ON THE NON-QUADRATICITY OF VALUES

OF THE q-EXPONENTIAL FUNCTION AND RELATED q-SERIES

CHRISTIAN KRATTENTHALER, IGOR ROCHEV, KEIJO VÄÄNÄNEN, AND WADIM ZUDILIN

To our great Peter Bundschuh on his 70th birthday

Abstract. We investigate arithmetic properties of values of the entire function

F (z) = F_{q}(z; λ) = ∑_{n=0}^{∞} \frac{z^{n}}{∏_{j=1}^{n}(q^{j} − λ)}, |q| > 1, λ ∉ q^{ℤ_{>0}},

that includes as special cases the Tschakaloff function (λ = 0) and the q-exp-onential function (λ = 1). In particular, we prove the non-quadraticity of the numbers F_{q}(α; λ) for integral q, rational λ and α ∉ −λq^{ℤ_{>0}}, α ≠ 0.

1. Introduction and main results

Consider the q-exponential function

(1.1)  E_{q}(z) = ∑_{n=0}^{∞} \frac{z^{n}}{∏_{j=1}^{n}(q^{j} − 1)},

which is an entire function in the complex z-plane for any q ∈ ℂ, |q| > 1. It is not difficult to adopt the classical proof of the irrationality of

e = ∑_{n=0}^{∞} \frac{1}{n!}

to the case of the number E_{q}(1) for an integer q > 1. Indeed, assuming, by contradiction, that E_{q}(1) = r/s for certain positive integers r and s, we see that the real

Date: August 12, 2008.

2000 Mathematics Subject Classification. Primary 11J72, 11J82; Secondary 11C20, 15A15, 33D15.

Key words and phrases. Irrationality, non-quadraticity, q-exponential function, Hankel determinant, cyclotomic polynomial.

The work of the first author was partially supported by the Austrian Science Foundation FWF, grant S9607-N13, in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory.” The work of the second author was supported by the Russian Foundation for Basic Research, grant no. 06-01-00518. The work of the fourth author was supported by a fellowship of the Max Planck Institute for Mathematics (Bonn). Part of this work was done during the first and fourth authors’ stay at the Erwin Schrödinger Institute for Physics and Mathematics, Vienna, during the programme “Combinatorics and Statistical Physics” in Spring 2008.

number

(1.2) r ∏_{j=1}^{k}(q^{j} − 1) − s ∑_{n=0}^{k} ∏_{j=n+1}^{k}(q^{j} − 1)

= s ∏_{j=1}^{k}(q^{j} − 1) · (E_{q}(1) − ∑_{n=0}^{k} \frac{1}{∏_{j=1}^{n}(q^{j} − 1)}) = s ∑_{n=k+1}^{∞} \frac{1}{∏_{j=k+1}^{n}(q^{j} − 1)}

is integral (according to the left-hand side representation) and positive (because of the right-hand side representation), hence it is at least 1, for any integer k ≥ 1. On the other hand,

s ∑_{n=k+1}^{∞} \frac{1}{∏_{j=k+1}^{n}(q^{j} − 1)} < \frac{s}{q^{k+1} − 1} ∑_{n=0}^{∞} \frac{1}{2^{n}}

= \frac{2s}{q^{k+1} − 1} → 0 as k → ∞,

leading to a contradiction.

The above proof is based on the simple observation that truncations of the series defining E_{q}(1) (see the intermediate term in (1.2)) provide rational approximations that are good enough to conclude the irrationality of the number in question. This argument has been generalized in various ways. For example, this truncation idea lies at the heart of Mahler’s method [13] of proving the algebraic independence of values of the series satisfying certain, quite restrictive, functional equations. In the same paper [13], K. Mahler posed a transcendence problem for values of the series that form a solution to more general functional equations. This problem remains unsolved until today, with the sole exception of values of quasi-modular functions [14]. In particular, only irrationality and linear independence results are known so far for values of the q-exponential function.

Recently, J.-P. Bézivin [1] proposed a new approach for the study of arithmetic properties of values of certain q-series. Among other things, he managed to prove the non-quadraticity of values of the so-called Tschakaloff function

(1.3)  T_{q}(z) = ∑_{n=0}^{∞} q^{−n(n+1)/2}z^{n}

at non-zero rational points if q = ρ/σ ∈ ℚ satisfies γ := log |ρ|/ log |σ| > 14. Furthermore, he proved the irrationality of these values if γ > 28/15 = 1.866 …, and thus extended considerably the possible values of q in the earlier irrationality results [18], [2]–[5], where γ > (3+\sqrt{5})/2 = 2.618 …. It is interesting that Bézivin’s approach was also an implicit generalization of the truncation idea. The method of [1] was applied to the q-exponential function by R. Choulet [6], who could not prove the non-quadraticity of its values, but improved the bound γ > 7/3 of the earlier irrationality result of Bundschuh [2] for E_{q}(z) to γ > 2. He also improved the above bound γ > 14 in Bézivin’s non-quadraticity result for T_{q}(z) to γ > 14/3 and the bound γ > 28/15 in the irrationality result to γ > 28/17.

ON THE NON-QUADRATICITY OF VALUES OF q-SERIES 3

The aim of this article is two-fold. First of all, we further generalize Bézivin’s method [1] to prove non-quadraticity results for values of the q-series

(1.4)  F (z) = F_{q}(z; λ) = ∑_{n=0}^{∞} \frac{z^{n}}{∏_{j=1}^{n}(q^{j} − λ)},  |q| > 1,

that include the Tschakaloff function and the q-exponential function as special cases (λ = 0 and λ = 1, respectively), and we further extend the values of q giving irrational values for F_{q}(z; λ). Secondly, in our proofs we use a more direct method than the p-adic approach used in [1] and [6]. This allows us to perceive the additional arithmetic information which can hardly be seen from the p-adic considerations.

We state our results in the following two theorems.

Theorem 1. Let q = ρ/σ ∈ ℚ with |q| > 1, and let α and λ satisfy α ≠ 0, λ ∉ q^{ℤ_{>0}} and α ∉ −λq^{ℤ_{>0}}. If

γ = \frac{log |ρ|}{log |σ|} > \begin{cases}\frac{126π^{2}}{47π^{2} − 72\sqrt{3} Im Li_{2}(e^{2π\sqrt{−1}/3})} = 3.27694460 … & if λ = 0, \\ \frac{27π^{2}}{5π^{2} − 18\sqrt{3} Im Li_{2}(e^{2π\sqrt{−1}/3})} = 9.43194241 … & if λ ≠ 0, \end{cases}

then α, λ, and μ = F_{q}(α; λ) in (1.4) cannot all belong to a quadratic extension of ℚ. In particular, if α and λ are rational then F_{q}(α; λ) is neither rational nor quadratic.

In the case that λ ≠ 0, the above result is entirely new, while its special case λ = 0 improves Choulet’s bound γ > 14/3 considerably.

The next theorem gives improvements for the above mentioned lower bounds of γ in the irrationality results.

Theorem 2. Under the hypotheses of Theorem 1, if

γ = \frac{log |ρ|}{log |σ|} > \begin{cases}\frac{252π^{2}}{173π^{2} − 72\sqrt{3} Im Li_{2}(e^{2π\sqrt{−1}/3})} = 1.53237645 … & if λ = 0, \\ \frac{27π^{2}}{16π^{2} − 9\sqrt{3} Im Li_{2}(e^{2π\sqrt{−1}/3})} = 1.80828115 … & if λ ≠ 0, \end{cases}

then α, λ, and μ = F_{q}(α; λ) in (1.4) cannot all be rational.

Since the function F_{q}(z; λ) satisfies the functional equation

F (qz) = (z + λ)F (z) + (1 − λ),

the irrationality of the values of F_{q}(z; λ) at non-zero rational points ∉ −λq^{ℤ_{>0}} follows from [17] if a rational number λ ∉ q^{ℤ_{>0}} and a rational number q satisfies γ > 7/3 for λ ∈ q^{ℤ_{≤0}} and γ > 2 + \sqrt{2} otherwise.

Sections 2–4 prepare for the proofs of these theorems. In Section 2, we review Bézivin’s construction, applied to our more general context. It involves in particular the introduction of a sequence (v_{n})_{n∈ℤ}, the Hankel determinant of which plays a fundamental role in the sequel. This determinant is a polynomial in q and two other variables. Propositions 1 and 2 in Section 3 address the power of q which appears in this Hankel determinant as a polynomial factor, while an asymptotic upper bound for the Hankel determinant is found in Proposition 3. Finally, Proposition 4 in Section 4 detects large amounts of cyclotomic factors (in q) in the Hankel determinant. All these ingredients are put together for the proofs of Theorems 1 and 2 in Section 5.

