A New Quantity Counted by OEIS Sequence A006012

TL;DR

This paper proves a conjecture linking a specific recursive sequence to the count of permutations avoiding four complex patterns, enhancing understanding of pattern-avoiding permutations.

Contribution

It establishes a proof that the recursive sequence from OEIS sequence A006012 counts permutations avoiding four particular generalized patterns.

Findings

Confirmed the sequence counts pattern-avoiding permutations

Connected recursive sequence to permutation pattern avoidance

Enhanced understanding of permutation pattern enumeration

Abstract

We prove an existing conjecture that the sequence defined recursively by $a_1=1, a_2=2, a_n=4a_{n-1}-2a_{n-2}$ counts the number of length-$n$ permutations avoiding the four generalized permutation patterns 1-32-4, 1-42-3, 2-31-4, and 2-41-3.

Full text

A New Quantity Counted by OEIS Sequence A006012

Yonah Biers-Ariel

Rutgers University

We prove a conjecture of Callan in [1] that OEIS sequence A006012 counts a certain kind of permutation. Call this sequence $(a_{n})_{n=1}^{\infty}$; then $a_{n}$ is defined by $a_{1}=1$, $a_{2}=2$, and $a_{n}=4a_{n-1}-2a_{n-2}$ (the actual sequence in the OEIS is offset by one, so $a_{0}=1$, $a_{1}=2$, and the recursion is the same). The conjecture states that $a_{n}$ is equal to the number of permutations of length $n$ for which no subsequence $abcd$ has the following two properties: $c=b+1$ and $\max\{a,c\}<\min\{b,d\}$.

We can rewrite this conjecture in the language of pattern avoidance, in particular, using the dashed notation for generalized pattern avoidance introduced in [2]. Therefore, we define a pattern to be a permutation $\pi_{1}...\pi_{k}$, some (or all) of whose elements may be separated by dashes. We say that a subsequence of a permutation is an occurrence of a pattern if (i) all the elements have the same relative order as the elements of the pattern, and (ii) if there is no dash between the $i\textsuperscript{th}$ and $i+1\textsuperscript{th}$ elements of the pattern, then the $i\textsuperscript{th}$ and $i+1\textsuperscript{th}$ element of the subsequence occur consecutively in the permutation. We say that a permutation avoids a pattern if it does not contain any occurrence of the pattern, and a permutation avoids a set of patterns if it does not contain any occurrence of any of them. If $A$ is a set of patterns, we will write Av$(A)$ for the set of permutations which avoid them, and $\text{Av}_{n}(A)$ for the set of length-$n$ permutations which avoid them. The following two examples should help clarify these definitions.

Example: The permutation 251346 contains the subsequence 5146 which is an occurrence of the pattern 3-1-24 because the elements of the subsequence occur in the same relative order as 3124, and the 4 and 6 are consecutive in the original permutation (the 5 and 1 are also consecutive - that is allowed but not necessary).

Example: The permutation 251346 avoids 32-1-4 (i.e. $251346\in\text{Av}_{n}(\{32\text{-}1\text{-}4\})\subseteq\text{Av}(\{32\text{-}1\text{-}4\})$).

Using this notation, we rewrite the conjecture as $a_{n}=|\text{Av}_{n}(\{1\text{-}32\text{-}4,1\text{-}42\text{-}3,2\text{-}31\text{-}4,2\text{-}41\text{-}3\}|$.

We will prove two propositions. The first is that if $A=\{1\text{-}32\text{-}4,1\text{-}42\text{-}3,2\text{-}31\text{-}4,2\text{-}41\text{-}3\}$ and $B=\{1\text{-}3\text{-}2\text{-}4,1\text{-}4\text{-}2\text{-}3,2\text{-}3\text{-}1\text{-}4,2\text{-}4\text{-}1\text{-}3\}$, then Av$(A)$ and Av$(B)$ are the same set. The second proposition is that $|\text{Av}_{n}(B)|$ follows the defining recurrence of $a_{n}$, i.e. $|\text{Av}_{1}(B)|=1,|\text{Av}_{2}(B)|=2,|\text{Av}_{n}(B)|=4|\text{Av}_{n-1}(B)|-2|\text{Av}_{n-2}(B)|$.

Proposition 1: Let $A=\{1\text{-}32\text{-}4,1\text{-}42\text{-}3,2\text{-}31\text{-}4,2\text{-}41\text{-}3\}$ and $B=\{1\text{-}3\text{-}2\text{-}4,1\text{-}4\text{-}2\text{-}3,2\text{-}3\text{-}1\text{-}4,2\text{-}4\text{-}1\text{-}3\}$. The sets Av$(A)$ and Av$(B)$ are the same.

