# Combinatorics

In previous posts we have established the asymptotic densities of the Motzkin numbers modulo $2^{k}, k \in \{1, 2, 3 \}$ and $5$. The main result of this note is that

Theorem 1.

$\lim_{N \to \infty} \frac {1}{N} \#\{n \leq N: M_n \equiv 0 \mod 3 \} = 1$.

Firstly, let $T(\, 01 )\,$ denote the set of natural numbers which have a base 3 expansion containing only the digits $0$ and $1$. The following theorem from [1] will be used to prove theorem 1.

Theorem 2. (Corollary 4.10 of [1]). The Motzkin numbers satisfy

$M_n \equiv -1 \mod 3 \quad$ if  $\quad n \in 3T (\, 01 )\, - 1$,

$M_{n} \equiv 1 \mod 3 \quad$  if  $\quad n \in 3T(\, 01 )\quad$ or $\quad n \in 3T(\, 01 )\, - 2$,

$M_n \equiv 0 \mod 3 \quad$ otherwise.

We will first examine the nature of the set $T(\, 01 \,)$. We have

Theorem 3. The asymptotic density of the set $T(\, 01 \,)$ is zero.

Proof:

Choose $k \in N: 3^{k} \leq N < 3^{k+1}$. Then $k = \lfloor log_3 (\, N )\, \rfloor$ and

$\frac {1}{N} \# \{ n \leq N : n \in T(01) \} \leq \frac {2^{k+1}}{N} \leq \frac {2^{k+1}}{3^{k}} \to 0$ as $N \to \infty$.

$\Box \,$ .

Proof of Theorem 1.

Since the asymptotic density of $T(\, 01 \,)$ is zero, so is the asymptotic density of the sets $3T(\, 01 \,) - k$ for $k \in \{0, 1, 2\}$. Therefore Theorem 2 implies that

$\lim_{N \to \infty} \frac {1}{N} \# \{n \leq N : M_n \equiv +-1 \mod 3 \} = 0$

and the result follows.

$\Box \,$ .

## References.

[1] Deutsch, E., & Sagan, B. E. (2006). Congruences for Catalan and Motzkin numbers and related sequences. Journal of Number Theory, 117(1), 191–215. http://doi.org/10.1016/j.jnt.2005.06.005

This note will establish formulae for the asymptotic densities in the natural numbers of two sets of numbers.  These sets are $S_{1}(\, q, r, s, t )\, = \{ (\, qi + r )\, q^{sj + t} : i, j \in \mathbb{N} \} \quad$

and

$S_{2}(\, q, r, s, t )\, = \{ (\, qi + r )\, q^{sj + t} : i, j \in \mathbb{N}, j > 0 \}$

for integers $\quad q, r, s, t$.

Theorem 1. Let $\quad q, r, s, t \in \mathbb{Z} \quad$ with $\quad q, r, s > 0 \quad$ and $\quad 0 < r < q \quad$. Then the asymptotic density of the set $S_{1} \quad$ is $\quad (\, q^{t + 1 - s} (\, q^{s} - 1 )\, )\, ^{-1} \quad$. In addition the asymptotic density of the set $\, S_{2} \,$ is $\, (\, q^{t + 1} (\, q^{s} - 1 )\, )\, ^{-1}$.

Proof.

We firstly look at $S_{1}$. We have, for fixed $j \geq 0$,

$\#\{ i \geq 0: (\, qi + r)\, q^{sj + t} \leq N \} = \frac {N}{q^{qj+t+1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\,$

where $0 \leq E( \, j, N, q, r, s, t ) < 1$ is an error term introduced by not rounding down to the nearest integer. So, letting

$U(N, s, t) =: \quad \lfloor \frac {\log_q (\, N )\, - t - 1}{s} \rfloor \quad \,$,

we have

$\#\{n < N: n = (\, qi + r)\, q^{sj + t} \, for some \, i, j \in \mathbb{N} \}$

$= \sum_{j \geq 0} (\, \frac {N}{q^{sj + t + 1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\, )\, = \sum_{j = 0}^{U} (\, \frac {N}{q^{sj+t + 1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\, )\,$

$= \frac {N}{q^{t + 1} } \sum_{j = 0}^{U} (\, \frac {1}{q^{s}} )\,^{j} - E^{'}(\, N, q, r, s, t )\,$

where the new error term $E^{'} (\, N, q, r, s, t )\,$ satisfies $0 < E^{'} (\, N, q, r, s, t )\, < 2(\, U + 1 )\,$.