2. Review of Bézivin’s construction

The general idea of Bézivin’s method [1] refers to a function (2.1)

F (z) = ∑_{n=0}^{∞} a_{n}(q)z^{n}, a_{0}(q) = 1, \frac{a_{n−1}(q)}{a_{n}(q)} = b_{n}(q) = b(q^{n}) for n = 1, 2, …,

where b( ·) is a polynomial (in general, a rational function) over a number field. Let α ∈ ℂ. One takes the coefficients v_{n} appearing in

(2.2)  \frac{F (αz) − F (α)}{z − 1} = ∑_{n=0}^{∞} v_{n}a_{n}(q)z^{n}

and forms the Hankel determinant

(2.3)  V_{n} = det_{0≤i,j≤n−1}(v_{i+j}).

Then one has to provide an analytic upper bound for |V_{n}| and, under the assumption that both α and μ = F (α) belong to a certain algebraic number field K, an arithmetic lower bound, in order to find them contradictory; this shows that the assumption on α and μ cannot be true.

Before going into the details of the construction, note that relation (2.2) may be written in the form

∑_{n=0}^{∞} a_{n}(q)α^{n}z^{n} − μ = (z − 1) ∑_{n=0}^{∞} v_{n}a_{n}(q)z^{n} = −v_{0} + ∑_{n=1}^{∞}(v_{n−1}a_{n−1}(q) − v_{n}a_{n}(q))z^{n},

yielding

(2.4)  v_{0} = μ − 1 and v_{n} = v_{n−1}b_{n}(q) − α^{n} for n = 1, 2, ….

Hence, by induction, we easily arrive at the formula

(2.5)  v_{n} = μ ∏_{j=1}^{n} b_{j}(q) − ∑_{k=0}^{n} α^{k} ∏_{j=k+1}^{n} b_{j}(q).

Remark 1. Since we shall make use of it later on, we point out that Formula (2.5) also holds for negative n (that is, if we extend the sequence (v_{n}) to all integers n by letting the recurrence (2.4) hold for all integers n) under the conventions

∑_{k=m}^{n−1} Expr(k) = \begin{cases}∑_{k=m}^{n−1} Expr(k) & n > m, \\ 0 & n = m \\ − ∑_{k=n}^{m−1} Expr(k) & n < m, \end{cases}

and

∏_{k=m}^{n−1} Expr(k) = \begin{cases}∏_{k=m}^{n−1} Expr(k) & n > m, \\ 1 & n = m \\ 1/ ∏_{k=n}^{m−1} Expr(k) & n < m. \end{cases}

Assuming that b( ·) in (2.1) is a polynomial of degree s, Formula (2.5) shows that, for positive integers n, V_{n} is a polynomial in μ, α, and q of degree at most n in μ, n(n − 1) in α, and

(2.6)  s ∑_{i=0}^{n−1} \frac{2i(2i + 1)}{2} = \frac{sn(n − 1)(4n + 1)}{6}

in q (cf. [1, Lemma 2.4]). Formula (2.5) may also be written as

v_{n} = ∏_{j=1}^{n} b_{j}(q) · (μ − ∑_{k=0}^{n} α^{k} ∏_{j=1}^{k} \frac{1}{b_{j}(q)})

= a_{n}(q)^{−1} · (∑_{k=0}^{∞} a_{k}(q)α^{k} − ∑_{k=0}^{n} a_{k}(q)α^{k})

= a_{n}(q)^{−1} · ∑_{k=n+1}^{∞} a_{k}(q)α^{k} = ∑_{k=n+1}^{∞} \frac{α^{k}}{∏_{j=n+1}^{k} b_{j}(q)},

showing that the v_{n}’s are nothing else but tails of the series μ = ∑_{k=0}^{∞} a_{k}(q)α^{k} (normalized by the factors a_{n}(q)^{−1}; cf. the intermediate part of (1.2)). This fact somehow explains why the determinant in (2.3) is expected to be ‘small’.

Our basic example (1.4) corresponds to the choice b_{n}(q) = q^{n} − λ, for a fixed algebraic number λ. In this case, we have a_{n}(q) = ∏_{k=1}^{n}(q^{k} − λ)^{−1}, and the Hankel determinant V_{n} is also a polynomial in λ of degree at most n(n − 1). The choice b_{n}(q) = q^{n} (that is, λ = 0), yielding the Tschakaloff function (1.3), was the illustrative example of the method in [1], while the choice b_{n}(q) = q^{n} − 1 (when λ = 1) results in the q-exponential function (1.1). In [6], Choulet treated both the Tschakaloff and q-exponential cases.

We replace the argument of Bézivin and Choulet by a more direct approach (see Sections 3–5 below); in particular, we do not require the non-trivial p-adic techniques used in [1] and [6], thus making our proofs more ‘concrete’ and elementary. An essential gain, which allows us to succeed in proving the non-quadraticity of the values of (1.4), is due to extraction of cyclotomic factors in the factorization of the Hankel determinant (2.3); this is explained in Section 4.

3. Determinant calculus

Define the (q-)order of a Laurent series f(q) = ∑_{n∈ℤ} c_{n}q^{n} as

ord f(q) = ord_{q} f(q) = min{n : c_{n} ≠ 0}.

The q-binomial coefficient [ \begin{matrix}m \\ k \end{matrix} ]_{q} is defined by

[\begin{matrix}m \\ k \end{matrix}]_{q} = \begin{cases}\frac{(1 − q^{m})(1 − q^{m−1}) ⋯ (1 − q^{m−k+1})}{(1 − q^{k})(1 − q^{k−1}) ⋯ (1 − q)} & if k ≥ 0, \\ 0 & if k < 0. \end{cases}

Moreover, we adopt the usual notation for shifted q-factorials, given by (a; q)_{m} := (1 − a)(1 − aq) ⋯ (1 − aq^{m−1}) if m > 0, and (a; q)_{0} := 1.

Specializing b_{j}(q) = q^{j} − λ in (2.4), where λ ≠ q^{ℤ_{>0}}, we consider the sequence defined by

(3.1)  v_{0} = μ − 1,  v_{n} = (q^{n} − λ)v_{n−1} − α^{n},

where

(3.2)  μ = ∑_{n=0}^{∞} \frac{α^{n}}{∏_{k=1}^{n}(q^{k} − λ)}.

We follow Remark 1 in requiring the recursive relation to be valid for all n ∈ ℤ. This does, in fact, not work if q^{n} − λ = 0 for some integer n ≤ 0. However, since the only places where we take recourse on the extension of (3.1) to negative integers is in Remark 2 and in the proof of Proposition 2, in a context where λ = 0, we do not have to worry about these exceptional cases.

Let N denote the backward shift operator acting (solely) on the index of the sequence (v_{n})_{n∈ℤ}, that is N v_{n} = v_{n−1}. Introduce the difference operator

(3.3)  D_{l} = (−λN; q)_{l} (αN; q)_{l} = ∏_{k=0}^{l−1}(I + (λ − α)q^{k}N − λαq^{2k}N^{2}),

where I is the identity operator.

Lemma 1. For n ∈ ℤ and l ≥ 0 we have

(3.4)  D_{l}v_{n} = q^{l(n−l)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}v_{n−l−s}.

Proof. By the q-binomial theorem (cf. [9, Ex. 1.2(vi)])

(3.5)  (1 + z)(1 + qz) ⋯ (1 + q^{m−1}z) = ∑_{ℓ=0}^{m} q^{\binom{ℓ}{2}} [\begin{matrix}m \\ ℓ \end{matrix}]_{q} z^{ℓ},

we can write

D_{l} = ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}N^{k_{1}+k_{2}}.

ON THE NON-QUADRATICITY OF VALUES OF q-SERIES 7

Hence, what we want to prove is

(3.6) ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}}

= q^{l(n−l)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}v_{n−l−s}

for n ∈ ℤ and l ∈ ℕ_{0}. For l = 0, 1, this equality can be readily verified.

We now assume that (3.6) is valid for some l ≥ 1 and all n. Substituting n − 1 and n − 2 instead of n, we get

(3.7) ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}−1}

= q^{l(n−l−1)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}v_{n−l−s−1},

respectively

(3.8) ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}−2}

= q^{l(n−l−2)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}v_{n−l−s−2}.

Next we form the linear combination

(3.9)  (3.6) + (λ − α)q^{l} · (3.7) − λαq^{2l} · (3.8).

