Equidistribution of Descents, Adjacent Pairs, and Place-Value Pairs on Permutations

Emeric Deutsch Department of Mathematics Polytechnic Institute of NYU

Brooklyn, NY 11201 USA

 deutsch@duke.poly.edu

Sergey Kitaev^{1} The Mathematics Institute School of Computer Science

Reykjavík University

IS-103 Reykjavík Iceland

      sergey@ru.is

Jeffrey Remmel^{2} Department of Mathematics

University of California, San Diego

La Jolla, CA 92093-0112

USA

  remmel@math.ucsd.edu

1The work presented here was supported by grant no. 090038011 from the Icelandic Research Fund.

2Partially supported by NSF grant DMS 0654060.

Abstract

An (X, Y)-descent in a permutation is a pair of adjacent elements such that the first element is from X, the second element is from Y, and the first element is greater than the second one. An (X, Y)-adjacency in a permutation is a pair of adjacent elements such that the first one is from X and the second one is from Y. An (X, Y)- place-value pair in a permutation is an element y in position x, such that y is in Y and x is in X. It turns out, that for certain choices of X and Y some of the three statistics above become equidistributed. Moreover, it is easy to derive the distribution formula for (X, Y)-place-value pairs thus providing distribution for other statistics under consideration too. This generalizes some results in the literature. As a result of our considerations, we get combinatorial proofs of several remarkable identities. We also conjecture existence of a bijection between two objects in question preserving a certain statistic.

1 Introduction

Let S_{n} denote the set of permutations of [n] = {1, …, n} and ℕ = {1, 2, …}. Also, 𝔼 and 𝕆 denote the set of even and odd numbers, respectively. For σ = σ_{1} ⋯ σ_{n} ∈ S_{n} and X, Y ⊆ ℕ define the following permutation statistics

des_{X,Y} (σ) = |{i : σ_{i} > σ_{i+1}, & σ_{i} ∈ X & σ_{i+1} ∈ Y }|,
adj_{X,Y} (σ) = |{i : σ_{i} ∈ X & σ_{i+1} ∈ Y }|,
val_{X,Y} (σ) = |{i : i ∈ X & σ_{i} ∈ Y }|,
exc_{X,Y} (σ) = |{i : σ_{i} > i & i ∈ X & σ_{i} ∈ Y }|,

and the following corresponding polynomials

D_{n}^{X,Y} (x) = ∑_{σ∈S_{n}} x^{des_{X,Y} (σ)} = ∑_{s=0}^{n−1} D_{n,s}^{X,Y} x^{s},

A_{n}^{X,Y} (x) = ∑_{σ∈S_{n}} x^{adj_{X,Y} (σ)} = ∑_{s=0}^{n−1} A_{n,s}^{X,Y} x^{s},

V_{n}^{X,Y} (x) = ∑_{σ∈S_{n}} x^{val_{X,Y} (σ)} = ∑_{s=0}^{n−1} V_{n,s}^{X,Y} x^{s},

E_{n}^{X,Y} (x) = ∑_{σ∈S_{n}} x^{exc_{X,Y} (σ)} = ∑_{s=0}^{n−1} E_{n,s}^{X,Y} x^{s}.

Objects counted by des_{X,Y} are called (X, Y)-descents in [2]. Similarly, we can talk of (X, Y)- adjacencies, (X, Y)-place-value pairs, and (X, Y)-excedances.

Foata’s first transformation [1] exchanging excedances and descents (to be used in the paper) can most easily be explained with an example. The permutation w = 61437258 has three excedances: 6, 4, and 7 in positions 1, 3, and 5, respectively. We write w in cycle form: (162)(34)(57)(8). Next, write each cycle with largest element last, and order the cycles by increasing largest element: (34)(216)(57)(8). Finally, reverse each cycle and erase the parentheses to get the outcome permutation 43612758 with the descents 43, 61, and 75.

Remark 1. Using Foata’s first transformation, one obtains that D_{n}^{X,Y} (x) = E_{n}^{Y,X}(x). Thus, we do not need to provide any arguments for the polynomial E_{n}^{X,Y} (x) and its coefficients, instead studying the other three polynomials.

