy, we may introduce its natural companion as a sum of a Toeplitz and a Hankel matrix:

T̃_{c,a}(b) = [u_{i−j} + u_{i+j+1}].

A truly parallel argument (as in the previous sections) produces a very similar determinantal evaluation:

Theorem 5.1. We have

det T̃_{c,a}(b) = ∏_{i=1}^{c} ∏_{j=1}^{a} \frac{(b + i − j)(b + 2i + j − 1)}{(b + 2i − j − 1)(i + j − 1)}.

To give some context, let f(z) be a complex L_{1}(C) function on the unit circle having Fourier coefficients

u_{k} = \frac{1}{2π} ∫_{0}^{2π} f(e^{iθ})e^{−ikθ}dθ, i = \sqrt{−1}.

Then, in the terminology of Widom [18], the determinant of a Toeplitz matrix with symbol f is given by

E_{n}(f) = det[u_{j−i}].

Toeplitz determinants find homes in many topics, such as statistical physics and random matrix theory. Different types of matrix ensembles lend themselves to classes of Toeplitz plus Hankel matrices of the form

det[u_{j−i} ± u_{j+i}],  det[u_{j−i} ± u_{j+i+1}],  etc.

From a more mathematical vantage point, such determinants arise in the decomposition of the determinants E_{n}(f). Namely, if the symbol f is an even function (in the sense f(z) = f(\frac{1}{z}) on the unit circle), then

E_{2n+1}(f) = \frac{1}{2} det[u_{j−i} − u_{j+i+2}]_{0}^{n−1} × det[u_{j−i} + u_{j+i}]_{0}^{n},
E_{2n}(f) = det[u_{j−i} − u_{j+i+1}]_{0}^{n−1} × det[u_{j−i} + u_{j+i}]_{0}^{n−1}.

Andrews and Stanton ([1], see p. 274) involved a somewhat analogous splitting up of a certain determinant that arose in plane partitions although they did not view them as such.

6. BONUS: CT identities and plane partitions

Borrowing notations from Section 5, the Toeplitz matrix [u_{i−j}] that appeared as a component in our matrix T_{c,a}(b) is in fact recognizable due to its relevance in the theory of plane partitions. Indeed, MacMahon [14] gave an elegant explicit formula for the number of plane partitions inside an a × b × c box

∏_{i=0}^{a−1} ∏_{j=0}^{b−1} ∏_{k=0}^{c−1} \frac{i + j + k + 2}{i + j + k + 1}.

One may readily associate a determinantal formulation [13] to this enumeration

det[u_{i−j}] = det [\binom{a + b}{a − i + j}]_{i,j=0}^{c−1} = ∏_{i=0}^{a−1} ∏_{j=0}^{b−1} ∏_{k=0}^{c−1} \frac{i + j + k + 2}{i + j + k + 1}.

We have found what appears to be new (for us, at least) matrix whose determinant matches the above product formula and enjoys the expected symmetry. Namely,

Lemma 6.1. It holds true that

det [\binom{i + j + a + b}{i + a}]_{i,j=0}^{c−1} = ∏_{i=0}^{a−1} ∏_{j=0}^{b−1} ∏_{k=0}^{c−1} \frac{i + j + k + 2}{i + j + k + 1}.

Proof. This is an immediate application of Dodgson’s Condensation [19]. □

Remark 2. The third-named author has previously employed the same approach towards furnishing a simple proof [20] of a determinant by MacMahon [14].

Proof. Here is a combinatorial proof for Lemma 6.1 given by C. Krattenthaler [12]. We thank him for allowing us to reproduce it.

Plane partitions in an a × b × c box are in bijection with families (P_{0}, P_{1}, …, P_{c−1}) of non-intersecting lattice paths, where P_{i} runs from (−i − a, i) to (−i, i + b), i = 0, 1, …, c − 1. This is explained in Section 3.3 of Bressoud’s book [2]. An example for such a family of paths for a = 6 and b = c = 3 is displayed in the figure below.

