School of Education, Culture and Communication
Division of Mathematics and Physics
BACHELOR’S DEGREE PROJECT IN MATHEMATICS
Counting Double-Descents and Double-Inversions in Permutations
by
Jonas Boberg
MAA322 — Examensarbete i matematik för kandidatexamen
DIVISION OF MATHEMATICS AND PHYSICS
MÄLARDALEN UNIVERSITY SE-721 23 VÄSTERÅS, SWEDEN
School of Education, Culture and Communication
Division of Mathematics and Physics
MAA322 — Bachelor’s Degree Project in Mathematics
Date of presentation:
3rd June 2021
Project name:
Counting Double-Descents and Double-Inversions in Permutations
Author: Jonas Boberg Version: 3rd June 2021 Supervisor: Fredrik Jansson Reviewer: Masood Aryapoor Examiner: Christopher Engström Comprising: 15 ECTS credits
Abstract
In this paper, new variations of some well-known permutation statistics are introduced and studied. Firstly, a double-descent of a permutation 𝜋 is defined as a position 𝑖 where 𝜋𝑖 ≥ 2𝜋𝑖+1. By proofs by induction and direct proofs, recursive and explicit expressions for the number of 𝑛-permutations with 𝑘 double-descents are presented. Also, an expression for the total number of double-descents in all 𝑛-permutations is presented. Secondly, a double-inversion of a permutation 𝜋 is defined as a pair (𝜋𝑖, 𝜋𝑗) where 𝑖 < 𝑗 but 𝜋𝑖 ≥ 2𝜋𝑗. The total number of double-inversions in all 𝑛-permutations is presented.
Acknowledgements
I would like to thank my supervisor Fredrik Jansson for assisting me to come up with this topic, and for guiding me throughout the work with this thesis. I would also like to thank the reviewer of this thesis, Masood Aryapoor, who assisted with helpful comments.
Contents
Acknowledgements 2
1 Introduction and Literature Review 4
1.1 Preliminaries . . . 4
2 The Descent and Major Index Statistics 4
2.1 Introducing Double-Descents and Major Indices of Degree 2 . . . 6
3 The Inversion Statistic 14
3.1 Introducing Double-Inversions . . . 15
1
Introduction and Literature Review
Considering the sequence of numbers or more precisely the permutation 2143, we would all agree on the fact that it is out of its natural order. The question of how much it is out of order might feel like a stranger one. However, combinatorists have come up with permutation statistics like inversions and descents that, among other things, make it possible to quantify this property.
Questions like these are a matter of the subject of combinatorics which in the early 1900s developed rapidly. One leading and well-known mathematician was MacMahon who in 1915 published Combinatory analysis [6] which is a comprehensive work in combinatorics. Another, perhaps less known, mathematician who had an important role in the development of combinatorics was Netto, who in 1901 published Lehrbuch der combinatorik [7]. Both of these works laid a foundation for today’s combinatorial knowledge.
More recently, Kitaev and Remmel published two papers where they study refinements of the descent statistic [3, 4]. They introduce and study an adjustment of the statistic where one of the integers in the pair causing the descent is either odd or even. They then study a generalization where one of the integers in the pair causing the descent is equivalent to 0 mod 𝑘 for any 𝑘 ≥ 2.
Simultaneously as Kitaev’s and Remmel’s work was written, Jansson wrote and published a work on variations of another permutation statistic, namely excedances [2]. He adjusted the permutation set to only look at excedances of even integers or of odd integers.
I will in this paper introduce new variations of some classical permutation statistics, namely descents, major indices and inversions, and study how these variations are distributed and how they behave.
1.1
Preliminaries
The following definitions and notations are needed as a basis of knowledge for this paper.
Definition 1. Let 𝑆 be a set with 𝑛 elements. Then, a permutation of 𝑆 is a linear ordering of
the elements in 𝑆.
• We denote the set of all natural numbers up to and including 𝑛 as [𝑛]. For example {1, 2, 3} = [3].
• The set of all permutations of the set [𝑛] is denoted S𝑛. Such a permutation is called an
𝑛-permutation.
2
The Descent and Major Index Statistics
A permutation statistic can be used to quantify and analyze a certain property of a set of permutations, and is defined as a function that converts a permutation into a natural number, that is 𝑓 : S𝑛 → N, [5]. Two well-known statistics in combinatorics are the descent and
the major index statistics. In this section, the definitions and some theorems regarding these statistics will be presented.
Definition 2. Let 𝜋 = 𝜋1𝜋2. . . 𝜋𝑛be a permutation. Then 𝑖 < 𝑛 is a descent if 𝜋𝑖 > 𝜋𝑖+1. The set of all descents in 𝜋 is denoted 𝐷 (𝜋) = {𝑖 : 𝜋𝑖 ≥ 𝜋𝑖+1}, and 𝑑𝑒𝑠(𝜋) = |𝐷 (𝜋)|.
Remark. The permutation statistic in this case is 𝑑𝑒𝑠(𝜋) which is a function from S𝑛to N. Definition 3. The major index of a permutation 𝜋, denoted 𝑚𝑎 𝑗 (𝜋) is the sum of all descents
in 𝜋.