We claim that the left-hand side of (3.9) is equal to the left-hand side of (3.6) with l replaced by l +1. To see this, we rewrite the left-hand side of λq^{l} ·(3.7) in the form

(3.10) λq^{l} ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}−1}

= ∑_{k_{1}=0}^{l+1} ∑_{k_{2}=0}^{l+1} q^{l−k_{1}+1+\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} − 1 \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}},

we rewrite the left-hand side of −αq^{l} · (3.7) in the form

(3.11) − αq^{l} ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}−1}

= ∑_{k_{1}=0}^{l+1} ∑_{k_{2}=0}^{l+1} q^{l−k_{2}+1+\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} − 1 \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}},

and we rewrite the left-hand side of −λαq^{2l} · (3.8) in the form

(3.12) − λαq^{2l} ∑_{k_{1}=0}^{l} ∑_{k_{2}=0}^{l} q^{\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}−2}

= ∑_{k_{1}=0}^{l+1} ∑_{k_{2}=0}^{l+1} q^{(l−k_{1}+1)+(l−k_{2}+1)+\binom{k_{1}}{2}+\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{1} − 1 \end{matrix}]_{q} [\begin{matrix}l \\ k_{2} − 1 \end{matrix}]_{q} λ^{k_{1}}(−α)^{k_{2}}v_{n−k_{1}−k_{2}}.

By summing the left-hand side of (3.6) and the right-hand sides of (3.10), (3.11), and (3.12), we obtain indeed the left-hand side of (3.6) with l replaced by l + 1, after little simplification.

We now turn our attention to the right-hand side of (3.9), that is, to

q^{l(n−l)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}(v_{n−l−s} + (λ − α)v_{n−l−s−1} − λαv_{n−l−s−2}).

By (3.6) with l = 1 and n replaced by n − l − s, this is equal to

q^{l(n−l)} ∑_{s=0}^{l} [\begin{matrix}l \\ s \end{matrix}]_{q} q^{\binom{l−s+1}{2}}(−α)^{s}q^{n−l−s−1}(qv_{n−l−s−1} − αv_{n−l−s−2}).

It is not difficult to transform this into the right-hand side of (3.6) with l replaced by l + 1. □

As a corollary, we get for l ≥ 0 and n ≥ 2l − 1

(3.13)  ord_{q} D_{l}v_{n} = l(n − l).

Moreover, we have

q^{−l(n−l)}D_{l}v_{n}|_{q=0} = (−α)^{l}v_{n−2l}|_{q=0}.

Remark 2. In the proof of Proposition 2 below, under the hypothesis λ = 0, we require the estimate

ord_{q} D_{l}v_{n} > l(n − l) if n < 2l − 1,

which complements (3.13) and also follows from Lemma 1. This estimate is valid for negative indices n as well: recall that the definition (3.1) of our sequence (v_{n})_{n∈ℤ} in the case λ = 0 and α ≠ 0 results in

v_{−n−1} = q^{n}(v_{−n} + α^{−n}),

whence ord_{q} v_{−n} = n − 1 for n ≥ 1, implying the desired estimate for negative indices.

Proposition 1. Let λ ≠ 0, and let the sequence v_{n} be given by (3.1). Then the Hankel determinant V_{n} = det_{0≤i,j≤n−1}(v_{i+j}), viewed as an analytic function (in fact, a polynomial) in q, α, λ, and μ, admits the representation

V_{n} = \begin{cases}α^{n(n−1)/2}λ^{n(n−2)/4}(λ − (λ + α)μ)^{n/2} · q^{e_{0}(n)} + O(q^{e_{0}(n)+1}) & if n is even, \\ α^{n(n−1)/2}λ^{(n−1)^{2}/4}(μ − 1)(λ − (λ + α)μ)^{(n−1)/2} · q^{e_{0}(n)} + O(q^{e_{0}(n)+1}) \\ if n is odd, \end{cases}

where

(3.14)  e_{0}(n) = \frac{n(n − 1)(n − 2)}{6} = \binom{n}{3}.

In particular, its q-order under any specialization of α, λ, and μ is at least e_{0}(n).

Proof. We act on the i-th row of the matrix (v_{i+j})_{0≤i,j≤n−1} by the operator D_{⌊i/2⌋}; doing this for i = n − 1, n − 2, …, 1, 0 (in this order!). By definition (3.3) we have a sequence of elementary row operations, hence the new matrix with entries a_{ij} = D_{⌊i/2⌋}v_{i+j}, 0 ≤ i, j ≤ n − 1, has the same determinant V_{n}. According to (3.4),

e_{ij} := ord_{q} a_{ij} = ⌊\frac{i}{2}⌋ (i + j − ⌊\frac{i}{2}⌋) = ⌊\frac{i}{2}⌋ (⌈\frac{i}{2}⌉ + j) ,

and for a permutation τ of {0, 1, …, n − 1} we have

∑_{i=0}^{n−1} e_{i,τ(i)} = ∑_{i=0}^{n−1} ⌊\frac{i}{2}⌋ ⌈\frac{i}{2}⌉ + ∑_{i=0}^{n−1} ⌊\frac{i}{2}⌋ τ(i).

We claim that the minimal value of the latter expression is equal to n(n−1)(n−2)/6 and is attained, e.g., for τ(0) > τ(1) > ⋯ > τ(n − 1). To see this, first observe that, if ⌊i_{1}/2⌋ > ⌊i_{2}/2⌋ and j_{1} > j_{2}, then

⌊\frac{i_{1}}{2} ⌋ j_{1} + ⌊\frac{i_{2}}{2} ⌋ j_{2} > ⌊\frac{i_{1}}{2} ⌋ j_{2} + ⌊\frac{i_{2}}{2} ⌋ j_{1}.

Hence we necessarily have τ(0) > τ(i) and τ(1) > τ(i) for all i ≥ 2. In other words, the 2-element set {τ(0), τ(1)} is {n−1, n−2}. Continuing in this manner, we obtain {τ(2), τ(3)} = {n − 3, n − 4}, {τ(4), τ(5)} = {n − 5, n − 6}, and so on. It follows that indeed, for any permutation τ, we have

∑_{i=0}^{n−1} e_{i,τ(i)} ≥ ∑_{i=0}^{n−1} ⌊\frac{i}{2}⌋ ⌈\frac{i}{2}⌉ + ∑_{i=0}^{⌊n/2⌋−1} i(n − 1 − 2i + n − 2 − 2i) = \binom{n}{3}.

Moreover, the coefficient of the minimal power q^{n(n−1)(n−2)/6} is equal to the determinant of the ‘anti-diagonal’ matrix

(3.15)  (0  \begin{matrix}v_{n−2} & v_{n−1} \\ v_{n−1} & v_{n} \end{matrix} ; ··^{·} ; \begin{matrix}(−α)^{l}v_{n−2l−2} & (−α)^{l}v_{n−2l−1} \\ (−α)^{l}v_{n−2l−1} & (−α)^{l}v_{n−2l} \end{matrix} ; _{·}^{·}^{·}  0 )

evaluated at q = 0. Here, if n is odd, the left lower angle of the matrix contains just the 1 × 1-matrix (−α)^{(n−1)/2}v_{0}. Let us compute the determinant of a 2 × 2-box in (3.15) assuming q = 0 throughout. For n ≥ 0, the explicit expression (2.5) for v_{n} with b_{j}(q) = q^{j} − λ = −λ yields

v_{n} = μ(−λ)^{n} − (−λ)^{n} \frac{1 − (−\frac{α}{λ})^{n+1}}{1 + \frac{α}{λ}} .

Hence,

det (\begin{matrix}(−α)^{l}v_{n−2l−2} & (−α)^{l}v_{n−2l−1} \\ (−α)^{l}v_{n−2l−1} & (−α)^{l}v_{n−2l} \end{matrix}) = α^{n−1}(−λ)^{n−2l−1} (μ (1 + \frac{α}{λ}) − 1) .

