Asymptotic densities of Motzkin numbers modulo 5

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 5.4 of [1]). The Motzkin number M_n is divisible by 5 if and only if n is one of the following forms

(\, 5i + 1 )\, 5^{2j}  - 2,\, (\, 5i + 2 )\, 5^{2j-1}  - 1,\, (\, 5i + 3 )\, 5^{2j-1}  - 2,\,  (\, 5i + 4 )\, 5^{2j}  - 1

where i, j \in \mathbb{N} and j \geq 1.

 

Theorem 2: The asymptotic density of numbers of the first form in theorem 1 is \frac {1}{120}. Numbers of the fourth form also have asymptotic density \frac {1}{120}. The asymptotic density of numbers of the second and third forms in theorem 1 is \frac {1}{24} each.

 

Corollary 3: The asymptotic density of \#\{n < N: M_{n} \equiv 0 \mod {5} \} is \frac {1}{10} \, .

 

Remark 4: Corollary 3 follows immediately from theorems 1 and 2 and the disjointness of the 4 forms of integers listed in theorem 1.

 

Remark 5: Numerical tests also show that roughly 22.5% of Motzkin numbers are congruent to each of 1, 2, 3, 4 \mod {5}.

 

Proof of Theorem 2.

Firstly consider numbers of the form (\, 5i + 1)\, 5^{2j} - 2\,. As we are interested in asymptotic density it is enough to look at numbers of the form (\, 5i + 1)\, 5^{2j}\,. We have, for fixed j \geq 1,

\#\{ i \geq 0: (\, 5i + 1)\, 5^{2j} \leq N \} = \frac {N}{5^{2j+1}} - \frac {1}{5} - E(\, j, N )\,

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

\#\{n < N: n = (\, 5i + 1)\, 5^{2j} \, for some  \, i, j \in \mathbb{N} \, with \, j \geq 1 \}

= \sum_{j \geq 1} (\,  \frac {N}{5^{2j+1}}  - \frac {1}{5} - E(\, j, N )\, )\, = \sum_{j = 1}^{\lfloor \frac{1}{2} (\, \log_{5} (\, N )\,- 1 )\,  \rfloor} (\, \frac {N}{5^{2j+1}}  - \frac {1}{5} - E(\, j, N )\, )\,

= \frac {N}{125} \sum_{j = 0}^{\lfloor \frac{1}{2} (\, \log_{5} (\, N )\,- 1 )\,  \rfloor - 1} (\, \frac {1}{25} )\,^{j} - E^{'}(\, N )\,

where the new error term E^{'} (\, N )\, satisfies 0 \leq E^{'} (\, N )\, < \frac{3}{5} (\, \log_{5} (\, N )\,. Then

\#\{n < N: n = (\, 5i + 1)\, 5^{2j} \, for some  \, i, j \in \mathbb{N} \, with \, j \geq 1 \}

= (\, \frac {N}{125} )\, \frac {1 - (\, \frac {1}{25} )\,^{\lfloor \frac{1}{2} (\, \log_{5} (\, N )\,- 1 )\,  \rfloor} }{1 - \frac {1}{25} } - E^{'}(\, N )\,.

Since \lim_{N \to \infty} \frac {E^{'}(\, N )\, }{N} = 0

\lim_{N \to \infty} \frac {1}{N} \#\{n < N: n = (\, 5i + 1)\, 5^{2j} \, for some  \, i, j \in \mathbb{N} \, with \, j \geq 1 \} = \frac {1}{120}.

A similar argument can be used to establish the asymptotic densities of the other 3 forms of numbers in the theorem.

\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

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