Example 1. Consider the permutation 𝜋 = 14352, then the descent set is 𝐷 (𝜋) = {2, 4}, and
therefore 𝑑𝑒𝑠(𝜋) = 2 and 𝑚𝑎 𝑗 (𝜋) = 2 + 4 = 6.
When encountering statistics like these, some intuitive combinatorial questions could be "how many descents are there in all 𝑛-permutations?", "how are the descents distributed over all 𝑛-permutations?" or "how are the major indices distributed over all 𝑛-permutations?". All of these questions have been answered in different ways, some of which will be presented below.
We will begin with a recursive formula for the number of 𝑛-permutations with 𝑘 descents.
Theorem 1. [1, Thm. 1.7] Let 𝑑𝑒𝑠(𝑛, 𝑘) be the number of 𝑛-permutations with 𝑘 descents,
then for all non-negative integers 𝑘 and all positive integers 𝑛 satisfying 𝑘 ≤ 𝑛 −1, the following
recursive formula holds
𝑑𝑒 𝑠(𝑛, 𝑘) = (𝑘 + 1) · 𝑑𝑒𝑠(𝑛 − 1, 𝑘) + (𝑛 − 𝑘) · 𝑑𝑒𝑠(𝑛 − 1, 𝑘 − 1),
where
𝑑𝑒 𝑠(1, 0) = 1.
Proof. Consider an (𝑛 − 1)-permutation with 𝑘 − 1 descents. To create a 𝑘𝑡 ℎ
descent when inserting 𝑛 the first option is to insert it at a position such that it does not break an already existing descent. There are (𝑛 − 2) − (𝑘 − 1) = 𝑛 − 𝑘 − 1 such positions. The second option is to insert 𝑛 in the first position. Therefore there are 𝑛 − 𝑘 positions for each permutation in 𝑑𝑒 𝑠(𝑛 − 1, 𝑘 − 1).
Now consider an (𝑛 − 1)-permutation with 𝑘 descents. Then we must either insert 𝑛 at a position such that it breaks an already existing descent, there are 𝑘 such positions, or we can insert it in the last position. So there are 𝑘 + 1 positions for each permutation in 𝑑𝑒𝑠(𝑛 − 1, 𝑘). Since every 𝑛-permutation arises from an (𝑛 − 1)-permutation in exactly one of these two
ways the proof is complete.
Remark. In [1], the expression (denoted 𝐴(𝑛, 𝑘)) provides the number of 𝑛-permutations with 𝑘 − 1 descents. The reason for the shift in 𝑘 is that a permutation with 𝑘 − 1 descents has 𝑘 ascending runs. However, since ascending runs are not mentioned in this paper, the expression in this theorem is adjusted so that it fits our notation.
This can also be expressed in an explicit formula as in the following theorem.
Theorem 2. [1, Thm. 1.11] For all non-negative integers 𝑛 and 𝑘 satisfying 𝑘 ≤ 𝑛 − 1, the
following explicit formula holds.
𝑑𝑒 𝑠(𝑛, 𝑘) = 𝑘+1 Õ 𝑖=0 (−1)𝑖 𝑛+ 1 𝑖 (𝑘 + 1 − 𝑖)𝑛 .
Remark. The proof of this theorem is found in [1].
Lastly, we will look at the sum of all major indices in all 𝑛-permutations.
Proposition 1. The sum of major indices in all 𝑛-permutations is
Õ
𝜋∈S𝑛
𝑚 𝑎 𝑗(𝜋) =
𝑛!𝑛(𝑛 − 1) 4 .
Proof. Consider a pair (𝜋𝑖, 𝜋𝑗) satisfying 1 ≤ 𝜋𝑗 < 𝜋𝑖 ≤ 𝑛. This pair will be at position
(𝑚, 𝑚 + 1), in (𝑛 − 2)! permutations for every 𝑚 such that 1 ≤ 𝑚 ≤ 𝑛 − 1. Therefore, the major index that this pair generates in all permutations is
(𝑛 − 2)! · 1 + (𝑛 − 2)! · 2 + (𝑛 − 2)! · 3 + . . . + (𝑛 − 2)! · (𝑛 − 1) = (𝑛 −2)! · (1 + 2 + 3 + . . . + (𝑛 − 1)) = (𝑛 −2)! · (𝑛 − 1)𝑛 2 .
Since there are a total of 1 + 2 + 3 + . . . + (𝑛 − 1) = 𝑛(𝑛−1)
2 pairs (𝑎, 𝑏) of integers satisfying
1 ≤ 𝑏 < 𝑎 ≤ 𝑛, the sum of all major indices is
Õ 𝜋∈S𝑛 𝑚 𝑎 𝑗(𝜋) = (𝑛 − 2)! · (𝑛 − 1)𝑛 2 · 𝑛(𝑛 − 1) 2 = 𝑛! 2 · 𝑛(𝑛 − 1) 2 = 𝑛!𝑛(𝑛 − 1) 4 .
Remark. In the sources used for this thesis, I did not find a proof for this proposition although it is most certainly proved somewhere.
2.1
Introducing Double-Descents and Major Indices of Degree 2
In this section we will make some adjustments on these two statistics and study them to find out if we can find the same type of results as those that are known for descents and major indices. Consider the following two definitions.
