Enumeration of Carlitz Multipermutations

TL;DR

This paper enumerates Carlitz multipermutations with specific repetitions, deriving recurrences and confirming a conjecture for the case of three copies, thereby advancing combinatorial enumeration techniques.

Contribution

It provides explicit enumeration formulas and recurrences for Carlitz multipermutations with k=2,3,4, including a proof and improvement of a conjectured recurrence for k=3.

Findings

Derived recurrences for k=2,3,4 cases

Proved and improved a conjectured recurrence for k=3

Enumerated Carlitz multipermutations for small k values

Abstract

A multipermutation with $k$ copies each of $1\ldots n$ is Carlitz if neighbours are different. We enumerate these objects for $k=2,3,4$ and derive recurrences. In particular, we prove and improve a conjectured recurrence for $k=3$, stated in OEIS, the Online Encyclopedia of Integer Sequences.

Tables

Table 1. Table 1. Number of Carlitz mutipermutations

$a_k\left(n\right)$	$n=0$	$n=1$	$n=2$	$n=3$	$n=4$	$n=5$	$n=6$
$k=1$	1	1	2	6	24	120	720
$k=2$	1	0	2	30	864	39 480	2 631 600
$k=3$	1	0	2	174	41 304	19 606 320	16 438 575 600

Table 2. Table 2. Number of ordered Carlitz mutipermutations

$a_k^{\prime }\left(n\right)$	$n=0$	$n=1$	$n=2$	$n=3$	$n=4$	$n=5$	$n=6$
$k=1$	1	1	1	1	1	1	1
$k=2$	1	0	1	5	36	329	3 655
$k=3$	1	0	1	29	1 721	163 386	22 831 355

Equations

$$A_2(2)$$

$$A_2(3)$$

$$A_2^{\prime }(2)$$

$$A_2^{\prime }(3)$$

$$a_{2}(n)={\sum}_{s+t=n}\Big{[}(-1)^{t}{n\choose s}\frac{(2s+t)!}{(2!)^{s}}\Big{]}$$

$$a_{3}(n)={\sum}_{s+t+u=n}\Big{[}(-1)^{t}{n\choose{s,t,u}}\frac{(3s+2t+u)!}{(3!)^{s}}\Big{]}$$

$$1\cdot \frac{9!}{6^3}-3\cdot \frac{8!}{6^2}+3\cdot \frac{7!}{6^2}+3\cdot \frac{7!}{6}-6\cdot \frac{6!}{6}+3\cdot \frac{5!}{6}-1\cdot \frac{6!}{1}+3\cdot \frac{5!}{1}-3\cdot \frac{4!}{1}+1\cdot \frac{3!}{1}=174$$

$$a_{4}(n)={\sum}_{s+t+u+v+w=n}\Big{[}(-1)^{t+w}{n\choose{s,t,u,v,w}}\frac{(4s+3t+2u+2v+w)!}{(4!)^{s}(2!)^{v+t}}\Big{]}$$

$$a_4(n)=\Phi ((\frac{t^3}{6}-t^2+t{)}^n),$$

$$p_{n+1}=(2n+1)p_n+p_{n-1},p_0=1,p_1=0,$$

$$\begin{array}[]{rcll}2p_{n+1}&=&(3n+3)q_{n}-2(3n+1)p_{n}+2p_{n-1},&p_{0}=1,\ p_{1}=0,\\ q_{n}&=&(3n+2)p_{n}+2q_{n-1},&q_{0}=0,\end{array}$$

$$p_{n+1}=\lambda p_n+\mu p_{n-1}+\nu p_{n-2},p_0=1,p_1=0,p_2=1$$

$$2p_n=3n{q}_{n-1}-2(3n-2)p_{n-1}+2p_{n-2}$$

$$\begin{array}[]{rcl}3p_{n+1}&=&(4n+1)q_{n}+3(10q_{n-1}-r_{n}+4r_{n-1}+(6n+7)p_{n}+p_{n-1})\\ 2q_{n}&=&(4n+6)r_{n}+6r_{n-1}-(16n+6)p_{n}\\ r_{n}&=&(4n+3)p_{n}+3q_{n-1},\ p_{0}=1,\ p_{1}=0,\ q_{0}=0,\ r_{0}=0\end{array}$$

Full text

1. Introduction

Leonard Carlitz [1] enumerated compositions with adjacent parts being different. We will count multipermutations of $1^{k},2^{k},\ldots,n^{k}$ with the same condition.

Definition 1.1**.**

A multipermutation is Carlitz if adjacent elements are different.

