A New Characterization of Catalan Numbers Related to Hankel Transforms and Fibonacci

Numbers

Belgacem Bouras Department of Mathematics College of Sciences Gabès University Tunisia

            Belgacem.Bouras@issatgb.rnu.tn
             Belgacem.Bouras.2003@voila.fr

Abstract

Cvetković, Rajković, and Ivković proved that the Hankel transform of the sequence of sums of two successive Catalan numbers is the sequence of Fibonacci numbers with odd indices. Later, Benjamin, Cameron, Quinn, and Yerger extended this result and proved that if we remove one term from this sequence of sums, then the Hankel transform is the sequence of Fibonacci numbers with even indices. In this paper, we prove that the Catalan numbers are the unique nonnegative integer sequence satisfying this property.

1 Introduction

The Hankel transform of a sequence {a_{n}}_{n=0}^{∞} is the sequence H({a_{n}}_{n=0}^{∞}) = {h_{n}}_{n=0}^{∞} defined as follows:

h_{n} = det[a_{i+j}]_{0≤i, j≤n−1}.  (1)

The Hankel transform was introduced and first studied by Layman [7], who showed that the Hankel transform of a sequence is equal to the Hankel transform of the binomial transform of this sequence and conjectured that the Hankel transform of the sequence of sums of two adjacent Catalan numbers is a subsequence of Fibonacci number sequence.

Recall that the Catalan number sequence {C_{n}}_{n=0}^{∞} is defined as follows:

C_{n} = \frac{\binom{2n}{n}}{n + 1}, n ≥ 0.

This sequence is the unique nonnegative integer sequence {a_{n}}_{n=0}^{∞} fulfilling (see [8])

det[a_{i+j}]_{0≤i, j≤n−1} = 1

and

det[a_{i+j}]_{1≤i, j≤n} = 1.

Cvetković, Rajković, and Ivković [6] proved the following formula, known as Layman’s conjecture,

det[C_{i+j+1} + C_{i+j+2}]_{0≤i, j≤n−1} = F_{2n+2}, n ≥ 1.  (2)

where F_{n}, n ≥ 0 is the n-th Fibonacci number defined by the following recurrence:

F_{0} = 0, F_{1} = 1, F_{n+2} = F_{n+1} + F_{n}, n ≥ 0.  (3)

The proof depends on special properties of the corresponding orthogonal polynomials. Benjamin, Cameron, Quinn, and Yerger [1] used a combinatorial approach and proved that

det[C_{i+j} + C_{i+j+1}]_{0≤i, j≤n−1} = F_{2n+1}, n ≥ 1.  (4)

In view of (4) and (2), it should be mentioned here that (4) and (2) are special cases of the more general determinant evaluation given in [3].

A natural question arises: Is the Catalan number sequence the unique nonnegative integer sequence fulfilling (4) and (2)? More precisely, is there any other nonnegative integer sequence {a_{n}}_{n=0}^{∞} fulfilling

det[a_{i+j} + a_{i+j+1}]_{0≤i, j≤n−1} = F_{2n+1}, n ≥ 1  (5)

and

det[a_{i+j+1} + a_{i+j+2}]_{0≤i, j≤n−1} = F_{2n+2}, n ≥ 1.  (6)

The aim of this paper is to answer this question. Namely, we prove the main theorem of this paper:

Theorem 1. The unique nonnegative integer sequence fulfilling (5) and (6) is the Catalan number sequence.

Our paper is organized as follows: In Section 2, we state some basic results concerning the theory of orthogonal polynomial. These results will be used in Section 3 to prove the main theorem.

2 Notation and preliminary results

Let P be the vector space of polynomials with complex coefficients and let P′ be its algebraic dual. We denote by ⟨u, f⟩ the action of u ∈ P′ on f ∈ P. In particular, we denote by u_{n} = ⟨u, x^{n}⟩ , n ≥ 0 the moments of u. If u ∈ P′ and Φ ∈ P, then the left multiplication of the functional u by the polynomial Φ, denoted by Φu, is the functional in P′ defined as follows:

⟨Φu, f⟩ = ⟨u, Φf⟩ , f ∈ P.  (7)

Definition 2. (see [5]) A sequence of polynomials {P_{n}}_{n≥0} is said to be a monic orthogonal polynomial sequence (MOPS) with respect to a linear functional u if

i) deg P_{n} = n and the leading coefficient of P_{n}(x) is equal to 1. ii) ⟨u, P_{n}P_{m}⟩ = r_{n}δ_{n,m}, n, m ≥ 0, r_{n} ≠ 0, n ≥ 0. A linear functional u is called regular if there exists a polynomial sequence {P_{n}}_{n≥0} orthogonal with respect to u. Throughout this paper, we will take all regular linear functionals u normalized, i.e., u_{0} = 1.