Then

$\#\{n < N: n = (\, qi + r)\, q^{sj + t} \, for some \, i, j \in \mathbb{N} \}$

$= (\, \frac {N}{q^{t+1}} )\, (\, 1 - (\, \frac {1}{q^{s}} )\, ^{U + 1} )\, (\, 1 - \frac {1}{q^{s} } )\, ^{-1} - E^{'}$

Since $\lim_{N \to \infty} \frac {E^{'}(\, N, q, r, s, t )\, }{N} = 0$ and $\lim_{N \to \infty} \frac {1}{N} (\, \frac {1}{q^{s} } )\, ^{U + 1} = 0$

$\lim_{N \to \infty} \frac {1}{N} \#\{n < N: n = (\, qi + r)\, q^{sj + t} \, for some \, i, j \in \mathbb{N} \}$

$= (\, q^{t + 1 - s} (\, q^{s} - 1 )\,)\, ^{-1}$.

The proof for $S_{2}$ is very similar. With $E, E^{'}$ and $U$ defined as above we have for fixed $j \geq 1$,

$\#\{ i \geq 0: (\, qi + r)\, q^{sj + t} \leq N \} = \frac {N}{q^{qj+t+1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\,$

and

$\#\{n < N: n = (\, qi + r)\, q^{sj + t} \, for some \, i, j \in \mathbb{N} \, with \, j \geq 1 \}$

$= \sum_{j \geq 1} (\, \frac {N}{q^{sj + t + 1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\, )\, = \sum_{j = 1}^{U} (\, \frac {N}{q^{sj+t + 1}} - \frac {r}{q} - E(\, j, N, q, r, s, t )\, )\,$

$= \frac {N}{q^{t + 1 + s} } \sum_{j = 0}^{U - 1} (\, \frac {1}{q^{s}} )\,^{j} - E^{'}(\, N, q, r, s, t )\,$

$= (\, \frac {N}{q^{t+1+s}} )\, (\, 1 - (\, \frac {1}{q^{s}} )\, ^{U} )\, (\, 1 - \frac {1}{q^{s} } )\, ^{-1} - E^{'}$

and

$\lim_{N \to \infty} \frac {1}{N} \#\{n < N: n = (\, qi + r)\, q^{sj + t} \, for some \, i, j \in \mathbb{N} \, with \, j \geq 1 \}$

$= (\, q^{t + 1 } (\, q^{s} - 1 )\,)\, ^{-1}$.

$\Box$.

The Motzkin numbers $M_n$ are defined by

$M_n := \sum_{k \geq 0} \binom {n}{2k} C_{k}$

where $C_{k}\,$ are the Catalan numbers. The following result is established in [1]

Theorem 1 (Theorem 5.5 of [1]). The nth Motzkin number $M_n$ is even if and only if $n = (4i + \epsilon)4^{j+1} - \delta$ for $i, j \in \mathbb{N}, \epsilon \in \{1, 3\}$ and $\delta \in \{1, 2\}$. Moreover, we have

$M_n \equiv 4 \mod 8$ if $(\, \epsilon, \delta )\, = (\, 1, 1 )\,$ or $(\, 3, 2 )\,$

$M_n \equiv 4y + 2 \mod 8$ if $(\, \epsilon, \delta )\, = (\, 1, 2 )\, or (\, 3, 1 )\,$

where $\, y \,$ is the number of digit 1’s in the base 2 representation of $\, 4i + \epsilon - 1$.

Remark 2. The 4 choices for $(\, \epsilon, \delta )\,$ in the above theorem give 4 disjoint sets of numbers $n = (4i + \epsilon)4^{j+1} - \delta$.

Theorem 3. Each of the 4 disjoint sets defined by the choice of $(\, \epsilon, \delta )\,$ in Theorem 1 has asymptotic density $\frac {1}{12}$ in the natural numbers.

Proof: Use the result for the set $S_{1}$ of Theorem 1 of the post here with $q = 4, r = \epsilon, s = 1, t = 1$.

$\Box \,$.

Corollary 4: The asymptotic density of

$\{ n < N: M_n \equiv 0 \mod 2 \}$ is $\frac {1}{3}$.

The asymptotic density of

$\{ n < N: M_n \equiv 4 \mod 8 \}$ is $\frac {1}{6}$.

The asymptotic density of each the sets

$\{ n < N: M_n \equiv 2 \mod 8 \}$ and $\{ n < N: M_n \equiv 6 \mod 8 \}$ is $\frac {1}{12}$.

Remark 5: The first 2 statements of Corollary 4 follows immediately from theorems 1 and 3. The third statement follows from the observation that the numbers of 1’s in the base 2 expansion of $i$ is equally likely to be odd or even and therefore the same applies the the number of 1’s in the base 2 expansion of $4i + \epsilon - 1$. Since the asymptotic density of the 2 sets combined is $\frac {1}{6}$ (from Theorems 1 and 3), each of the two sets has asymptotic density of $\frac {1}{12}$.

## Reference

[1] Eu, S.-P., Liu, S.-C., & Yeh, Y.-N. (2008). Catalan and Motzkin numbers modulo 4 and 8. European Journal of Combinatorics, 29(6), 1449–1466. http://doi.org/10.1016/j.ejc.2007.06.019