For $k=1$, these are just the $n!$ ordinary permutations, but for $k>1$ there are few results. OEIS has entries A114938 for $k=2$, where an expression and a three-term recurrence is given, and A193638 for $k=3$, but with no formula and only a conjectured recurrence.

Let $A_{k}(n)$ be the set of Carlitz multipermutations of $1^{k},2^{k},\ldots,n^{k}$ and let $a_{k}(n)=|A_{k}(n)|$. The simplest examples are

[TABLE]

The numbers grow very fast. An upper bound is of course $(kn)!/(k!)^{n}$, the number of all multipermutations.

We see that $A_{2}(2)$ has two elements, but only one pattern, $xyxy$. If we identify elements with the same pattern, we get a smaller set $A_{k}^{\prime}(n)$. Every pattern may be realized in $n!$ ways as a multipermutation, so $a_{k}^{\prime}(n)=a_{k}(n)/n!$ as the examples show.

[TABLE]

As representative we choose the ordered multipermutation, where the elements appear in order. For any pattern, such as $zyzxyxyxz$, the order condition determines what numeral each letter represents, in this case $121323231$.

Sometimes, it seems more natural to work with $a_{k}^{\prime}(n)$, sometimes $a_{k}(n)$ is more convenient. OEIS has entries A278990 for $a_{2}^{\prime}(n)$, with formula and a three-term recurrence, and A190826 for $a_{3}^{\prime}(n)$ with no formula and an only conjectured recurrence.

2. Inclusion-exclusion formulas

Computing $a_{2}(n)$ by inclusion-exclusion is Example 2.2.3 in [3]. We show the method for $a_{2}(3)=30$.

To begin with, there are $6!/2^{3}=90$ multipermutations of $1\,1\,2\,2\,3\,3$. We subtract all containing the subpattern $1\!1$, i.e. multipermutations of the five symbols $1\!1\,2\,2\,3\,3$. These are $5!/2^{2}$. The same goes for $2\!2$ and $3\!3$ so we subtract $\binom{3}{1}5!/2^{2}=90$. Patterns with both $1\!1$ and $2\!2$ were subtracted twice, so we add $4!/2^{1}$ for every such pair, totalling $\binom{3}{2}4!/2^{1}=36$. Finally, patterns with all three $1\!1$, $2\!2$, $3\!3$ must be subtracted, that is $\binom{3}{3}3!/2^{0}=6$.

The general formula looks like this.

Proposition 2.1**.**

[TABLE]

The sum is to be taken over nonnegative $s,t$ that add upp to $n$. Here $s$ counts symbols that are separate, like $..x..x..$, and $t$ counts sybols that appear together, like $..xx..$, so there are $2s+t$ blocks to permute and $s$ indistinguishable pairs.

The case $k=3$ is trickier as we now have three subpatterns to consider. If $s$ of the symbols appear separated, like $.x.x.x.$, $t$ of the symbols appear two-plus-one, like $.xx.x.$, and $u$ of the symbols appear united, like $.xxx.$, inclusion-exclusion will produce a surprisingly simple formula. A more thorough treatment is given in Martin’s thesis [7].

Theorem 2.2**.**

[TABLE]

Proof.

A direct application of inclusion-exclusion would be possible if we knew how many multipermutations contain $1\!1$, how many contain $1\!1$ and $2\!2$ etc. The $t=2,u=0$ counts permutations of blocks, some of length 1 and some of length 2, for example $1\!1$ and $2\!2$. This will produce all desired multipermutations, but some of them will be counted twice, for $1\!11$ is the same sequence as $11\!1$. So we must subtract permutations where the ones are united, and this explains the term $t=1,u=1$. But now again we must add permutations with both $1\!1\!1$ and $2\!2\!2$ and this explains the term $t=0,u=2$. ∎

Let us try to compute $a_{3}(3)=174$ with the formula.

[TABLE]

It is easy to write down similar formulas for $k\geq 4$. We just give $k=4$ as an example. The proof has no new twists, so we omit it. Just note that $v$ and $w$ count $xx..xx$ resp. $xxxx$.

Theorem 2.3**.**

[TABLE]

We were able to give each term a combinatorial interpretation but the formulas are not new. Ira Gessel [2] used rook polynomials to derive more general expressions than these and Jair Taylor [4] proved the same formulas directly from the generating function. Their elegant version of Th.2.3 is

[TABLE]

where $\Phi(t^{n})=n!$, so after expansion each power of $t$ is replaced with a factorial.