Definition 4. Let 𝜋 = 𝜋1𝜋2. . . 𝜋𝑛 be a permutation. Then 𝑖 < 𝑛 is a double-descent if
𝜋𝑖 ≥ 2𝜋𝑖+1. The set of all double-descents in 𝜋 is denoted 𝐷2(𝜋) = {𝑖 : 𝜋𝑖 ≥ 2𝜋𝑖+1}, and 𝑑𝑒 𝑠2(𝜋) = |𝐷2(𝜋) |.
Definition 5. The major index of degree 2 of 𝜋, denoted 𝑚𝑎 𝑗2(𝜋), is the sum of all
double-descents in 𝜋.
Example 2. In the permutation 𝜋 = 15342, we see that 2 and 4 are descents and therefore
We can easily understand that there, in general, are fewer double-descents than normal ones. But can we answer the same questions we just looked at for the case of normal descents and major indices?
To find the desired expressions we will firstly present a general integer lemma that will be necessary for the later proofs. Consider the permutations of {1, 2, 3}, S3, we can quite quickly
realise that only the element 1 can generate a double-descent since 3 2 · 2. Further, the element 1 always generate a double-descent since 2 · 1 always is less than or equal to any other integer greater than or equal to 2. Looking closer at how many pairs generate double-descents we get the following lemma.
Lemma 1. The number of pairs (𝑖, 𝑗 ) where 1 ≤ 𝑖, 𝑗 ≤ 𝑛 and 𝑖 ≠ 𝑗 such that 𝑖 ≥ 2 𝑗 is
j
𝑛2 4
k .
Proof. This will follow a recursive pattern. We denote the number of pairs with the desired property 𝑎𝑛. Then 𝑎𝑛 = 𝑎𝑛−1 + 𝑏𝑛 where 𝑏𝑛 is the number of pairs (𝑛, 𝑗 ) such that 𝑛 ≥ 2 𝑗 .
Consider an even 𝑛, then 𝑏𝑛 = |{(𝑛,1), (𝑛, 2), . . . , (𝑛, 𝑛 2)}| =
𝑛
2. Now consider an odd 𝑛, then
𝑏𝑛= |{(𝑛,1), (𝑛, 2), . . . , (𝑛, 𝑛 2− 1 2)}| = 𝑛−1
2 . We then get that 𝑏𝑛−1+ 𝑏𝑛is one of the following:
𝑛even: 𝑏𝑛−1+ 𝑏𝑛= (𝑛 − 1) − 1 2 + 𝑛 2 = (𝑛 − 2) 2 + 𝑛 2 = 2𝑛 − 2 2 = 𝑛 −1 𝑛odd: 𝑏𝑛−1+ 𝑏𝑛 = 𝑛− 1 2 + 𝑛− 1 2 = 2(𝑛 − 1) 2 = 𝑛 −1.
So we see that 𝑎𝑛increases by 𝑛 − 1 when 𝑛 increases by 2, that is 𝑎𝑛 = 𝑎𝑛−2+ 𝑏𝑛−1+ 𝑏𝑛 =
𝑎𝑛−2+ 𝑛 − 1. We also know that 𝑎1 = 0 and 𝑎2 =1, so we get the following inhomogeneous recurrence relation: 𝑎𝑛 = 𝑎𝑛−2+ 𝑛 − 1 𝑎1=0 𝑎2=1. Assuming 𝑎ℎ 𝑛 = 𝐴1𝑟 𝑛
+ 𝐴2𝑟𝑛to solve the homogeneous equation
( 𝐴1+ 𝐴2)𝑟𝑛 − ( 𝐴1+ 𝐴2)𝑟𝑛−2 = ( 𝐴1+ 𝐴2) (𝑟𝑛 − 𝑟𝑛−2 ) = 0, we get 𝑟 = ±1 so 𝑎ℎ 𝑛 = 𝐴1+ 𝐴2(−1) 𝑛 .
To find the particular solution we assume 𝑎𝑛𝑝 = 𝐴𝑛
2+ 𝐵𝑛 +𝐶 and get the following equation,
𝐴𝑛2+ 𝐵𝑛 + 𝐶 − ( 𝐴(𝑛 − 2)2+ 𝐵(𝑛 − 2) + 𝐶)
= 𝐴𝑛2+ 𝐵𝑛 + 𝐶 − ( 𝐴𝑛2− 4𝐴𝑛 + 4𝐴 + 𝐵𝑛 − 2𝐵 + 𝐶) =4𝐴𝑛 − 4𝐴 + 2𝐵 = 𝑛 − 1,
which has the following solutions, ( 4𝐴 = 1 −4𝐴 + 2𝐵 = −1 =⇒ ( 𝐴=1/4 𝐵=0.
So we have 𝑎𝑛 = 𝑎 ℎ 𝑛+ 𝑎 𝑝 𝑛 = 𝐴1+ 𝐴2(−1) 𝑛 + 𝑛2
4. We can now use the initial values to solve for
𝐴1and 𝐴2. ( 𝑎1 = 1 4+ 𝐴1− 𝐴2=0 𝑎2 =1 + 𝐴1+ 𝐴2 =1 =⇒ ( 𝐴1= −1 8 𝐴2= 1 8.