Therefore the desired coefficient of q^{n(n−1)(n−2)/6} in V_{n} is equal to

∏_{l=0}^{n/2−1}(α^{n−1}(−λ)^{n−2l−2}(λ − (λ + α)μ))

if n is even, and it is equal to

(−α)^{(n−1)/2}(μ − 1) ∏_{l=0}^{(n−1)/2−1}(α^{n−1}(−λ)^{n−2l−2}(λ − (λ + α)μ))

if n is odd. □

Proposition 2. Let λ = 0, and let the sequence v_{n} be given by (3.1). Then the Hankel determinant V_{n} = det_{0≤i,j≤n−1}(v_{i+j}), viewed as an analytic function (in fact, a polynomial) in q, α, and μ, admits the representation

V_{n} = (−1)^{n(n+2)/8}α^{n(5n−2)/8}μ^{n/2} · q^{e_{0}(n)} + O(q^{e_{0}(n)+1})  if n is even, (3.16)
V_{n} = (−1)^{(n−1)(n−3)/8}α^{(n−1)(5n+1)/8}K_{(n−1)/2} · q^{e_{0}(n)} + O(q^{e_{0}(n)+1}) if n is odd, (3.17)

where the sequence K_{n} = K_{n}(α, μ) is defined in (3.20) below, and

(3.18)  e_{0}(n) = \begin{cases}\frac{n(n − 2)(5n − 2)}{24} & if n is even, \\ \frac{n(n − 1)(5n − 7)}{24} & if n is odd. \end{cases}

In particular, the q-order of V_{n} under any specialization of α and μ is at least e_{0}(n).

Proof. This time we act on the i-th row of the matrix (v_{i+j})_{0≤i,j≤n−1}, for i = n − 1, n − 2, …, 1, 0, by the operator D_{l_{i}}, where l_{i} = min{i, ⌊n/2⌋}. Again, these are elementary row transformations because D_{l} = (αN; q)_{l} in (3.3) for λ = 0. For the entries a_{ij} = D_{l_{i}}v_{i+j} of the resulting matrix, whose determinant is V_{n}, we have e_{ij} := ord_{q} a_{ij} ≥ l_{i}(i + j − l_{i}), with equality occurring when i + j ≥ 2l_{i} − 1 (cf. Remark 2). This fact and the fact that the sequence (l_{i}) is non-decreasing imply, for any permutation τ of {0, 1, …, n − 1}, that

(3.19)  ∑_{i=0}^{n−1} e_{i,τ(i)} ≥ ∑_{i=0}^{n−1} l_{i}(i + τ(i) − l_{i}) ≥ ∑_{i=0}^{n−1} l_{i}(n − 1 − l_{i}) = e_{0}(n),

for e_{0}(n) defined in (3.18). To get equality in (3.19), the following two conditions should be satisfied: (a) for each i we have i + τ(i) ≥ 2l_{i} − 1, implying τ(i) ≥ 2 ⌊n/2⌋ − i − 1 for i ≥ ⌊n/2⌋, and (b) for i < ⌊n/2⌋ we have τ(i) = n − i − 1.

In the case of even n, condition (a) gives us τ(i) ≥ n − i − 1 for i ≥ n/2, which in view of condition (b) is possible if and only if τ(i) = n − i − 1 for each i = 0, 1, …, n − 1. Therefore, the unique anti-diagonal product (−1)^{n(n−1)/2} ∏_{i=0}^{n−1} a_{i,n−1−i} provides the lowest power q^{n(n−2)(5n−2)/24} in the determinant det(a_{ij})_{0≤i,j≤n−1}, implying (3.16).

If n is odd, conditions (a) and (b) take the form

\begin{aligned}τ(i) & = n − i − 1 & for i = 0, 1, …, \frac{n−3}{2}, \\ τ(i) & ≥ n − i − 2 & for i = \frac{n−1}{2}, …, n − 1. \end{aligned}

In this case the coefficient of the lowest power q^{n(n−1)(5n−7)/24} in V_{n} is equal to the determinant of the matrix

( \begin{matrix}0 & \begin{aligned}& (−α)^{0}v_{n−1} \\ & (−α)^{1}v_{n−3} \\ · & · & ^{·} \\ (−α)^{(n−3)/2}v_{2} \end{aligned} \\ \begin{aligned}0 & & (−α)^{⌊n/2⌋}v_{−1} & (−α)^{⌊n/2⌋}v_{0} \\ & · ·^{·} & (−α)^{⌊n/2⌋}v_{0} & (−α)^{⌊n/2⌋}v_{1} \\ (−α)^{⌊n/2⌋}v_{−1} & · ·^{·} & & ⋮ \\ (−α)^{⌊n/2⌋}v_{0} & … & … & (−α)^{⌊n/2⌋}v_{⌊n/2⌋} \end{aligned} & 0 \end{matrix} )

evaluated at q = 0. It is clear that the non-vanishing of the coefficient will follow from the non-vanishing of the determinant

det (\begin{matrix}0 & & v_{−1} & v_{0} \\ & ··· & v_{0} & v_{1} \\ v_{−1} & ··· & & ⋮ \\ v_{0} & … & … & v_{(n−1)/2} \end{matrix})|_{q=0} = det (\begin{matrix}0 & & μ & μ − 1 \\ & ··· & μ − 1 & −α \\ μ & ··· & & ⋮ \\ μ − 1 & … & … & −α^{(n−1)/2} \end{matrix})

= (−1)^{(n−1)(n−3)/8}K_{(n−1)/2},

where

(3.20)  K_{n} = K_{n}(α, μ) = det (\begin{matrix}μ − 1 & −α & −α^{2} & … & −α^{n} \\ μ & μ − 1 & −α & … & −α^{n−1} \\ & μ & μ − 1 & … & −α^{n−2} \\ & & ... & ... & ⋮ \\ 0 & & & μ & μ − 1 \end{matrix}) .

Summarizing we have (3.17), and the proposition follows. □

Remark 3. Generically, the power e_{0}(n) in (3.18) is exact. Indeed, the determinant K_{n} is not identically zero, since it is trivially non-zero for α = 0 and μ ≠ 1. In fact, we can also write down explicit formulas for K_{n}, since the sequence satisfies the linear recurrence

\begin{matrix}K_{n+2} = (μ − 1 − αμ)K_{n+1} + αμ^{2}K_{n} for n = 0, 1, 2, …, \\ K_{0} = μ − 1, K_{1} = (μ − 1)^{2} + αμ. \end{matrix}

To find an (asymptotic) upper bound for our Hankel determinant

det_{0≤i,j≤n−1}(v_{i+j}),

where

(3.21)  v_{n} = ∑_{k=n+1}^{∞} \frac{α^{k}}{∏_{j=n+1}^{k}(q^{j} − λ)} for n = 0, 1, 2, …,

we will use the difference operator

(3.22)  D̃_{l} = (αq^{−1}N; q^{−1})_{l} = ∏_{k=1}^{l}(I − αq^{−k}N).

Remark 4. The operators (3.3) and (3.22) are directly related by

D_{l}v_{n} = q^{ln−\binom{l}{2}}D̃_{l}v_{n−l},

where l ≥ 0 and n ∈ ℤ. This is seen by applying the q-binomial theorem (3.5) to (3.22), and by comparing the result with (3.4). Equivalently,

D̃_{l}v_{n} = q^{\binom{l}{2}−l(n+l)}D_{l}v_{n+l}.

Lemma 2. Let q, α, λ be complex numbers with |q| > 1, α ≠ 0, and λ not a (positive) power of q. Then, for 0 ≤ l ≤ n, we have

|D̃_{l}v_{n}| ≤ \begin{cases}|q|^{\binom{l}{2}−nl}C_{1}^{n+1} & if λ ≠ 0, \\ |q|^{−nl}C_{2}^{n+1} & if λ = 0, \end{cases}

where C_{1} and C_{2} are positive real numbers not depending on n and l.

Proof. Let first λ ≠ 0. From (3.21), it is easy to see that we can find a real number C > 1 (depending on q, α and λ, but not on n), such that |v_{n}| ≤ C^{n+1}. Making use of the q-binomial theorem (3.5), of the fact that

|[\begin{matrix}m \\ k \end{matrix}]_{q}| ≤ [\begin{matrix}m \\ k \end{matrix}]_{|q|}

(following from the polynomiality in q of [ \begin{matrix}m \\ k \end{matrix}]_{q}, the coefficients being non-negative integers), and of our observation in Remark 4, we have

|D̃_{l}v_{n}| = |q|^{\binom{l}{2}−l(n+l)}|D_{l}v_{n+l}|

≤ |q|^{−\binom{l+1}{2}−ln} ∑_{k_{1}=0}^{l} |q|^{\binom{k_{1}}{2}} [\begin{matrix}l \\ k_{1} \end{matrix}]_{|q|} |λ|^{k_{1}} ∑_{k_{2}=0}^{l} |q|^{\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{2} \end{matrix}]_{|q|} |α|^{k_{2}}|v_{n+l−k_{1}−k_{2}}|

≤ |q|^{−\binom{l+1}{2}−ln}C^{n+l+1} ∏_{j=0}^{l−1}(1 + |λ| |q|^{j})(1 + |α| |q|^{j})

≤ |q|^{\binom{l}{2}−l−ln}C^{n+l+1}(1 + |λ|)^{l}(1 + |α|)^{l},

which, in view of l ≤ n, is exactly in line with the first assertion of the lemma.