•  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•} • • •  • • •

By the Lindström–Gessel–Viennot theorem [5], this leads to the determinant

det_{0≤i,j≤c−1} (\binom{a + b}{a − i + j})

for the enumeration of both objects being counted here. We may prepend and append “forced” path pieces, see the thick path portions in the figure below.

•  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}  _{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•}  ^{•} • • •  • • •

In this manner, we obtain families (Q_{0}, Q_{1}, …, Q_{c−1}) of non-intersecting lattice paths, where Q_{i} runs from (−i − a, 0) to (0, i + b), i = 0, 1, …, c − 1. The corresponding Lindström–Gessel–Viennot determinant [5] is

det_{0≤i,j≤c−1} (\binom{i + j + a + b}{i + a}) .

The proof is complete. □

In the context of Selberg’s integrals, there are equivalently-stated constant term identities, and we prove (a special case of) one due to Morris [15] associated with the root system A_{n}.

Proposition 1. We have

CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{a} (1 + t_{i}^{−1})^{b} ∏_{i≠j}^{0,c−1} (1 − t_{j}t_{i}^{−1})^{m} = ∏_{ℓ=0}^{c−1} \frac{(a + b + ℓm)! ((ℓ + 1)m)!}{(a + ℓm)! (b + ℓm)! m!}.

Proof. We are only proving the special case m = 1. Start with the determinant from Lemma 6.1 where the method of constant term extraction is brought to bear:

det [\binom{i + j + a + b}{i + a}] = det [CT_{t_{i}} (t_{i}^{i+a} (1 + \frac{1}{t_{i}} )^{i+j+a+b})]

= det [CT_{t_{i}} ((1 + t_{i})^{i+a} (1 + \frac{1}{t_{i}} )^{b} (1 + \frac{1}{t_{i}} )^{j})]

= CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{i+a} (1 + \frac{1}{t_{i}} )^{b} · det [(1 + \frac{1}{t_{i}} )^{j}]

= CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{i+a} (1 + \frac{1}{t_{i}} )^{b} · ∏_{i<j}^{0,c−1}(t_{j}^{−1} − t_{i}^{−1})

= \frac{1}{c!} CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{a} (1 + \frac{1}{t_{i}} )^{b} · ∏_{i<j}^{0,c−1}(t_{j} − t_{i})(t_{j}^{−1} − t_{i}^{−1})

= \frac{1}{c!} CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{i+a} (1 + \frac{1}{t_{i}} )^{b} · ∏_{0≤i≠j≤c−1} (1 − \frac{t_{j}}{t_{i}} ) .

From Lemma 6.1, again, we gather that

CTt⃗ ∏_{i=0}^{c−1}(1 + t_{i})^{i+a} (1 + \frac{1}{t_{i}} )^{b} · ∏_{i≠j}^{0,c−1} (1 − \frac{t_{j}}{t_{i}} ) = c! · ∏_{i=0}^{a−1} ∏_{j=0}^{b−1} ∏_{k=0}^{c−1} \frac{i + j + k + 2}{i + j + k + 1}.

Further algebraic manipulation confirms the desired conclusion. □

Question. Is there a q-analogue of Conjecture 1?

References

[1] G. E. Andrews, D. W. Stanton. Determinants in Plane Partition Enumeration, Europ. J. Combinatorics (1998) 19, 273282.

[2] D. M. Bressoud. Proofs and confirmations - The story of the alternating sign matrix conjecture, Cambridge University Press, Cambridge, 1999.

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Department of Mathematics, Tulane University, New Orleans, LA 70118, USA

E-mail address: tamdeber@tulane.edu

Johann Radon Institute for Computational and Applied Mathematics, Austrian Academy of Sciences, Altenberger Strasse 69, 4040 Linz, Austria

E-mail address: christoph.koutschan@oeaw.ac.at

Department of Mathematics, Rutgers University (New Brunswick),, Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA

E-mail address: DoronZeil@gmail.com