So we get the equation
𝑎𝑛 = 𝑎ℎ 𝑛+ 𝑎 𝑝 𝑛 = 𝑛2 4 − 1 8 + (−1)𝑛 8 . Now assume 𝑛 = 2𝑘, we see that
𝑎𝑛= 𝑛2 4 = (2𝑘)2 4 = 𝑘 2
is clearly an integer. Assuming 𝑛 is odd, 𝑛 = 2𝑘 + 1, the expression can be simplified as follows,
𝑎𝑛 = 𝑛2 4 − 1 4 = (2𝑘 + 1) 2 4 − 1 4 = (4𝑘 2+ 4𝑘 + 1) 4 − 1 4 = 4𝑘 2+ 4𝑘 4 + 1 4 − 1 4 = 𝑘2+ 𝑘,
which is also an integer. Therefore we can rewrite the expression to be
𝑎𝑛= 𝑛2 4 .
Remark. These pairs are all of the possible double-descents in an 𝑛-permutation.
We now know how many pairs there are, in an 𝑛-permutation, that can be double-descents. Knowing this, we can calculate the total number of double-descents in all 𝑛-permutations.
Theorem 3. The total number of double-descents in all 𝑛−permutations is
Õ 𝜋∈S𝑛 𝑑𝑒 𝑠2(𝜋) = 𝑛2 4 · (𝑛 − 1)!.
Proof. We know from Lemma 1 that there are j
𝑛2 4
k
unique pairs that can be double-descents. Each of these pairs are at position (𝑚, 𝑚 + 1) in (𝑛 − 2)! different 𝑛-permutations for all 𝑚
satisfying 1 ≤ 𝑚 ≤ 𝑛 − 1 since the 𝑛 − 2 integers that are not a part of the pair can be arranged in (𝑛 − 2)! ways. Therefore, the total number of double-descents is
𝑛2 4 · (𝑛 − 2)! · (𝑛 − 1) = 𝑛2 4 · (𝑛 − 1)!. We can also derive an expression for the sum of all major indices of degree 2 in all 𝑛−permutations with a proof very similar to the one of Theorem 3.
Proposition 2. The sum of major indices of degree 2 in all 𝑛−permutations is
Õ 𝜋∈S𝑛 𝑚 𝑎 𝑗2(𝜋) = 𝑛! 2 · 𝑛2 4 .
Proof. Again, we know from Lemma 1 that there are j
𝑛2 4
k
unique pairs that can be double-descents. Each of these pairs are at position (𝑚, 𝑚+1) in (𝑛−2)! different 𝑛-permutations for all 𝑚satisfying 1 ≤ 𝑚 ≤ 𝑛 − 1. Therefore, the major index of degree 2 that every double-descent pair generates in all permutations is
(𝑛 − 2)! · 1 + (𝑛 − 2)! · 2 + (𝑛 − 2)! · 3 + . . . + (𝑛 − 2)! · (𝑛 − 1) = (𝑛 −2)! · (1 + 2 + 3 + . . . + (𝑛 − 1)) = (𝑛 −2)! · (𝑛 − 1)𝑛 2 . So the sum of all major indices of degree 2 is
Õ 𝜋∈S𝑛 𝑚 𝑎 𝑗2(𝜋) = (𝑛 − 2)! · (𝑛 − 1)𝑛 2 · 𝑛2 4 = 𝑛! 2 · 𝑛2 4 . To get expressions for the distribution of the double-descents we will need another integer lemma. Similarly to Lemma 1, we will now derive an expression for the number of possible pairs that include 𝑛 but do not generate double-descents.
Lemma 2. There are 𝑛+1
2 − 1 positive integers 𝑝 < 𝑛 such that 𝑛 < 2𝑝.
Proof. Consider an odd integer 𝑛, then the integers 𝑛 − 1, 𝑛 − 2, . . . ,𝑛 2+
1
2satisfies the relation.
This makes a total of 𝑛 − 𝑛 2− 1 2 − 1 = 𝑛 2 − 1 2 = 𝑛 2 = 𝑛+1
2 − 1 such integers. Consider an
even integer 𝑛, then the integers 𝑛 − 1, 𝑛 − 2, . . . ,𝑛
2+ 1 satisfies the relation. This makes a total
of 𝑛 − 𝑛 2 − 1 = 𝑛 2− 1 = 𝑛+1 2 − 1 such integers.
With this lemma, we will now be able to find formulas for expressing the number of 𝑛-permutations with 𝑘 double-descents. However, we will begin by looking at a special case of that, namely the number of 𝑛-permutations with no double-descents.
Proposition 3. The number of 𝑛−permutations with no double-descents is 1 · 1 · 2 · 2 · 3 · · · | {z } 𝑛digits = j 𝑛 2 k ! · 𝑛+ 1 2 !.
Proof. We begin by noticing that
1·1·2·2·3 · · · = 2 2 − 1 + 1 3 2 − 1 + 1 4 2 − 1 + 1 5 2 − 1 + 1 6 2 − 1 + 1 · · · .
By induction where the first base case is 𝜋 = 1, there is only one way to construct a new permutation ˜𝜋with no double-descents and that is to insert 𝑛 = 2 in the last position to create
˜
𝜋 =12 which is the second base case. Here, there are two possible positions to insert 3. Either in the last position or between the elements 1 and 2 since if we were to insert the 3 at the first position we would create a double-descent.