In the case λ = 0, we can proceed in the same way. The only difference is that the above sum over k_{1} reduces to just the summand for k_{1} = 0. Hence, we obtain

|D̃_{l}v_{n}| ≤ |q|^{−\binom{l+1}{2}−ln} ∑_{k_{2}=0}^{l} |q|^{\binom{k_{2}}{2}} [\begin{matrix}l \\ k_{2} \end{matrix}]_{|q|} |α|^{k_{2}}|v_{n+l−k_{2}}|

≤ |q|^{−\binom{l+1}{2}−ln}C^{n+l+1} ∏_{j=0}^{l−1}(1 + |α| |q|^{j})

≤ |q|^{−l−ln}C^{n+l+1}(1 + |α|)^{l},

which is exactly in line with the second assertion of the lemma. □

Proposition 3. Let the sequence v_{n} be given by (3.21). Then, as n tends to ∞, the Hankel determinant V_{n} = det_{0≤i,j≤n−1}(v_{i+j}) is asymptotically

(3.23)  |V_{n}| ≤ \begin{cases}|q|^{−n^{3}/3} exp(O(n^{2})) & if λ ≠ 0, \\ |q|^{−n^{3}/2} exp(O(n^{2})) & if λ = 0. \end{cases}

Proof. Acting by the operator D̃_{i} on the i-th row of the matrix (v_{i+j})_{0≤i,j≤n−1} (this results in elementary row operations according to (3.22)) we get the matrix (a_{ij})_{0≤i,j≤n−1} with entries a_{ij} = D̃_{i}v_{i+j}, whose determinant is equal to V_{n}.

Let now λ ≠ 0. Writing 𝔖_{n} for the symmetric group on {0, 1, 2, …, n − 1}, we have

|V_{n}| ≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |D̃_{i}v_{i+τ(i)}|

≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |q|^{\binom{i}{2}−(i+τ(i))i}C_{1}^{i+τ(i)+1}

≤ exp(O(n^{2})) ∏_{i=0}^{n−1} |q|^{\binom{i}{2}−(n−1)i},

where, to go from the next-to-last to the last line, we used again the fact that the permutation achieving the maximum is the permutation sending i to n − i − 1, i = 0, 1, …, n − 1. This implies the first claim of the proposition.

On the other hand, if λ = 0, then we have

|V_{n}| ≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |D̃_{i}v_{i+τ(i)}|

≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |q|^{−(i+τ(i))i}C_{2}^{i+τ(i)+1}

≤ exp(O(n^{2})) ∏_{i=0}^{n−1} |q|^{−(n−1)i},

implying the second claim of the proposition. □

Remark 5. Using an analytic method for the (entire) generating series ∑_{n=0}^{∞} v_{n}z^{n}, Choulet [6, Lemmas 3.3 and 3.4] proves estimates that may be informally summarized in our settings as follows:

(3.24)  ord_{q} V_{n} ≥ \begin{cases}\frac{5}{24}n^{3} + O(n^{2}) & if λ = 0, \\ \frac{1}{6}n^{3} + O(n^{2}) & if λ ≠ 0, \end{cases}  as n → ∞,

and

(3.25)  |V_{n}| ≤ \begin{cases}|q|^{− \frac{1}{2} n^{3}+O(n^{2})} & if λ = 0, \\ |q|^{− \frac{1}{3} n^{3}+O(n^{2})} & if λ ≠ 0, \end{cases}  as n → ∞.

Therefore, our Propositions 1–2 sharpen the estimate (3.24) of Choulet, while our Proposition 3 provides a different proof of (3.25). (Strictly speaking, Choulet did not arrive at the better estimate in (3.25) for the case λ = 0 himself; this had been done earlier by Bézivin [1] using elementary considerations.) On the other hand, it is Choulet’s method that suggested to us the form of the difference operators (3.3) and (3.22).

4. Cyclotomic factorization

We now turn to the general Hankel determinant (2.3) with the sequence (v_{n}) defined in (2.4) and (2.5). Let us fix the notation

Φ_{l}(q) = _{∏_{1≤j≤l} ; gcd(j,l)=1}(q − e^{2πj\sqrt{−1}/l}), l = 1, 2, …,

for the cyclotomic polynomials.

Proposition 4. For any integer l in the range 1 ≤ l < n/2, the Hankel determinant V_{n} = det_{0≤i,j≤n−1}(v_{i+j}), where the v_{n}’s are given in (2.5) with b_{j}(q) = b(q^{j}) for a polynomial b( ·) of arbitrary degree, is divisible by Φ_{l}(q)^{e_{l}(n)}, where

(4.1)  e_{l}(n) = ∑_{i=0}^{n−1} (⌊\frac{i + l}{3l}⌋ + ⌊\frac{i}{3l}⌋) .

Remark 6. From (4.1), it is straightforward to compute a compact formula for e_{l}(n), namely,

e_{l}(n) = \frac{(n − j)(n + j − 2l)}{3l} + \begin{cases}0 & if 0 ≤ j < 2l, \\ j − 2l & if 2l ≤ j < 3l, \end{cases} where n ≡ j (mod 3l).

(In other words, j/(3l) is the fractional part of n/(3l).) In particular,

e_{1}(n) = ⌊\frac{(n − 1)^{2}}{3} ⌋ = \begin{cases}\frac{(n − 1)^{2}}{3} & if n ≡ 1 (mod 3), \\ \frac{n(n − 2)}{3} & otherwise, \end{cases} e_{2}(n) = ⌊\frac{(n − 2)^{2}}{6} ⌋.

Proof. We have to do some preparatory work first, before we are in the position to embark on the “actual” proof of the divisibility assertion in the proposition. The central part of this preparatory work is the identity (4.10).

Changing notation slightly, recall that, for n ≥ 1, we have

v_{n} = b(q^{n})v_{n−1} − α^{n},

with b(x) = x − λ. (In fact, the subsequent arguments hold for any function b(x). It is therefore that we shall write b( .) in the sequel instead of its explicit form which is of relevance in our context.) Hence, for n ≥ l ≥ 1, we have

(4.2)  v_{n} = v_{n−l} ∏_{k=0}^{l−1} b(q^{n−k}) − ∑_{j=0}^{l−1} α^{n−j} ∏_{k=0}^{j−1} b(q^{n−k}).

Writing

P_{j}(n, q) = ∏_{k=0}^{j−1} b(q^{n−k})

for non-negative integers j, the recurrence (4.2) takes the form

(4.3)  v_{n} = v_{n−l}P_{l}(n, q) − ∑_{j=0}^{l−1} α^{n−j}P_{j}(n, q).

Fix a positive integer l and a primitive l-th root of unity ζ. For integers j, t, n with j, t ≥ 0, set

(4.4)  P_{j}^{(t)}(n) = \frac{d^{t}}{dq^{t}} P_{j}(n, q)|_{q=ζ}.

In particular, we have

(4.5)  P_{0}^{(t)}(n) = \begin{cases}1 & if t = 0, \\ 0 & if t ≥ 1. \end{cases}

For j ≥ 1, we have

(4.6)  \frac{d^{t}}{dq^{t}} P_{j}(n, q) = ∑_{t_{0}+···+t_{j−1}=t} \frac{t!}{t_{0}! ⋯ t_{j−1}!} ∏_{k=0}^{j−1} \frac{d^{t_{k}}}{dq^{t_{k}}} b(q^{n−k}).

Applying the Faà di Bruno formula (cf. [7, Sec. 3.4]; but see also [8, 11]) we get (4.7)

\frac{d^{t}}{dq^{t}} b(q^{n−k}) = ∑_{m_{1}+2m_{2}+···+tm_{t}=t} \frac{t!}{m_{1}! ⋯ m_{t}!}b^{(m)}(q^{n−k}) ∏_{ν=1}^{t} (\binom{n − k}{ν}q^{n−k−ν})^{m_{ν}} ,

where m = m_{1} + ⋯ + m_{k}.