Proof: We will show that any permutation containing an occurrence of an element of $B$ must also contain an occurrence of an element of $A$ (the converse is immediately clear). Let $\pi$ be a permutation. First, note that a subpermutation $\pi_{a}\pi_{b}\pi_{c}\pi_{d}$ of $\pi$ is an occurrence of a pattern in $A$ if and only if $c=b+1$ and $\max\{\pi_{a},\pi_{c}\}<\min\{\pi_{b},\pi_{d}\}$ (in fact, this is the definition Callan provides in the OEIS). Similarly, a subpermutation $\pi_{a}\pi_{b}\pi_{c}\pi_{d}$ of $\pi$ is an occurrence of a pattern in $B$ if and only if $\max\{\pi_{a},\pi_{c}\}<\min\{\pi_{b},\pi_{d}\}$.

Choose any element of $B$, and suppose that $\pi$ contains an occurrence of this element. As noted above, this means that we can find $a<b<c<d$ such that $\max\{\pi_{a},\pi_{c}\}<\min\{\pi_{b},\pi_{d}\}$. Let $e$ be the largest index less than $c$ such that $\pi_{e}>\max\{\pi_{a},\pi_{c}\}$, i.e. $e=\max\{i:i<c,\pi_{i}>\max\{\pi_{a},\pi_{c}\}\}$. Because $b$ is an element of $\{i:i<c,\pi_{i}>\max\{\pi_{a},\pi_{c}\}\}$, it follows that $e$ exists and $a<b\leq e<e+1\leq c<d$. Now, we claim that $\pi_{a}\pi_{e}\pi_{e+1}\pi_{d}$ is an occurrence of a pattern in $A$. Obviously $e+1=e+1$, and so it remains to check that $\max\{\pi_{a},\pi_{e+1}\}<\min\{\pi_{e},\pi_{d}\}$. Because $\max\{\pi_{a},\pi_{c}\}<\min\{\pi_{b},\pi_{d}\}$, we conclude that $\pi_{a}<\pi_{d}$ and by the choice of $e$, we also have $\pi_{a}<\pi_{e}$. Now, either $e+1=c$, in which case $\pi_{e+1}=\pi_{c}$, or else $\pi_{e+1}<\max\{\pi_{a},\pi_{c}\}$ because otherwise we would have chosen $e+1$ as the $\max\{i:i<c,\pi_{i}>\max\{\pi_{a},\pi_{c}\}\}$ instead of $e$. It follows that $\pi_{e+1}\leq\max\{\pi_{a},\pi_{c}\}<\pi_{d},\pi_{e}$ for the same reasons as $\pi_{a}$. Therefore, $\max\{\pi_{a},\pi_{e+1}\}<\min\{\pi_{e},\pi_{d}\}$ and $\pi_{a}\pi_{e}\pi_{e+1}\pi_{d}$ is an occurrence of a pattern in $A$. We conclude that the permutations avoiding the patterns of $A$ are the same as the permutations avoiding the patterns of $B$.

Proposition 2: The number of permutations of length $n$ avoiding all patterns in $B$ (and hence in $A$) satisfies the recurrence $a_{1}=1,a_{2}=2,a_{n}=4a_{n-1}-2a_{n-2}$.

Proof: Since $\text{Av}_{1}(B)=\{1\}$ and $\text{Av}_{2}(B)=\{12,21\}$, the initial conditions hold. Our strategy will be as follows: given $\text{Av}_{n}(B)$, define four maps which, when all of them are applied to all the permutations of $\text{Av}_{n-1}(B)$, will generate all of the permutations of $\text{Av}_{n}(B)$. Then, we will count how many permutations of $\text{Av}_{n}(B)$ are double counted in this way, and find that there are two for every element of $\text{Av}_{n-2}(B)$, thereby establishing the recurrence.

Note that, for a permutation to avoid all patterns of $A$, it must be the case that either 1 and 2 occur consecutively (not necessarily in that order) or either 1 or 2 is the last element of the permutation. This observation motivates the following definitions of the four maps $f_{\text{before}},f_{\text{after}},f_{\text{end}},f_{\text{bump}}$. Let $f_{\text{before}}$ be the function that inputs a permutation and outputs that permutation with all elements increased by 1 and a 1 inserted immediately before the new 2. Let $f_{\text{after}}$ be the function that also inputs a permutation and outputs that permutation with all the elements increased by 1 and a 1 inserted immediately after the new 2. Similarly, let $f_{\text{end}}$ be the function that inputs a permutation, increases all its elements by 1 and puts a 1 at the end of it, and let $f_{\text{bump}}$ be the function that inputs a permutation, increases all its elements by 1, replaces the new 2 with a one and puts a 2 at the end. The following example gives a concrete illustration of the four functions.