Assuming there are 1 · 1 · 2 · 2 · 3 · · · | {z }
𝑝−1 digits
number of ( 𝑝 − 1)-permutations with no
double-descents. We can then insert 𝑝 in each of these permutations in any position so that 𝑝 does not create a double-descent with the element after it. From Lemma 2 we know that the number of positions that satisfies this is
j
𝑝+1 2
k
− 1 for all 𝑝. And since we can also insert 𝑝 in the last position there are
j
𝑝+1 2
k
− 1 + 1 valid positions for 𝑝 to be inserted. So by induction the number of 𝑛−permutations 𝜋 with no double-descents is 1 · 1 · 2 · 2 · 3 · · ·
| {z }
𝑛digits
. And we can clearly
see that 1 · 1 · 2 · 2 · 3 · · · | {z } 𝑛digits = j 𝑛 2 k ! · 𝑛+ 1 2 !
since every other integer in the product responds to the first factorial, and the other integers respond to the second factorial. This completes the proof.
Remark. An explicit expression for the number of 𝑛-permutations with 𝑘 double-descents is presented in Theorem 5, so this expression follows from that theorem. However, this proposition is included in this paper since the proof is more intuitive than the proof of the later theorem.
We will now use the same recursive principle but for any arbitrary 𝑘 number of double-descents.
Theorem 4. Let 𝑑𝑒𝑠2(𝑛, 𝑘) be the number of 𝑛-permutations with 𝑘 double-descents. Then,
for positive integers 𝑛 and non-negative integers 𝑘 satisfying 𝑘 ≤ b𝑛 2c 𝑑𝑒 𝑠2(𝑛, 𝑘) = 𝑛+ 1 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘) + j 𝑛 2 k − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘 − 1), where 𝑑𝑒 𝑠2(1, 0) = 1.
Proof. Consider an (𝑛 − 1)-permutation 𝜋 with 𝑘 double-descents. To construct an 𝑛-permutation with 𝑘 double-descents, the first option is to insert 𝑛 between position 𝑖 and position (𝑖 + 1) in 𝜋 such that the number at position 𝑖 + 1 (which after the insertion is on position (𝑖 + 2)) is greater than 𝑛
2. We know from Lemma 2 that there are
𝑛+1
2 − 1 such
positions. The second option is to insert 𝑛 between the numbers on position 𝑖 − 1 and 𝑖 where 𝑖 − 1 was a double-descent before inserting 𝑛, and we know that there are 𝑘 such positions. Note that these two options can not overlap. Lastly, we can also insert 𝑛 in the last position. This results in 𝑛+ 1 2 − 1 + 𝑘 + 1 = 𝑛+ 1 2 + 𝑘
ways to construct such a permutation for each permutation in 𝑑𝑒𝑠2(𝑛 − 1, 𝑘).
Now consider an (𝑛 − 1)-permutation 𝜋 with 𝑘 − 1 double-descents. To construct an 𝑛-permutation with 𝑘 double-descents, we need to insert 𝑛 in a position 𝑖 such that it creates a new double-descent. There are 𝑛
2 positive integers 𝑝 such that 2𝑝 ≤ 𝑛. However, if 𝑖 is
already a double-descent, we cannot insert the 𝑛 at position 𝑖 and since there are a total of 𝑘 − 1 double-descents in 𝜋, we have a total of
j 𝑛 2 k − (𝑘 − 1) = j 𝑛 2 k − 𝑘 + 1
ways to construct a new permutation for each permutations in 𝑑𝑒𝑠(𝑛 − 1, 𝑘 − 1).
Since these are the only ways to construct an 𝑛-permutation we get the recursive relation
𝑑𝑒 𝑠2(𝑛, 𝑘) = 𝑛+ 1 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘) + j 𝑛 2 k − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘 − 1). Now, we would of course like to find an explicit expression for this number, 𝑑𝑒𝑠2(𝑛, 𝑘).
Looking at the number of permutations of the first 𝑛 even positive integers with 𝑘 excedances (which in a permutation 𝜋 is an 𝑖 such that 𝜋𝑖 > 𝑖) which is found in [2, Table 1], it seems
like there is a bijection between these numbers. More precisely, it seems like the number of 𝑛-permutations with 𝑛 double-descents is the same as the number of permutations of the first 𝑛even positive integers with 𝑛 − 𝑘 excedances. This motivates the formula for the following theorem.
Theorem 5. Let 𝑛 be a positive integer and 𝑘 be a non-negative integer satisfying 𝑘 ≤ 𝑛 2 , then 𝑑𝑒 𝑠2(𝑛, 𝑘) = j 𝑛 2 k ! 𝑛+ 1 2 ! 𝑛 2 𝑛− 𝑘 − 𝑛+1 2 𝑛+1 2 𝑛− 𝑘 −𝑛 2 = 𝑛 2 ! 2 𝑛+1 2 ! 2 𝑛 2 − 𝑘 ! 𝑛+1 2 − 𝑘 !(𝑘!)2 .
Proof. Constructing this explicit expression, we use induction on the recursive expression but split the problem up in to odd and even 𝑛. Since the expression has two variables, we have to
use induction over both 𝑛 and 𝑘. The base case is that 𝑑𝑒𝑠2(1, 0) = 1 which holds so we now
use induction over 𝑛 and assume 𝑛 is odd, we then know that
𝑑𝑒 𝑠2(𝑛 + 1, 𝑘) = 𝑛+ 1 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛, 𝑘) + 𝑛+ 1 2 − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛, 𝑘 − 1).