It is straightforward to see that Equations (4.4)–(4.7) imply that, for any nonnegative integers j and t, the quantity P_{j}^{(t)}(n), as a function in n, is an l-quasi-polynomial of degree at most t, where an l-quasi-polynomial of degree at most t is a sequence of complex numbers (Q(n))_{n∈ℤ} of the form

Q(n) = ∑_{ν=0}^{t} a_{ν}(n)n^{ν},

the sequence {a_{ν}(n)}_{n∈ℤ} being l-periodic for each ν (cf. [16, Sec. 4.4]). We denote the set of all l-quasi-polynomials of degree at most t by Q_{t} = Q_{t}(l). Furthermore note that if Q(n) ∈ Q_{t}, with t > 0, then Q(n) − Q(n − l) ∈ Q_{t−1}. These facts will be used repeatedly.

Our next observation is that the sequence (P_{l}^{(0)}(n))_{n∈ℤ} is constant, where P_{l}^{(0)}(n) = P_{l}^{(0)}(0) = ∏_{k=0}^{l−1} b(ζ^{k}). We let

B = B_{l} = ∏_{k=0}^{l−1} b(ζ^{k}).

Clearly, B is independent of the particular choice of the primitive l-th root of unity ζ.

Now we introduce the difference operators

F = I − BN^{l}

and

G = I − α^{l}N^{l},

where I and N have the same meaning as earlier. Clearly, the operators F and G commute.

It is easy to check, using elementary facts of difference calculus that, for any Q(n) ∈ Q_{t}, we have

(4.8)  G^{t+1}(Q(n)α^{n}) = 0.

For non-negative integers m and n, let

v_{n}^{(m)} = \frac{∂^{m}v_{n}}{∂q^{m}} |_{q=ζ}.

By differentiating both sides of (4.3) m times, and by subsequently substituting q = ζ, we obtain for n ≥ l ≥ 1 the equation

(4.9)  v_{n}^{(m)} = ∑_{ν=0}^{m} \binom{m}{ν}P_{l}^{(m−ν)}(n)v_{n−l}^{(ν)} − ∑_{j=0}^{l−1} P_{j}^{(m)}(n)α^{n−j}.

We now claim that for an arbitrary Q(n) ∈ Q_{t}, for non-negative integers t and m, and for any integer n ≥ (2t + 3m + 2)l, we have

(4.10)  F^{t+2m+1}G^{t+m+1}(Q(n)v_{n}^{(m)}) = 0.

We prove this claim by a double induction: the external induction is over m, while the inner induction is over t.

We start by proving (4.10) for m = 0, by doing an induction over t. We put m = 0 in (4.9), and rewrite the resulting equation in the form

F(v_{n}^{(0)}) = − ∑_{j=0}^{l−1} P_{j}^{(0)}(n)α^{n−j}.

We apply the operator G on both sides. By (4.8), this implies

FG(v_{n}^{(0)}) = 0,

which in turn implies

FG(Q(n)v_{n}^{(0)}) = 0

for any l-periodic function Q(n), since the operators F and G are both polynomials in N^{l}. This is exactly (4.10) for m = t = 0.

Now let us assume that (4.10) is proved for m = 0 and for t − 1 instead of t. If Q(n) ∈ Q_{t} with t > 0, then (4.9) implies

F(Q(n)v_{n}^{(0)}) = B(Q(n) − Q(n − l))v_{n−l}^{(0)} − ∑_{j=0}^{l−1} Q(n)P_{j}^{(0)}(n)α^{n−j}.

After application of F^{t}G^{t+1} on both sides, we obtain

F^{t+1}G^{t+1}(Q(n)v_{n}^{(0)}) = BGF^{t}G^{t}((Q(n) − Q(n − l))v_{n−l}^{(0)})

− ∑_{j=0}^{l−1} F^{t}G^{t+1}(Q(n)P_{j}^{(0)}(n)α^{n−j}).

The summands in the sum over j vanish because of (4.8), while the first expression on the right-hand side vanishes because of the induction hypothesis. (Recall that Q(n) − Q(n − l) ∈ Q_{t−1}.) This proves (4.10) for m = 0 and arbitrary t.

Now we assume that (4.10) is proved for 0, 1, …, m−1 instead of m and arbitrary t. For m > 0, we write (4.9) as

Fv_{n}^{(m)} = ∑_{ν=0}^{m−1} \binom{m}{ν}P_{l}^{(m−ν)}(n)v_{n−l}^{(ν)} − ∑_{j=0}^{l−1} P_{j}^{(m)}(n)α^{n−j}.

We apply F^{2m}G^{m+1} on both sides, to obtain

F^{2m+1}G^{m+1}(v_{n}^{(m)}) = ∑_{ν=0}^{m−1} \binom{m}{ν}F^{m−ν−1}F^{(m−ν)+2ν+1}G^{(m−ν)+ν+1}(P_{l}^{(m−ν)}(n)v_{n−l}^{(ν)})

− ∑_{j=0}^{l−1} F^{2m}G^{m+1}(P_{j}^{(m)}(n)α^{n−j}).

Again, the summands in the sum over j vanish because of (4.8), while the first expression on the right-hand side vanishes because of the induction hypothesis. This establishes (4.10) for m and t = 0.

In order to prove (4.10) for m and arbitrary t, we do again an induction over t. We already know that (4.10) is true for t = 0. Let us assume that (4.10) is true for t − 1 instead of t. We multiply both sides of (4.9) by Q(n), and we apply F on both sides. The resulting equation can then be written in the form

F(Q(n)v_{n}^{(m)}) = B(Q(n) − Q(n − l))v_{n−l}^{(m)} + ∑_{ν=0}^{m−1} \binom{m}{ν}Q(n)P_{l}^{(m−ν)}(n)v_{n−l}^{(ν)}

− ∑_{j=0}^{l−1} Q(n)P_{j}^{(m)}(n)α^{n−j}.

After application of F^{t+2m}G^{t+m+1} on both sides, we arrive at

F^{t+2m+1}G^{t+m+1}(Q(n)v_{n}^{(m)}) = BGF^{t+2m}G^{t+m}((Q(n) − Q(n − l))v_{n−l}^{(m)})

+ ∑_{ν=0}^{m−1} \binom{m}{ν}F^{m−ν−1}F^{(t+m−ν)+2ν+1}G^{(t+m−ν)+ν+1}(Q(n)P_{l}^{(m−ν)}(n)v_{n−l}^{(ν)})

− ∑_{j=0}^{l−1} F^{t+2m}G^{t+m+1}(Q(n)P_{j}^{(m)}(n)α^{n−j}).

Again, by (4.8) and the induction hypothesis, the right-hand side in this identity vanishes. Thus, (4.10) is completely proved.

We are now ready to treat the Hankel determinant V_{n}. In fact, we need (4.10) only for t = 0. (For the proof, it was however necessary to play with t.) What (4.10) for t = 0 says is that, for m ≥ 1 and n ≥ (3m − 1)l, the polynomial (in q) w_{m,n} = (I − BN^{l})^{2m−1}(I − α^{l}N^{l})^{m}v_{n} satisfies

\frac{∂^{j}w_{m,n}}{∂q^{j}} |_{q=ζ} = 0, 0 ≤ j ≤ m − 1,

and, for any choice of the primitive l-th root of unity ζ. Hence, we have

(4.11)  Φ_{l}(q)^{m} | w_{m,n}.

This reasoning also shows that for the polynomial

w̃_{m,n} = (I − BN^{l})^{2m}(I − α^{l}N^{l})^{m}v_{n} = (I − BN^{l})w_{m,n} = w_{m,n} − Bw_{m,n−l}

we have

(4.12)  Φ_{l}(q)^{m} | w̃_{m,n},

as long as n ≥ 3ml.

For n − 1 ≥ i ≥ 2l we apply the operator (I − BN^{l})^{2l_{i}−1}(I − α^{l}N^{l})^{l_{i}}, where l_{i} = ⌊(i + l)/(3l)⌋ to the i-th row of the Hankel matrix (v_{i+j})_{0≤i,j≤n−1}, and subsequently, for n − 1 ≥ j ≥ 3l, we apply the operator (I − BN^{l})^{2m_{j}}(I − α^{l}N^{l})^{m_{j}}, where m_{j} = ⌊j/(3l)⌋ to the j-th column. The resulting matrix has entries

\begin{cases}w̃_{ε(i,j),i+j} & if i < 2l and j ≥ 3l, \\ w_{ε(i,j),i+j} & otherwise, \end{cases}

where

ε(i, j) = l_{i} + m_{j} = ⌊(i + l)/(3l)⌋ + ⌊j/(3l)⌋,

with the convention that w_{0,n} = v_{n}. As earlier, since the above operations correspond to row and column operations, the determinant of the resulting matrix is still equal to V_{n}.