According to Favard-Shohat theorem, a sequence of monic orthogonal polynomials satisfies a three-term recurrence relation (see [5]):

\begin{matrix}P_{0}(x) = 1, P_{1}(x) = x − β_{0}, \\ P_{n+2} = (x − β_{n+1})P_{n+1}(x) − γ_{n+1}P_{n}(x), n ≥ 0, \end{matrix}  (8)

with

(β_{n}, γ_{n}) ∈ ℂ × (ℂ \ {0}), n ≥ 0.

3 Proof of the main theorem

First of all, recall that a linear functional L is regular if and only if (see [5])

Δ_{n}(L) ≠ 0,  (9)

where Δ_{n}(L) is the Hankel determinant of order n + 1 of L defined as follows:

Δ_{−1}(L) = 1, Δ_{n}(L) = | \begin{matrix}L_{0} & L_{1} & ⋯ & L_{n} \\ L_{1} & L_{2} & ⋯ & L_{n+1} \\ ⋮ & ⋮ & & ⋮ \\ L_{n} & L_{n+1} & ⋯ & L_{2n} \end{matrix} | , n ≥ 0.  (10)

In this condition, the coefficient γ_{n} in (8) is given as follows (see [2, 5]):

γ_{n+1} = \frac{Δ_{n+1}(L)Δ_{n−1}(L)}{Δ_{n}^{2}(L)}, n ≥ 0.  (11)

Conversely, the Hankel determinant Δ_{n}(L) can be written in terms of γ_{n} as follows (see [2]):

Δ_{−1}(L) = Δ_{0}(L) = 1, Δ_{n}(L) = ∏_{k=1}^{n} τ_{k}, n ≥ 0,  (12)

where

τ_{k} = ∏_{i=1}^{k} γ_{i}, k ≥ 1.  (13)

Let us recall a result established in [4] concerning the so-called kernel polynomials. Given a MOPS {P_{n}}_{n≥0} orthogonal with respect to a linear functional L. The linear functional L^{∗} defined as follows:

L^{∗} = λ(x − c)L, c ∈ ℂ,

where λ is a normalization factor, is regular if and only if

P_{n}(c) ≠ 0, n ≥ 0.

In this case the MOPS corresponding to L^{∗}, denoted by {P_{n}^{∗}(c, x)}_{n≥0}, satisfies

P_{n}^{∗}(c, x) = \frac{1}{x − c}(P_{n+1}(x) − \frac{P_{n+1}(c)}{P_{n}(c)}P_{n+1}(x)).

The sequence {P_{n}^{∗}(c, x)}_{n≥0} is called the sequence of kernel polynomials of K-parameter c corresponding to {P_{n}}_{n≥0} . The recurrence coefficients of {P_{n}^{∗}(c, x)}_{n≥0} denoted by β_{n}^{∗} and γ_{n+1}^{∗} are expressed in terms of those of {P_{n}}_{n≥0} as follows:

β_{n}^{∗} = β_{n+1} + \frac{P_{n+2}(c)}{P_{n+1}(c)} − \frac{P_{n+1}(c)}{P_{n}(c)},  (14)

γ_{n+1}^{∗} = γ_{n+1} \frac{P_{n+2}(c)P_{n}(c)}{P_{n+1}^{2}(c)}, n ≥ 0.  (15)

Let {a_{n}}_{n=0}^{∞} be a non-negative integer sequence fulfilling (5) and (6). Then for n = 1, we get a_{0} + a_{1} = F_{3} = 2 ≠ 0 and a_{1} + a_{2} = F_{4} = 3 ≠ 0. Define the linear functionals u, v and v^{∗} by

u_{n} = \frac{C_{n} + C_{n+1}}{C_{0} + C_{1}} , n ≥ 0,  (16)

v_{n} = \frac{a_{n} + a_{n+1}}{a_{0} + a_{1}} , n ≥ 0,  (17)

v_{n}^{∗} = \frac{a_{n+1} + a_{n+2}}{a_{1} + a_{2}} , n ≥ 0.  (18)

From (10), (16) and (4), we have

Δ_{n}(u) = \frac{F_{2n+3}}{(C_{0} + C_{1})^{n+1}} , n ≥ 0.  (19)

Using the recurrence (3), we can easily see that F_{n} ≠ 0, n ≥ 0. Therefore, the condition (9) is satisfied by u. Hence, u is regular. Similarly, from (10), (5), (6), (17) and the relation (18), we get