We assume that the expression
𝑑𝑒 𝑠2(𝑛, 𝑘) = 𝑛+1 2 ! 2 𝑛 2 ! 2 𝑛+1 2 − 𝑘 ! 𝑛 2 − 𝑘 !(𝑘!)2 = 𝑛+1 2 ! 2 𝑛−1 2 ! 2 𝑛+1 2 − 𝑘 ! 𝑛−1 2 − 𝑘 !(𝑘!)2
holds (since we are assuming that 𝑛 is odd). This results in the following induction step.
𝑑𝑒 𝑠2(𝑛 + 1, 𝑘) = 𝑛+ 1 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛, 𝑘) + 𝑛+ 1 2 − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛, 𝑘 − 1) = 𝑛+ 1 2 + 𝑘 𝑛−1 2 ! 2 · 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 !𝑛−1 2 − 𝑘 !(𝑘!)2 + 𝑛+ 1 2 − 𝑘 + 1 𝑛−1 2 ! 2 · 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 + 1 ! 𝑛−1 2 − 𝑘 + 1 !((𝑘 − 1)!)2 = 𝑛+1 2 + 𝑘 𝑛−1 2 ! 2 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 !𝑛−1 2 − 𝑘 !(𝑘!)2 + 𝑛−1 2 ! 2 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 !𝑛+1 2 − 𝑘 !((𝑘 − 1)!)2 = 𝑛+1 2 − 𝑘 𝑛+1 2 + 𝑘 𝑛−1 2 ! 2 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 !2(𝑘!)2 + 𝑛−1 2 ! 2 𝑛+1 2 ! 2 (𝑘2) 𝑛+1 2 − 𝑘 !2(𝑘!)2 = 𝑛+1 2 − 𝑘 𝑛+1 2 + 𝑘 𝑛+1 2 ! 4 𝑛+1 2 2 𝑛+1 2 − 𝑘 ! 2 (𝑘!)2 + 𝑛+1 2 ! 4 (𝑘2) 𝑛+1 2 2 𝑛+1 2 − 𝑘 ! 2 (𝑘!)2 = 𝑛+1 2 ! 4 𝑛+1 2 2 − 𝑛+1 2 ! 4 𝑘2+ 𝑛+1 2 ! 4 𝑘2 𝑛+1 2 2 𝑛+1 2 − 𝑘 !2(𝑘!)2 = 𝑛+1 2 ! 4 𝑛+1 2 − 𝑘 ! 2 (𝑘!)2 ,
which is the expression for 𝑑𝑒𝑠2(𝑛 + 1, 𝑘) when 𝑛 + 1 is even.
We now do the same thing but assume 𝑛 is even, we know
𝑑𝑒 𝑠2(𝑛 + 1, 𝑘) = 𝑛+ 2 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛, 𝑘) + 𝑛 2 − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛, 𝑘 − 1), and we assume 𝑑𝑒 𝑠2(𝑛, 𝑘) = 𝑛 2 ! 2 · 𝑛+1 2 ! 2 𝑛+1 2 − 𝑘 ! 𝑛 2 − 𝑘 !(𝑘!)2 = 𝑛+1 2 ! 4 𝑛+1 2 − 𝑘 !2(𝑘!)2 .
The induction step is as follows:
𝑑𝑒 𝑠2(𝑛 + 1, 𝑘) = 𝑛+ 2 2 + 𝑘 · 𝑑𝑒𝑠2(𝑛, 𝑘) + 𝑛 2 − 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛, 𝑘 − 1) = 𝑛 2 + 𝑘 + 1 " 𝑛 2! 4 𝑛 2− 𝑘 ! 2 (𝑘!)2 # + 𝑛 2− 𝑘 + 1 " 𝑛 2! 4 𝑛 2 − 𝑘 + 1 ! 2 ( (𝑘 − 1)!)2 # = " 𝑛 2 + 𝑘 + 1 𝑛 2− 𝑘 + 1 𝑛 2! 4 𝑛 2− 𝑘 ! 𝑛 2− 𝑘 + 1 !(𝑘!)2 # + " 𝑛 2! 4 𝑘2 𝑛 2 − 𝑘 ! 𝑛 2 − 𝑘 + 1 !(𝑘!)2 # = 𝑛 2+ 1 + 𝑘 𝑛 2+ 1 − 𝑘 𝑛 2! 4 + 𝑛 2! 4 𝑘2 𝑛 2 − 𝑘 ! 𝑛 2 − 𝑘 + 1 !(𝑘!)2 = 𝑛 2! 4 𝑛 2+ 1 2 − 𝑛 2! 4 𝑘2+ 𝑛 2! 4 𝑘2 𝑛 2− 𝑘 ! 𝑛 2− 𝑘 + 1 !(𝑘!)2 = 𝑛 2! 2 𝑛 2+ 1 ! 2 𝑛 2− 𝑘 ! 𝑛 2− 𝑘 + 1 !(𝑘!)2 ,
which is the expression for 𝑑𝑒𝑠2(𝑛 + 1, 𝑘) when 𝑛 + 1 is odd. Finally we use induction on 𝑘.