In this paper, we use the following notation for any X ⊆ ℕ and integer n ≥ 1:

X_{n} = [n] ∩ X, x_{n} = |X_{n}|, X_{n}^{c} = [n] − X, and x_{n}^{c} = |X_{n}^{c}|.

Collecting some data on the polynomials, we noticed several equidistributions among the statistics, and nice formulas associated with them, for particular choices of sets X and Y. We collect those observations in Table 1, where a_{n,k} denotes the number of permutations in S_{n} with k occurrences of the corresponding statistic.

Many of formulas listed in Table 1 are known (see, e.g., [4]). Others are new but quite easy to prove. Our idea to establish the equidistribution results is to prove general recurrence relations for the statistics for arbitrary choice of sets X and Y. Then we will get the equidistributions in Table 1 as a simple corollary to the fact that the recurrences for the statistics in a given block are the same for a particular choice of X and Y. For example, we will show that whenever X and Y are disjoint subsets of ℕ, then A_{n,s}^{X,Y} = V_{n,s}^{X,Y} for all n and s. Indeed, the recursions that we develop will allow us to give a bijective proof of this fact. Other equidistribution results follow from simple bijections. For example, it is easy to see that for any X and Y, A_{n,s}^{X,Y} = A_{n,s}^{Y,X} since if σ_{i}σ_{i+1} is an (X, Y)-adjacency in σ = σ_{1} ⋯ σ_{n}, then σ_{i+1}σ_{i} is a (Y, X)-adjacency in the reverse of σ, σ^{r} = σ_{n}σ_{n−1} ⋯ σ_{1}.

Several of our formulas are quite easy to prove for one of our three statistics. For example, it is always easy to compute V_{n,s}^{X,Y}.

Theorem 2. For any X, Y ⊆ ℕ, n ≥ 1, and 0 ≤ s ≤ n,

V_{n,s}^{X,Y} = s!(x_{n} − s)!(x_{n}^{c})!\binom{x_{n}}{s}\binom{y_{n}}{s}\binom{y_{n}^{c}}{x_{n} − s}.  (1)

Proof. To count the number of permutations of length n with s occurrences of val_{X,Y}, we can first pick s positions from X_{n} in \binom{x_{n}}{s} ways for the places where we will have values of Y occurring in the places corresponding to X_{n}. Then we pick s values from Y_{n} in \binom{y_{n}}{s} ways, and permute the values in s! ways to arrange the s occurrences of values in Y_{n} in the places in X_{n}. In the remaining x_{n} − s places in X_{n}, we must choose values from Y_{n}^{c}. We thus have \binom{y_{n}^{c}}{x_{n}−s} ways to choose those values and (x_{n} − s)! ways to rearrange them. Finally we have x_{n}^{c}! ways to arrange the elements in places outside of X_{n}.

Similarly, it is easy to count A_{n,s}^{X,X} for any set X ⊆ ℕ. That is, we have the following theorem.