3. Recurrences

For many purposes, recurrences are superior to the explicit formulas of last section. We will show how to get recurrences for $a_{k}^{\prime}(n)$, the number of ordered Carlitz multipermutations. Recall that $a_{k}^{\prime}(n)=a_{k}(n)/n!$.

The OEIS [5] gives conjectural three-term recurrences for $a_{2}(n)$ and $a_{2}^{\prime}(n)$, a four-term recurrence for $a_{3}(n)$ and a five-term recurrence for $a_{3}^{\prime}(n)$. All these conjectures will be proved below.

Theorem 3.1**.**

The sequence $p_{n}$, recursively defined by

[TABLE]

counts ordered Carlitz words of $1^{2},\ldots,n^{2}$.

Proof.

As $p_{n}=a_{2}^{\prime}(n)$, $p_{2}=1$ counts the word 1212 and $p_{3}=5$ counts the words 010212,012012,012102,012120,012021, using symbols 012. The first four words are of the type 0..0̂., that is the zero may be removed without violating the Carlitz property, but the fifth word is of the type 0..x0x..

Now, we count words in $0^{2},1^{2},\ldots,n^{2}$. according to type.

0..0̂. is counted by $2np_{n}$ (insert 0̂ anywhere).

0..x0x for $x>1$ is counted by $p_{n}$ (transform 1..1 $\mapsto$ 0..x0x).

0101.. is counted by $p_{n-1}$ (prefix 0101). ∎

In our example, 1212 $\mapsto$ 02x0x2, which is the same pattern as 012021.

Theorem 3.2**.**

The sequences $p_{n},q_{n}$, recursively defined by

[TABLE]

count ordered Carlitz words of $1^{3},\ldots,n^{3}$ resp. of $0^{2},1^{3},\ldots,n^{3}$.

Proof.

$p_{2}=1$ counts the word 121212 and $q_{2}=8$ counts the words 010̂21212,…,01212120̂,01202121,01212021. The first six words of the type 0..0̂. are counted by $3np_{n}$, the last two 0..x0x..x. and 0..x..x0x. with $x>1$ by $2p_{n}$. Finally, 0101..1. and 01..101. are counted by $2q_{n-1}$. This proves the recurrence for $q_{n}$.

We now count $p_{n+1}$ by cases according to type of $\tt 0$. As there are two noninitial zeros, the cases will sum to $2p_{n+1}$.

0..0..0̂. is counted by $(3n-1)q_{n}$ (insert 0̂ in empty slot).

0..0..x..x0x. for $x>1$ is counted by $2(q_{n}-p_{n}-q_{n-1})$, for our transformation 1..0..1..1. $\mapsto$ 0..0..x..x0x. does not work for 101..1. (counted by $p_{n}$) or for 10..1..1. (counted by $q_{n-1}$) .

0101..1..0 and the equinumerous 01..101..0 split into subcases depending on the position of the other zero.

010101.. is counted by $p_{n-1}$.

01010..1., 0101..01. and 0101..10. are counted by $3q_{n-1}$.

0101..0..1. and 0101..1..0. are counted by $2p_{n}$.

Collecting terms and replacing $2q_{n-1}$ with $q_{n}-(3n+2)p_{n}$ we get the recurrence for $p_{n+1}$. ∎

Corollary 3.3**.**

The recursively defined sequence

[TABLE]

where $\lambda=(9n^{2}\!+\!9n\!+\!8)/2\ +2/n$, $\mu=(6n\!+3)-4/n$, $\nu=-2-2/n$ counts ordered Carlitz words of $1^{3},\ldots,n^{3}$

Proof.

Lowering indices in Th.3.2 we get

[TABLE]

Adding $-2-\frac{2}{n}$ times this to the $2p_{n+1}$-recurrence and then using the $q_{n}$-recurrence , we get the desired four-term recurrence. ∎

The five-term recurrence in OEIS entry A190826 was found by Richard J. Mathar using an ansatz with twenty unknown coefficients [6]. It is of course easily derived by adding two versions of our four-term recurrence, one of them with lowered indices.

The four-term recurrence in OEIS entry A193638 was found by Alois P. Heintz. It is now a corollary obtained by multiplication with $(n+1)!$.

Recurrences for $a_{k}^{\prime}(n)$ with $k>3$ may be derived in exactly the same way. We state the result for $k=4$ here and leave the details to the reader.

Theorem 3.4**.**

The sequences $p_{n},q_{n},r_{n}$, recursively defined by

[TABLE]

count ordered Carlitz words of $1^{4},\ldots,n^{4}$ resp. of $0^{3},1^{4},\ldots,n^{4}$, and of $0^{2},1^{4},\ldots,n^{4}$.