We know that 𝑑𝑒 𝑠2(𝑛, 𝑘 + 1) = 𝑛+ 1 2 + 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘 + 1) + j 𝑛 2 k − 𝑘 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘), and we assume 𝑑𝑒 𝑠2(𝑛, 𝑘) = 𝑛 2 ! 2 · 𝑛+1 2 ! 2 𝑛 2 − 𝑘! 𝑛+1 2 − 𝑘 !(𝑘!)2 .
We then get the following induction step: 𝑑𝑒 𝑠2(𝑛, 𝑘 + 1) = 𝑛+ 1 2 + 𝑘 + 1 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘 + 1) + j 𝑛 2 k − 𝑘 · 𝑑𝑒𝑠2(𝑛 − 1, 𝑘) = 𝑛+ 1 2 + 𝑘 + 1 𝑛 2 ! 2 𝑛−1 2 !2 𝑛 2 − 𝑘 − 1 ! 𝑛−1 2 − 𝑘 − 1 !((𝑘 + 1)!)2 + j 𝑛 2 k − 𝑘 𝑛 2 ! 2 𝑛−1 2 ! 2 𝑛 2 − 𝑘 ! 𝑛−1 2 − 𝑘 !(𝑘!)2 = 𝑛+1 2 + 𝑘 + 1 𝑛 2 ! 2 𝑛−1 2 ! 2 𝑛−1 2 − 𝑘 𝑛 2 − 𝑘 − 1 !𝑛−1 2 − 𝑘 !((𝑘 + 1)!)2 + 𝑛 2 ! 2 𝑛−1 2 ! 2 (𝑘 + 1)2 𝑛 2 − 𝑘 − 1 ! 𝑛−1 2 − 𝑘 !((𝑘 + 1)!)2 = 𝑛 2 ! 2 𝑛−1 2 ! 2 𝑛+1 2 + 𝑘 + 1 𝑛−1 2 − 𝑘 + (𝑘 + 1)2 𝑛 2 − 𝑘 − 1 !𝑛−1 2 − 𝑘 !((𝑘 + 1)!)2 = 𝑛 2 ! 2 𝑛−1 2 ! 2 𝑛+1 2 + (𝑘 + 1) 𝑛+1 2 − (𝑘 + 1) + (𝑘 + 1)2 𝑛 2 − 𝑘 − 1 ! 𝑛−1 2 − 𝑘 !((𝑘 + 1)!)2 = 𝑛 2 ! 2 𝑛+1 2 ! 2 𝑛 2 − 𝑘 − 1 !𝑛+1 2 − 𝑘 − 1 !((𝑘 + 1)!)2 ,
which concludes the proof.
This confirms that the number of 𝑛-permutations with 𝑘 double-descents is the same as the number of permutations of the first 𝑛 even positive integers with 𝑛 − 𝑘 excedances. However, no bijection between these two sets has been found.
3
The Inversion Statistic
In this section we will study a third statistic, namely inversions which is commonly used as a way of describing how much out of order a permutation is. The more inversions in a permutation the more it is our of order. It is defined in the following way.
Definition 6. Let 𝜋 = 𝜋1𝜋2. . . 𝜋𝑛 be a permutation, then the pair (𝜋𝑖, 𝜋𝑗) is an inversion if 𝑖 < 𝑗 and 𝜋𝑖 > 𝜋𝑗. The set of all inversions in 𝜋 is defined as 𝐼 (𝜋) = {(𝜋𝑖, 𝜋𝑗) : 𝑖 < 𝑗 , 𝜋𝑖 > 𝜋𝑗} and the permutation statistic is 𝑖𝑛𝑣 (𝜋) = |𝐼 (𝜋)|.
In 1838, Terquem presented a paper [8] that answered the question of how many inversions there are in all 𝑛-permutations. Later, another, arguably easier, proof was found for the same question [9]. That proof is the following.
Theorem 6. The total number of inversions in all 𝑛-permutations is
𝑛!𝑛(𝑛 − 1) 4 .
Proof. Let 𝜋 = 𝜋1𝜋2. . . 𝜋𝑛be a permutation and ¯𝜋 = 𝜋𝑛𝜋𝑛−1. . . 𝜋1. Then if (𝜋𝑖, 𝜋𝑗) is in 𝐼 (𝜋), then it is not in 𝐼 ( ¯𝜋) and vice versa. This holds for all pairs (𝜋𝑖, 𝜋𝑗) and since there are a total of 𝑛! · (1 + 2 + . . . + (𝑛 − 1)) pairs in all 𝑛-permutations and half of all pairs are inversions, then the total number of inversions is
𝑛!𝑛(𝑛 − 1) 4 .
3.1
Introducing Double-Inversions
Similar to the double-descent statistic, we now introduce the double-inversion statistic.
Definition 7. Let 𝜋 = 𝜋1𝜋2. . . 𝜋𝑛be a permutation, then the pair (𝜋𝑖, 𝜋𝑗) is a double-inversion if 𝑖 < 𝑗 and 𝜋𝑖 ≥ 2𝜋𝑗. The set of all double-inversions in 𝜋 is defined as 𝐼2(𝜋) = {(𝜋𝑖, 𝜋𝑗) :
𝑖 < 𝑗 , 𝜋𝑖 ≥ 2𝜋𝑗} and the permutation statistic is 𝑖𝑛𝑣2(𝜋) = |𝐼2(𝜋) |.