Stat. Description, related polynomial, and enumeration
S_{1} # of even descent-tops (D_{n,k}^{𝔼,ℕ}). E.g., S_{1}(215436) = 2.
S_{2} # of even excedance values (E_{n,k}^{ℕ,𝔼}). E.g., S_{2}(215436) = 1.
S_{3} # of even entries in even positions (V_{n,k}^{𝔼,𝔼}). E.g., S_{3}(215436) = 2.
a_{2n,k} = [n!\binom{n}{k}]^{2}; a_{2n+1,k} = n!(n + 1)!\binom{n}{k}\binom{n+1}{k+1}.
S_{4} # of odd descent-bottoms (D_{n,k}^{ℕ,𝕆}). E.g., S_{4}(215436) = 2.
S_{5} # of odd excedance positions (E_{n,k}^{𝕆,ℕ}). E.g., S_{5}(215436) = 2.
S_{6} # of even entries in odd positions (V_{n,k}^{𝕆,𝔼}). E.g., S_{6}(215436) = 1.
S_{7} # of odd entries in even positions (V_{n,k}^{𝔼,𝕆}). E.g., S_{7}(215436) = 1.
S_{8} # of (odd,even) pairs (A_{n,k}^{𝕆,𝔼}). E.g., S_{8}(21\underline{54} \underline{36}) = 2.
S_{9} # of (even, odd) pairs (A_{n,k}^{𝔼,𝕆}). E.g., S_{9}(\underline{21}5\underline{43}6) = 2.
a_{2n,k} = [n!\binom{n}{k}]^{2}; a_{2n+1,k} = n!(n + 1)!\binom{n}{k}\binom{n+1}{k}.
S_{10} # of odd descent-tops (D_{n,k}^{𝕆,ℕ}). E.g., S_{10}(215436) = 1.
S_{11} # of odd excedance values (E_{n,k}^{ℕ,𝕆}). E.g., S_{11}(215436) = 1.
S_{12} # of (odd,odd) pairs (A_{n,k}^{𝕆,𝕆}). E.g., S_{12}(2\underline{15}436) = 1.
a_{2n,k} = (n!)^{2}\binom{n−1}{k}\binom{n+1}{k+1}; a_{2n+1,k} = n!(n + 1)!\binom{n}{k}\binom{n+1}{k}.
S_{13} # of even descent-bottoms (D_{n,k}^{ℕ,𝔼}). E.g., S_{13}(215436) = 1.
S_{14} # of even excedance positions (E_{n,k}^{𝔼,ℕ}). E.g., S_{14}(215436) = 0.
a_{2n,k} = (n!)^{2}\binom{n−1}{k}\binom{n+1}{k+1}; a_{2n+1,k} = n!(n + 1)!\binom{n}{k}\binom{n+1}{k+1}.
S_{15} # of odd entries in odd positions (V_{n,k}^{𝕆,𝕆}). E.g., S_{15}(215436) = 2.
a_{2n,k} = [n!\binom{n}{k}]^{2}; a_{2n+1,k} = n!(n + 1)!\binom{n}{k−1}\binom{n+1}{k}.
S_{16} # of (even,even) pairs (A_{n,k}^{𝔼,𝔼}). E.g., S_{16}(215436) = 0.
a_{2n,k} = (n!)^{2}\binom{n−1}{k}\binom{n+1}{k+1}; a_{2n+1,k} = n!(n + 1)!\binom{n−1}{k}\binom{n+2}{k+2}.

Table 1: 16 statistics under consideration classified into 6 statistic groups.

Theorem 3. For any X ⊆ ℕ, n ≥ 1, and 0 ≤ s ≤ n − 1,

A_{n,s}^{X,X} = (x_{n})!(x_{n}^{c})!\binom{x_{n} − 1}{s}\binom{x_{n}^{c} + 1}{x_{n} − s}.  (2)

Proof. Fix n ≥ 1. First we pick a permutation σ of X ∩ [n] and a permutation τ of [n] − X. Clearly, we have (x_{n})!(x_{n}^{c})! ways to pick σ and τ. We are now interested in finding the number of permutations of γ of S_{n} such that γ restricted to the elements in X ∩ [n] yields the permutation σ, γ restricted to the elements in [n] − X yields the permutation τ, and adj_{X,X}(γ) = s. Next in σ_{1}σ_{2} ⋯ σ_{x_{n}}, we think of choosing s spaces from the x_{n} − 1 spaces between the elements of σ to create the adjacencies that will appear in such a γ. For example, if n = 12, s = 2, X = 𝔼, σ = 4 2 10 8 6 12, and we pick spaces 2 and 5, then our choice partitions σ into four blocks, 4, 2 − 10, 8 and 6 − 12. Our idea is to insert these blocks into the spaces that either lie immediately before an element of τ or immediately after the last element of τ. We label these spaces from left to right. For example, suppose τ = 5 1 7 9 3 11 and we pick spaces 2, 4, 5, and 7. Then we would insert the block 4 immediately before 1, the block 2 − 10 immediately before 9, the block 8 immediately before 3, 6 − 12 immediately after 11 to obtain the permutation

5 4 1 7 2 10 9 8 3 11 6 12.

Clearly there are \binom{x_{n}−1}{s} ways to choose the spaces to obtain our s adjacencies. This will leave us with x_{n} − s blocks. Then there are \binom{x_{n}^{c}+1}{x_{n}−s} to choose the spaces for τ where we insert the blocks.