Δ_{n}(v) = \frac{F_{2n+3}}{(a_{0} + a_{1})^{n+1}} , n ≥ 0  (20)

and

Δ_{n}(v^{∗}) = \frac{F_{2n+4}}{(a_{1} + a_{2})^{n+1}} , n ≥ 0.  (21)

For the same reason, v and v^{∗} are regular. We will denote by {P_{n}}_{n≥0} and {P_{n}^{∗}(c, x)}_{n≥0} the MOPS corresponding to v and v^{∗} respectively. From (17) and (18), we have

v^{∗} = \frac{a_{0} + a_{1}}{a_{1} + a_{2}} xv.

So, the sequence {P_{n}^{∗}(c, x)}_{n≥0} is the sequence of kernel polynomials of K-parameter 0 corresponding to {P_{n}}_{n≥0} . Hence, using (12), (13) and (15), with c = 0, we get

Δ_{n}(v^{∗}) = \frac{P_{n+1}(0)}{(P_{1}(0))^{n+1}} Δ_{n}(v), n ≥ 0.

Taking into account (20) and (21), we obtain

F_{2n+4} = (\frac{a_{1} + a_{2}}{a_{0} + a_{1}} )^{n+1} \frac{P_{n+1}(0)}{(P_{1}(0))^{n+1}} F_{2n+3}, n ≥ 0.  (22)

Suppose that {P_{n}}_{n≥0} satisfies the recurrence (8). We have

P_{1}(0) = −β_{0}.

But, from Definition 2 and the relation (8), we have

0 =< v, P_{1}P_{0} >=< v, P_{1} >=< v, x > −β_{0} < v, 1 >= v_{1} − β_{0}.

Thus

β_{0} = v_{1} = \frac{a_{1} + a_{2}}{a_{0} + a_{1}} .

Therefore

P_{1}(0) = −\frac{a_{1} + a_{2}}{a_{0} + a_{1}} .  (23)

Substitution of (23) in (22) gives

F_{2n+4} = (−1)^{n+1}P_{n+1}(0)F_{2n+3}, n ≥ 0

or equivalently

P_{n+1}(0) = (−1)^{n+1} \frac{F_{2n+4}}{F_{2n+3}} , n ≥ 0.  (24)

From (11) and (20), we get

γ_{n+1} = \frac{F_{2n+5}F_{2n+1}}{F_{2n+3}^{2}} , n ≥ 0.  (25)

Taking x = 0 in (8) and using (24) and (25), we get

β_{n+1} = \frac{F_{2n+3}F_{2n+6}}{F_{2n+4}F_{2n+5}} + \frac{F_{2n+2}F_{2n+5}}{F_{2n+3}F_{2n+4}} , n ≥ 0.

On account of the recurrence relation (3), we get

β_{n+1} = \frac{F_{2n+4}}{F_{2n+3}} + \frac{F_{2n+3}}{F_{2n+5}} , n ≥ 0.  (26)

We have proved that the recurrence coefficients of the MOPS corresponding to any sequence {a_{n}}_{n=0}^{∞} satisfying (5) and (6) are given by (25) and (26). Since the sequence {C_{n}}_{n=0}^{∞} itself satisfies (5) and (6) then the recurrence coefficients of its MOPS are also given by (25) and (26). Consequently, by the uniqueness of the linear functional corresponding to a MOPS, we have

u = v

or, equivalently,

u_{n} = v_{n}, n ≥ 0.

Therefore, owing to (16) and (17), we obtain

a_{n} + a_{n+1} = C_{n} + C_{n+1}, n ≥ 0.  (27)

On the other hand, we have

a_{0} + a_{1} = F_{3} = 2.

Taking into account the fact that {a_{n}}_{n=0}^{∞} is a nonnegative sequence, we get a_{0} = 1 = C_{0} and a_{1} = 1 = C_{1}. By induction and using (27), we can easily prove that,

a_{n} = C_{n}, n ≥ 0.

Which completes the proof.

4 Acknowledgments

We are grateful to the anonymous referee for careful reading the text and essential remarks.

References

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[8] R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

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2010 Mathematics Subject Classification: Primary: 33C45 ; Secondary 42C05. Keywords: Hankel transform, Catalan number, recurrence relation, integer sequence.

(Concerned with sequences \underline{A000159}, \underline{A014523}, and \underline{A103433}.)

Received July 11 2012; revised versions received October 23 2012; December 28 2012; February 3 2013. Published in Journal of Integer Sequences, March 2 2013.

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