Example: Let $\pi=31542$. Then $f_{\text{before}}(\pi)=412653,\allowbreak f_{\text{after}}(\pi)=421653,\allowbreak f_{\text{end}}(\pi)=426531,\allowbreak\text{ and }f_{\text{bump}}(\pi)=416532$. Note that $\pi\in\text{Av}(B)$, and so are all its images.

We claim that (i) these four functions all map elements of $\text{Av}_{n-1}(B)$ to elements of $\text{Av}_{n}(B)$ and (ii) $f_{\text{before}}(\text{Av}_{n-1}(B))\cup f_{\text{after}}(\text{Av}_{n-1}(B))\cup f_{\text{end}}(\text{Av}_{n-1}(B))\cup f_{\text{bump}}(\text{Av}_{n-1}(B))\supseteq\text{Av}_{n}(B)$ (by claim (i), we could replace the ‘$\supseteq$’ in claim (ii) with ‘$=$’). To verify the first claim, choose some $\sigma\in\text{Av}_{n-1}(B)$, and consider each function in turn. If $f_{\text{before}}(\sigma)$ or $f_{\text{after}}(\sigma)$ contains an occurrence of a pattern in $A$, then this occurrence must use no more than 1 of the elements 1 and 2 (because they are consecutive in both $f_{\text{before}}(\sigma)$ and $f_{\text{after}}(\sigma)$ but can’t be in any pattern in $A$). Therefore, either this occurrence fails to use 1 and would have already been an occurrence of the pattern in $\sigma$, or else it fails to use 2, in which case it could have used 2 instead of 1 and been an occurrence of the pattern in $\sigma$. Thus, no such occurrence is possible in $f_{\text{before}}(\sigma)$ or $f_{\text{after}}(\sigma)$. In addition, if $f_{\text{end}}(\sigma)$ or $f_{\text{bump}}(\sigma)$ contains an occurrence of a pattern in $A$, then this occurrence cannot use the last element because that element is either a 1 or a 2, and patterns in $A$ only end with $3$ or $4$. So, this occurrence would already be an occurrence of the pattern in $\sigma$, and therefore cannot exist.

To verify the second claim, chose some $\pi\in\text{Av}_{n}(B)$. As previously noted, either 1 and 2 occur consecutively in $\pi$, or else either 1 or 2 is the final element of $\pi$. Let $\pi^{\prime}$ be $\pi$ with the 1 removed and each element decreased by 1. We have introduced no new patterns, and so $\pi^{\prime}\in\text{Av}_{n-1}(B)$. Suppose that 1 occurs immediately before 2 in $\pi$, then $f_{\text{before}}(\pi^{\prime})=\pi$. If the 1 occurs immediately after 2 in $\pi$, then $f_{\text{after}}(\pi^{\prime})=\pi$. If the 1 occurs at the end of $\pi$, then $f_{\text{end}}(\pi^{\prime})=\pi$. If the 2 occurs at the end of $\pi$, then we will need to define $\pi^{\prime\prime}$ which is $\pi$ with the 1 removed, the 2 moved the position where the 1 used to be, and each element decreased by 1. Again, we have introduced no new patterns, and so $\pi^{\prime\prime}\in\text{Av}_{n-1}(B)$, and $f_{\text{bump}}(\pi^{\prime\prime})=\pi$.

If these four functions all had disjoint ranges, we could conclude that $a_{n}=4a_{n-1}$. Unfortunately, some permutations are counted twice. Each $f$ outputs a certain kind of permutation: $f_{\text{before}}$ outputs permutations where 1 immediately precedes 2, $f_{\text{after}}$ outputs permutations where 2 immediately precedes 1, $f_{\text{end}}$ outputs permutations where 1 occurs at the end, and $f_{\text{bump}}$ outputs permutations where 2 occurs at the end. If a permutation fulfills two of these criteria it will be double-counted. Such permutations must be counted once by either $f_{\text{before}}$ or $f_{\text{after}}$ and again by either $f_{\text{end}}$ or $f_{\text{bump}}$ because no permutation can be counted by both $f_{\text{before}}$ and $f_{\text{after}}$ or both $f_{\text{end}}$ and $f_{\text{bump}}$. Thus, the final two elements of such permutations are 1 and 2 (not necessarily in that order). Let $g:\text{Av}_{n}(B)\rightarrow\text{Av}_{n-2}(B)$ be defined as the function which takes a permutation, removes from it the elements 1 and 2, and reduces all other elements by 2. If we restrict $g$ to those permutations which end in either 12 or 21, $g$ becomes a 2-to-1 map from the double-counted permutations of $\text{Av}_{n}(B)$ to the permutations of $\text{Av}_{n-2}(B)$, and so the number of double-counted permutations is twice $a_{n-2}$. It follows that $a_{n}=4a_{n-1}-2a_{n-2}$.

The author would like to thank his (intended) advisor Doron Zeilberger for introducing him to the conjecture and fixing typos in the original draft.