Ideally, we would like to construct the same type of theorems as we did for the double-descent case. However, double-inversions appears to be more difficult to work with. But with the knowledge we got from Lemma 1 we can still derive the following theorem.
Theorem 7. The total number of double-inversions in all 𝑛-permutations is
Õ 𝜋∈S𝑛 𝑖𝑛𝑣2(𝜋) = 𝑛! 2 · 𝑛2 4 .
Proof. From Lemma 1 we know that the number of unique pairs that, in an 𝑛−permutation, can be double-inversions is
j
𝑛2 4
k
. Consider one of these pairs, (𝜋𝑖, 𝜋𝑗), then in half of all the
permutations we have that 𝑖 < 𝑗 (and in the other half 𝑖 > 𝑗 ). Therefore the pair will be a double-inversion in half of all 𝑛−permutations. This holds for all pairs, and therefore the total number of double-inversions in all 𝑛−permutations is
𝑛! 2 · 𝑛2 4 .
What more can be said regarding the number of 𝑛-permutations with 𝑘 double-inversions? Unlike the double-descent case, there seems to be a symmetry in the number of 𝑛-permutations with 𝑘 double-inversions. This is confirmed in the following proposition.
Proposition 4. Let 𝑖𝑛𝑣2(𝑛, 𝑘) be the number of 𝑛-permutations with 𝑘 double-inversions. Also
let 𝑛 be a positive integer and 𝑘 be a non-negative integer such that 𝑘 ≤
j
𝑛2 4
k
. Then the following symmetry holds.
𝑖𝑛𝑣2(𝑛, 𝑘) = 𝑖𝑛𝑣2 𝑛, 𝑛2 4 − 𝑘 .
Proof. Consider an 𝑛-permutation 𝜋 = 𝜋1𝜋2· · · 𝜋𝑛with 𝑘 number of double-inversions, and let ¯
𝜋 = 𝜋𝑛𝜋𝑛−1· · · 𝜋1. For every pair (𝜋𝑖, 𝜋𝑗), we have either (𝜋𝑖, 𝜋𝑗) ∈ 𝐼 (𝜋) or (𝜋𝑖, 𝜋𝑗) ∈ 𝐼 ( ¯𝜋) but
not both. Similarly, if this pair is a double-inversion, either (𝜋𝑖, 𝜋𝑗) ∈ 𝐼2(𝜋) or (𝜋𝑖, 𝜋𝑗) ∈ 𝐼2( ¯𝜋).
Since, by Lemma 1, an 𝑛-permutation can have a maximum of j 𝑛2 4 k double-inversions, ¯𝜋will have j𝑛2 4 k
− 𝑘 double-inversions. Since this holds for any pair of 𝑛-permutations 𝜋 and ¯𝜋, the number of permutations with 𝑘 double-inversions is the same as the number of 𝑛-permutations with j 𝑛2 4 k − 𝑘 double-inversions.
4
Future Work
We have now found expressions for the distribution of the double-descent statistic. What are the next steps we can do to widen our knowledge in this area?
Firstly, another way of representing the distribution of double-descents in all 𝑛-permutation is with a generating function. Having a generating function is efficient since it is a way to get all number of 𝑛-permutations with 𝑘 double-descents for a fixed 𝑛.
Secondly, we only touched on the double-inversion statistic. Finding the distribution of double-inversions over all 𝑛-permutation would be a big step. This seems to be more difficult than finding the distribution of double-descents since double-inversions behave less reliably than double-descents.
Lastly, one could study what happens if, instead of defining a double-descent with a non-strict inequality, a non-strict double-descent could be defined with a non-strict inequality. One could also look at a more general case where the factor 2 in the definitions of double-descents and double-inversions is replaced by any 𝑚 > 2, and study how this new variation is distributed. Both of these variations would result in similar recursive expressions as the one in Theorem 4, since it is only the result of Lemma 2 that would differ in the argumentation.
References
[1] Miklós Bóna. Combinatorics of permutations. CRC Press, 2012.
[2] Fredrik Jansson. Variations on the excedance statistic in permutations. PhD thesis, Master’s thesis, Chalmers Tekniska Högskola, 2006.
[3] Sergey Kitaev and Jeffrey Remmel. Classifying descents according to equivalence mod k.
arXiv preprint math/0604455, 2006.
[4] Sergey Kitaev and Jeffrey Remmel. Classifying descents according to parity. Annals of
Combinatorics, 11(2):173–193, 2007.
[5] Madeleine Leander. Combinatorics of stable polynomials and correlation inequalities. PhD thesis, Department of Mathematics, Stockholm University, 2016.
[6] Percy Alexander MacMahon. Combinatory Analysis, volume 1-2. Cambridge University Press, 1915.
[7] Eugen Netto. Lehrbuch der combinatorik, volume 7. BG Teubner, 1901.
[8] MO Terquem. Solution d’un probléme de combinaison. J. de Math. Pures et Appl, 3:559–560, 1838.
[9] Thotsaporn Thanatipanonda. Inversions and major index for permutations. Mathematics
