The Catalan numbers have no forbidden residue modulo primes

TL;DR

This paper proves that for any prime p ≥ 5, Catalan numbers modulo p cover all residues infinitely often, and any integer can be expressed as a product of central binomial coefficients modulo p.

Contribution

It establishes that Catalan numbers modulo primes p ≥ 5 are surjective onto all residues and characterizes the automaton structure of their residues.

Findings

Catalan numbers modulo p attain all residues for p ≥ 5.

All residues occur infinitely often in the sequence.

Any positive integer can be expressed as a product of central binomial coefficients modulo p.

Abstract

Let $C_n$ be the $n$th Catalan number. For any prime $p \geq 5$ we show that the set $\{C_n : n \in \mathbb{N} \}$ contains all residues mod $p$. In addition all residues are attained infinitely often. Any positive integer can be expressed as the product of central binomial coefficients modulo $p$. The directed sub-graph of the automata for $C_n \mod p$ consisting of the constant states and transitions between them has a cycle which visits all vertices.

Equations

$$C_n:=\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right).$$

$$\{(\genfrac{}{}{0ex}{}{2d}{d})\mathrm{mod}p:0\le d\le \frac{p-1}{2}\}.$$

$${\Lambda }_{d,d}(1\ast Q^{p-1})=\left(\genfrac{}{}{0ex}{}{2d}{d}\right).$$

$$c_{1}c_{2}=\Lambda_{d_{1},d_{1}}\biggr{(}\;\Lambda_{d_{2},d_{2}}(\,1*Q^{p-1}\,)\;\biggr{)}.$$

$$\left(\genfrac{}{}{0ex}{}{q+1}{\frac{q+1}{2}}\right)\in S$$

$$\left(\genfrac{}{}{0ex}{}{q+1}{\frac{q+1}{2}}\right)=qr$$

$$q=\left(\genfrac{}{}{0ex}{}{q+1}{\frac{q+1}{2}}\right)r^{-1}\in S.$$

$$n=\prod _i\left(\genfrac{}{}{0ex}{}{2d_i}{d_i}\right)\mathrm{mod}p$$

$$c_2{c}_1^{-1}=\prod _i\left(\genfrac{}{}{0ex}{}{2d_i}{d_i}\right)\mathrm{mod}p.$$

$${\Lambda }_{d,d}(c\ast Q^{p-1})=c{\Lambda }_{d,d}(1\ast Q^{p-1})=c\left(\genfrac{}{}{0ex}{}{2d}{d}\right),$$

$$\biggr{(}\,\prod_{i}\Lambda_{d_{i},d_{i}}\biggr{)}\,(\,c_{1}*Q^{p-1}\,)=c_{1}\prod_{i}\binom{2d_{i}}{d_{i}}=c_{2}\mod p.$$

Full text

1 Introduction

The Catalan numbers are defined by

[TABLE]

This note is an addendum to our paper [1]. In that paper we analysed the Catalan numbers modulo primes $\geq 5$ using automata. Refer to that paper and to [6] for details of how automata can be used to study Catalan numbers and other sequences.

A set $S$ is said to have a forbidden residue $r$ modulo $p$ if no element of $S$ is $\;\equiv r\mod p$. We show below that the Catalan numbers have no forbidden residue modulo any prime. Garaeva, Luca and Shparlinski [3] established this result for sufficiently large primes. They also showed that in a certain sense the distribution of $C_{n}\mod p$ amongst the non-zero residue classes is roughly equal. They also proved that the set {$C_{n}:n\leq p^{13/2}(\log p)^{6}\,\}$ already includes all residue classes modulo $p$. Our results do not say anything about how quickly $C_{n}$ covers all residue classes or about how often proportionally each residue class is attained. We do show that each residue class is attained infinitely often. The result for $C_{n}\mod p$ differs from the situation for powers of primes. Eu, Liu and Yeh [2] showed that $3$ is a forbidden residue for $C_{n}$ modulo $4$ and $\{\,3,7\,\}$ are forbidden residues for $C_{n}\mod 8$. Liu and Yeh in [5] calculated $C_{n}\mod 16$ and $\mod 64$ and thereby determined the forbidden residues in each case. They also showed that $C_{n}$ has forbidden residues $\mod 2^{k}$ for any $k$. Forbidden residues for $C_{n}$ modulo $\{32,64,128,256,512\}$ were calculated by Rowland and Yassawi using automata in [6]. Kauers, Krattenthaler and Müller calculated the generating function for $C_{n}\mod 4096$ in terms of a special function and so, in theory, could determine forbidden residues in this case. Similarly Krattenthaler and Müller [4] determined the generating function for $C_{n}\mod 27$ in terms of a special function.