In view of (4.11) and (4.12), it remains to observe that, for an arbitrary permutation τ,

∑_{i=0}^{n−1} ε(i, τ(i)) = ∑_{i=0}^{n−1} (⌊\frac{i + l}{3l}⌋ + ⌊\frac{τ(i)}{3l}⌋) = ∑_{i=0}^{n−1} (⌊\frac{i + l}{3l}⌋ + ⌊\frac{i}{3l}⌋) = e_{l}(n),

This completes the proof of the proposition. □

Propositions 1, 2 and 4 may be summarized as follows: The Hankel determinant (2.3) of the sequence (3.1) admits the factorization

(4.13)  V_{n} = Δ_{n}(q) · Ṽ_{n},

where

(4.14)  Δ_{n}(q) = q^{e_{0}(n)} ∏_{l≥1} Φ_{l}(q)^{e_{l}(n)}

and the exponents e_{0}(n), e_{1}(n), e_{2}(n), … are given by very simple formulas. Since

\frac{log |Φ_{l}(q)|}{log |q|} = φ(l) + O(1) as l → ∞

and

e_{l}(n) = O(\frac{n^{2}}{l} ) as n → ∞ uniformly in l ≥ 1,

the asymptotic behaviour of Δ_{n}(q) as n → ∞ is governed by the degree of the polynomial,

(4.15)  \frac{log |Δ_{n}(q)|}{log |q|} ∼ deg Δ_{n} = e_{0}(n) + ∑_{l=1}^{∞} e_{l}(n)φ(l) as n → ∞.

The following lemma enables us to determine the asymptotic behaviour of the sum on the right-hand side of (4.15) as n → ∞.

20 C. KRATTENTHALER, I. ROCHEV, K. VÄÄNÄNEN, AND W. ZUDILIN

Lemma 3. Let a and c be real numbers with 0 ≤ c < a. Then, as n → ∞,

∑_{l≥1} φ(l) ∑_{i=0}^{n} ⌊\frac{i + cl}{al}⌋ = \frac{n^{3}}{π^{2}} ∑_{m=1}^{∞} \frac{1}{(am − c)^{2}} + O(n^{2} log^{2} n).

Proof. By interchanging summations, we have

(4.16) ∑_{l=1}^{∞} φ(l) ∑_{i=0}^{n} ⌊\frac{i + cl}{al}⌋ = ∑_{i=0}^{n} ∑_{l=1}^{∞} ⌊\frac{i + cl}{al}⌋ φ(l).

Writing Σ(x) = ∑_{l≤x} φ(l), the expression (4.16) can be rewritten in the form

∑_{i=0}^{n} ∑_{m=1}^{∞} m (Σ (\frac{i}{am − c}) − Σ (\frac{i}{a(m + 1) − c})) = ∑_{i=0}^{n} ∑_{m=1}^{∞} Σ (\frac{i}{am − c}) .

Using Mertens’ classical asymptotic formula (cf. [10, p. 268, Theorem 330])

Σ(x) = \frac{3x^{2}}{π^{2}} + O(x log x) as x → ∞,

we obtain the following asymptotic estimate for (4.16):

∑_{i=⌈a−c⌉}^{n} ∑_{m=1}^{⌊(i+c)/a⌋} (\frac{3i^{2}}{π^{2}(am − c)^{2}} + O (\frac{i log n}{m})) + O(1)

= ∑_{i=⌈a−c⌉}^{n} (\frac{3i^{2}}{π^{2}} ∑_{m=1}^{∞} \frac{1}{(am − c)^{2}} + O(i log^{2} n)) + O(1),

which immediately implies the assertion of the lemma. □

Using Formula (4.1) for e_{l}(n) and the asymptotics from Lemma 3, we obtain

(4.17) ∑_{l=1}^{∞} e_{l}(n)φ(l) ≈ 0.05301135n^{3} as n → ∞,

where the exact value of the constant in (4.17) is

\frac{1}{54} + \frac{1}{π^{2}} ∑_{m=1}^{∞} \frac{1}{(3m − 1)^{2}} = \frac{5}{54} − \frac{Im Li_{2}(e^{2π\sqrt{−1}/3})}{π^{2}\sqrt{3}}.

Summarizing our findings from Propositions 1 and 2, as well as from (4.15) and (4.17), we have the following result for the asymptotic degree of Δ_{n}(q). Proposition 5. Let Δ_{n}(q) be defined as in (4.14). Then, as n → ∞, we have

\frac{log |Δ_{n}(q)|}{log |q|} ∼ deg_{q} Δ_{n}(q) ∼ Bn^{3},

where

B = \frac{5}{54} − \frac{Im Li_{2}(e^{2π\sqrt{−1}/3})}{π^{2}\sqrt{3}} + \begin{cases}\frac{5}{24} & if λ = 0, \\ \frac{1}{6} & if λ ≠ 0. \end{cases}

Remark 7. If λ is a root of unity, expected formulas for the exponents of the cyclotomic factors of V_{n} obey a different law. To write them down in the (q-exponential) case λ = 1, we represent the polynomial Δ_{n} from the factorization (4.13) in the form

Δ_{n} = q^{e_{0}(n)} ∏_{l≥1}(q^{l} − 1)^{ẽ_{l}(n)}.

Then

(4.18)  ẽ_{l}(n) = 2 max{0, n − 2l},

which, together with (3.14), implies that, in the case λ = 1, we have

deg Δ_{n} = \begin{cases}\frac{n(n − 1)^{2}}{4} & if n is odd, \\ \frac{n^{2}(n − 2)}{4} & if n is even. \end{cases}

It is of definite interest to prove also these formulas for the cyclotomic exponents.

5. Arithmetic ingredients

In this section we provide the proof of Theorems 1 and 2. It rests on Propositions 3–5, and two additional auxiliary results, given in Lemmas 4 and 5 below. The first one says that, under certain arithmetic constraints on the complex parameters α, λ and q, if we generalize our sequence v_{n} to v_{n}(x), where v_{0}(x) = x − 1 and

(5.1)  v_{n}(x) = v_{n−1}(x) · (q^{n} − λ) − α^{n} for n = 1, 2, …,

then the corresponding Hankel determinant

(5.2)  V_{n}(x) = det_{0≤i,j≤n−1}(v_{i+j}(x))

is non-zero infinitely often, while the second establishes a (crude) asymptotic upper bound for it. The reader should note that v_{n}(x) becomes our previous v_{n} defined in (3.1) if x = μ, where μ is given by (3.2). Hence, if x = μ, the Hankel determinant V_{n}(x) becomes our earlier Hankel determinant V_{n}.

Lemma 4. Let α, λ, q, x be complex numbers with α ≠ 0, λ ∉ q^{ℤ_{>0}}, and α ∉ −λq^{ℤ_{>0}}. Then there are infinitely many positive integers n such that V_{n}(x) ≠ 0.

Proof. Writing the relation (5.1) for the generating series

G(z) = G_{x}(z) = ∑_{n=0}^{∞} v_{n}(x)z^{n}

we arrive at

G(z) = v_{0}(x) + z(qG(qz) − λG(z)) − ∑_{n=1}^{∞} α^{n}z^{n}

= z(qG(qz) − λG(z)) + x − \frac{1}{1 − αz} .

Equivalently,

(5.3)  (1 + λz)G(z) − qzG(qz) = x − \frac{1}{1 − αz} .

We claim that this equation does not have a rational function solution unless α ∈ −λq^{ℤ_{>0}}. Indeed, if z = 1/β, z = 1/(qβ), …, z = 1/(q^{k−1}β) are poles of G(z), for some k, then z = 1/(qβ), z = 1/(q^{2}β), …, z = 1/(q^{k}β) are poles of G(qz). Hence, the only way that this is possible in (5.3) is that the factor 1 + λz cancels the pole z = 1/β of G(z), while the term −1/(1−αz) on the right-hand side cancels the pole z = 1/(q^{k}β) of G(qz).

By a result of Kronecker (see [12, pp. 566–567] or [15, Division 7, Problem 24]), the fact that the series G(z) is not a rational function of z implies that infinitely many terms of the sequence V_{n}(x), where n = 1, 2, …, do not vanish. □

Lemma 5. Let μ̄, ᾱ, λ̄, q be complex numbers with ᾱ ≠ 0 and |q| > 1. Define the sequence (v̄_{n})_{n≥0} by (5.1) with x = μ̄, α replaced by ᾱ, and λ replaced by λ̄, and let V̄_{n} = det_{0≤i,j≤n−1}(v̄_{i+j}) be the corresponding Hankel determinant. Then we have

|V̄_{n}| ≤ |q|^{\frac{2}{3} n^{3}+o(n^{3})}, as n → ∞.