Another difference between the $\mod p$ case and when higher powers of $p$ are involved is that in the $\mod p$ situation all residues are attained infinitely often. Rowland and Yassawi showed that some residues for $C_{n}\mod\{8,16\}$ are attained only finitely many times.

2 Results

Let $p$ be prime and let the set $S$ be defined as the multiplicative closure in \biggr{(}\,\frac{\mathbb{Z}}{p\mathbb{Z}}\,\biggr{)}^{\times} of the set of elements

[TABLE]

All elements of $S$ are non-zero as $2d\leq p-1$ for $d\leq\frac{p-1}{2}$. We showed in [1] that $S$ is contained in the set of constant states of the automaton for $C_{n}\mod p$ and hence that the elements of $S$ appear as residues of $C_{n}\mod p$ for some $n$. In explanation of this remark, as shown in [1] we have for $d\leq\frac{p-1}{2}$

[TABLE]

Then if $c_{1}=\binom{2d_{1}}{d_{1}}$ and $c_{2}=\binom{2d_{2}}{d_{2}}$ we have

[TABLE]

Therefore $c_{1}c_{2}$ is also a constant state of the automaton for $C_{n}\mod p$ and therefore also a residue of $C_{n}\mod p$ for some $n$.

Lemma 2.1**.**

The set $S$ contains all non-zero residues modulo $p$.

Proof.

Since $S$ is multiplicatively closed it is enough to show that $S$ contains all primes $q<p-1$. We observe that if $c\in S$ then $c^{-1}\mod p=c^{p-2}\mod p$ is also in $S$. We proceed by induction on the set of primes. Firstly, $1=\binom{0}{0}$ and $2=\binom{2}{1}\in S$. Let $q<p-1$ be prime. Then $\frac{q+1}{2}\leq\frac{p-1}{2}$ and

[TABLE]

with

[TABLE]

where $r$ is the product of primes strictly less than $q$. Then by induction $r\in S$ and by the observation above $r^{-1}\in S$. So

[TABLE]

∎

Corollary 2.2**.**

For any prime $p\geq 5$, the Catalan numbers have no forbidden residue modulo $p$.

Proof.

Lemma 2.1 shows that all non-zero residues appear in $C_{n}\mod p$. In addition, the values of $\;n:C_{n}\equiv 0\mod p\;$ are plentiful, having asymptotic density $1$ (see [1]). ∎

Corollary 2.3**.**

For $p$ prime any $n\in\mathbb{N}$ can be written as

[TABLE]

for suitable choices of $\{d_{i}\in\mathbb{N}\}$.

Proof.

The same inductive argument as in Lemma 2.1 can be used to prove the corollary. The choice of $\{d_{i}\}$ is not necessarily unique. ∎

The set of states and transitions for the automata of $C_{n}\mod p$ is a directed graph with the states as vertices and transitions as directed edges. The sub-graph $G$ consisting of the non-zero constant states and transitions between them is also a directed graph.

Corollary 2.4**.**

The directed graph $G$ formed by the non-zero constant states and transitions has a cycle which visits all vertices in $G$.

Proof.

Let $c_{1}$ and $c_{2}$ be two constant states. It is enough to show that there is a directed state path from $c_{1}$ to $c_{2}$. Firstly, $c_{2}c_{1}^{-1}$ is also a constant state by the multiplicative closure of the set $S$. From corollary 2.3 there are $\{d_{i}\}$ such that

[TABLE]

Then since for constants $c$

[TABLE]

we have modulo $p$

[TABLE]

Since the application of each $\Lambda_{d_{i},d_{i}}$ corresponds to a transition between states, the product of the $\Lambda_{d_{i},d_{i}}$ corresponds to a directed path from $c_{1}$ to $c_{2}$. ∎

Observation 2.5**.**

Each residue is attained infinitely often. Firstly from [1] numbers $n$ which have base $p$ representations containing only digits from the set $\{0,1,...,\frac{p-1}{2}\}$ have a state path which ends at a non-zero constant state. Since $C_{n}\mod p$ is the value of the end state of the state path for $n$, it is non-zero $\mod p$. So at least one non-zero constant state (and so at least one non-zero residue) is attained infinitely often. Secondly, the existence of a cycle in the directed graph of the constant states shows that all non-zero constant states are visited infinitely often.