Proof. From (2.5) with α replaced by ᾱ, μ replaced by μ̄, and with b_{j}(q) = q^{j} − λ̄, we see that

|v̄_{n}| ≤ |q|^{\frac{1}{2} n^{2}+o(n^{2})}.

Hence, we have

|V̄_{n}| ≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |v̄_{i+τ(i)}| ≤ n! max_{τ∈𝔖_{n}} ∏_{i=0}^{n−1} |q|^{\frac{(i+τ(i))^{2}}{2} +o((i+τ(i))^{2})}

≤ n! ∏_{i=0}^{n−1} |q|^{\frac{(2i)^{2}}{2} +o(n^{2})} ≤ |q|^{\frac{2}{3} n^{3}+o(n^{3})},

as desired. □

We are now finally in the position to prove Theorems 1 and 2. Our proof simplifies the p-adic approach of Bézivin [1] and Choulet [6].

Proof of Theorems 1 and 2. Let q = ρ/σ ∈ ℚ, |q| > 1 and ρ > 1. Furthermore, let γ = (log ρ)/(log |σ|) (γ = ∞ if q ∈ ℤ). Let us now assume that all the numbers α, λ and μ = F_{q}(α; λ) are algebraic and write K = ℚ(α, λ, μ), and d = [K : ℚ].

In considering V_{n} we write, as before in (4.13), V_{n} = Δ_{n}Ṽ_{n} and note that, as n → ∞, we have

|V_{n}| ≤ |q|^{−An^{3}+o(n^{3})}, (5.4)

deg_{q} Δ_{n}(q) = Bn^{3} + o(n^{3}), |Δ_{n}(q)| = |q|^{Bn^{3}+o(n^{3})}, (5.5) where

(5.6)

A = \frac{1}{2}, B = \frac{65}{216} − \frac{Im Li_{2}(e^{2π\sqrt{−1}/3})}{π^{2}\sqrt{3}}  if λ = 0,

A = \frac{1}{3}, B = \frac{7}{27} − \frac{Im Li_{2}(e^{2π\sqrt{−1}/3})}{π^{2}\sqrt{3}}  if λ ≠ 0,

by Propositions 3 and 5. On the other hand, by Lemma 5, for all K-conjugates V_{n}^{[i]} of V_{n} we have

|V_{n}^{[i]}| ≤ |q|^{Cn^{3}+o(n^{3})}, i = 1, 2, …, d,

where C = 2/3. (Of course, for i = 1, that is, the case where V_{n}^{[i]} = V_{n}, we have a better estimate in (5.4).) Clearly, Δ_{n}(q) remains invariant under conjugation, whence, by (5.4) and (5.5), we have

(5.7)  |Ṽ_{n}| = |Ṽ_{n}^{[1]}| ≤ |q|^{−(A+B)n^{3}+o(n^{3})} as n → ∞,

and, for i = 2, 3, …, d,

(5.8)  |Ṽ_{n}^{[i]}| ≤ |q|^{(C−B)n^{3}+o(n^{3})} as n → ∞.

We know that V_{n} is a polynomial in q, α, λ and μ with integer coefficients, hence also Ṽ_{n}. Since the degree of V_{n} in each of μ, λ, α is at most n^{2} (see the paragraph containing (2.6)), the same is also true for Ṽ_{n}. On the other hand, by (2.6), we know that the degree in q of V_{n} is at most 2n^{3}/3 + o(n^{3}), whence we are able to find a positive integer Ω(n), log Ω(n) = o(n^{3}), such that

(5.9)  σ^{(C−B)n^{3}}Ω(n)Ṽ_{n} ∈ ℤ_{K},

where ℤ_{K} denotes the ring of integers of K. If V_{n} ≠ 0, then the product of all K-conjugates of the K-integer in (5.9) is a non-zero integer. Therefore, using (5.7) and (5.8),

1 ≤ | ∏_{i=1}^{d} σ^{(C−B)n^{3}}Ω(n)Ṽ_{n}^{[i]}|

≤ |σ|^{(C−B)dn^{3}} exp(o(n^{3})) |Ṽ_{n}| ∏_{i=2}^{d} |Ṽ_{n}^{[i]}|

≤ |σ|^{(C−B)dn^{3}}|q|^{−(A+B)n^{3}+(C−B)(d−1)n^{3}} exp(o(n^{3}))

≤ |σ|^{(A+C)n^{3}}ρ^{−(A+C−d(C−B))n^{3}+o(n^{3})}

≤ ρ^{{(A+C)/γ−(A+C−d(C−B))}n^{3}+o(n^{3})}.

If

(5.10)  \frac{1}{γ} (A + C) − (A + C − d(C − B)) < 0,

then the above inequality implies that V_{n} = 0 for all large n, contradicting Lemma 4. The reader should note that (5.10) can only hold if

(5.11)  A + C − d(C − B) > 0 or, equivalently, d < \frac{A + C}{C − B} ,

in which case, we have

(5.12)  γ > \frac{A + C}{A + C − d(C − B)}.

From (5.6) it follows that the only values of the degree d = [K : ℚ] satisfying (5.11) are d = 1 and d = 2. Theorems 1 and 2 follow then from (5.12) with d = 2 (in Theorem 1) and d = 1 (Theorem 2) by using the values of A and B from (5.6), and C = 2/3. □

Remark 8. We have a strong feeling that the method used in this work potentially makes it possible to deduce irrationality measures for the values of F_{q}(α; λ) in the cases when the number in question is irrational by Theorem 2. The only problem, which we are not able to overcome, is to establish the required density of nonvanishing of the determinant V_{n}(x) in (5.2) for a given x. More precisely, in our proof of Theorems 1 and 2 we use the fact (see Lemma 4) that V_{n}(x) ≠ 0 infinitely often, and this is (more than) sufficient for a quantitative irrationality, respectively, non-quadraticity result. We expect that a stronger assertion is true, which would then indeed yield irrationality measures for values of F_{q}(α; λ). Namely, for a given x ∈ ℂ and the sequence v_{n}(x) defined in (5.1), there should exist two positive constants c_{1} and c_{2}, c_{1} < c_{2}, such that for any m ≥ 1 one can find an index n in the range c_{1}m < n < c_{2}m, for which the Hankel determinant V_{n}(x) in (5.2) does not vanish. (In fact, we need this statement only for rational values of x, but this does not seem to be easier than the general case.) The belief in this statement rests upon the fact that the sequence v_{n}(x) is ‘highly structured’ (for instance, it is a solution of the simple recurrence relation (2.4) or (5.1) with general x; cf. [1] and [6]); hence V_{n}(x) should admit a certain structure as well. In fact, it was pointed out by the anonymous referee that in the case λ = 0 of the Tschakaloff function, V_{n} = V_{n}(μ) is nonzero for all n if q > 1 and α > 0. This follows from Lemma 2.2 in [1], which provides in this case the expression

V_{n} = α^{n^{2}−n} ∑_{1≤j_{1}<···<j_{n}} w_{j_{1}} ⋯ w_{j_{n}}(\frac{α}{q})^{j_{1}+···+j_{n}}(V (s_{j_{1}}, …, s_{j_{n}}))^{2},

where w_{j} = q^{−j(j+1)/2}, s_{j} = q^{−j}, and V (s_{j_{1}}, …, s_{j_{n}}) is the Vandermonde determinant built on s_{j_{1}}, …, s_{j_{n}}. The proof of this result is based on the tail expression (3.21) of v_{n} = v_{n}(μ) and does not work for V_{n}(x) with x general. This fact clearly supports our expectations above, although it is not enough for irrationality measures.

Acknowledgments

The authors are indebted to the anonymous referee for pointing out some subtle insufficiencies in an earlier version of the paper, and for several helpful comments concerning the discussion in Remark 8.

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Fakultät für Mathematik, Universität Wien, Nordbergstrasse 15, A-1090 Vienna, AUSTRIA

WWW-address: http://www.mat.univie.ac.at/~kratt

Department of Mechanics and Mathematics, Moscow Lomonosov State University, Vorobiovy Gory, GSP-1, 119991 Moscow, RUSSIA

E-mail address: justrip@rambler.ru

Department of Mathematical Sciences, University of Oulu, P. O. Box 3000, 90014 Oulu, FINLAND

E-mail address: kvaanane@sun3.oulu.fi

Max-Planck-Institut für Mathematik, Vivatsgasse 7, D-53111 Bonn, GERMANY

WWW-address: http://wain.mi.ras.ru/