Hall and Remmel [2] gave direct combinatorial proofs of a pair of formulas for D_{n,s}^{X,Y} which combined with our equidistribution results, gives formulas for the other polynomials under consideration. We state these results here together with an example of using them.

Theorem 4. For any X, Y ⊆ ℕ, n ≥ 1, and 0 ≤ s ≤ n − 1,

D_{n,s}^{X,Y} = |X_{n}^{c}|! ∑_{r=0}^{s}(−1)^{s−r}\binom{|X_{n}^{c}| + r}{r}\binom{n + 1}{s − r} ∏_{x∈X_{n}}(1 + r + α_{X,n,x} + β_{Y,n,x}),  (3)

Theorem 5. For any X, Y ⊆ ℕ, n ≥ 1, and 0 ≤ s ≤ n − 1,

D_{n,s}^{X,Y} = |X_{n}^{c}|! ∑_{r=0}^{|X_{n}|−s}(−1)^{|X_{n}|−s−r}\binom{|X_{n}^{c}| + r}{r}\binom{n + 1}{|X_{n}| − s − r} ∏_{x∈X_{n}}(r + β_{X,n,x} − β_{Y,n,x}), (4)

where for any set A and any j, 1 ≤ j ≤ n, we define

α_{A,n,j} = |A^{c} ∩ {j + 1, j + 2, …, n}| = |{x : j < x ≤ n & x ∉ A}|, and
β_{A,n,j} = |A^{c} ∩ {1, 2, …, j − 1}| = |{x : 1 ≤ x < j & x ∉ A}|.

Example 1. Suppose X = {2, 3, 4, 6, 7, 9}, Y = {1, 4, 8}, and n = 6. Thus X_{6} = {2, 3, 4, 6}, X_{6}^{c} = {1, 5}, Y_{6} = {1, 4}, Y_{6}^{c} = {2, 3, 5, 6}, and we have the following table of values of α_{X,6,x}, β_{Y,6,x}, and β_{X,6,x}.

\begin{matrix}x & 2 & 3 & 4 & 6 \\ α_{X,6,x} & 1 & 1 & 1 & 0 \\ β_{Y,6,x} & 0 & 1 & 2 & 3 \\ β_{X,6,x} & 1 & 1 & 1 & 2 \end{matrix}

Equation (3) gives

D_{6,2}^{X,Y} = 2! ∑_{r=0}^{2}(−1)^{2−r}\binom{2 + r}{r}\binom{7}{2 − r}(2 + r)(3 + r)(4 + r)(4 + r)
= 2 (1 · 21 · 2 · 3 · 4 · 4 − 3 · 7 · 3 · 4 · 5 · 5 + 6 · 1 · 4 · 5 · 6 · 6)
= 2(2016 − 6300 + 4320)
= 72,

while (4) gives

D_{6,2}^{X,Y} = 2! ∑_{r=0}^{2}(−1)^{2−r}\binom{2 + r}{r}\binom{7}{2 − r}(1 + r)(0 + r)(−1 + r)(−1 + r)
= 2 (1 · 21 · 1 · 0 · (−1) · (−1) − 3 · 7 · 2 · 1 · 0 · 0 + 6 · 1 · 3 · 2 · 1 · 1)
= 2(0 − 0 + 36)
= 72.

The paper is organized as follows. In Section 2 we find general recurrence relations for D_{n,k}^{X,Y}, A_{n,k}^{X,Y}, and V_{n,k}^{X,Y}, and use them to explain the facts in Table 1. In Section 3 we generalize several of the results that appear in Table 1, and use this to obtain combinatorial proofs of several remarkable identities. Finally, in Section 4, we discuss some directions for further research.

2 Recurrence relations for D_{n,k}^{X,Y}, A_{n,k}^{X,Y}, and V_{n,k}^{X,Y}

In this section, we derive recurrence relations for D_{n,k}^{X,Y}, A_{n,k}^{X,Y}, and V_{n,k}^{X,Y}. We notice that the recurrences we get for A_{n,k}^{X,Y} and V_{n,k}^{X,Y} are almost identical, except for the case when the element n + 1 ∈ X ∩ Y — the recurrences differ by “1+.” However, assuming X ∩ Y = ∅, we do not have this case, leading, in particular, to the explanation of all of the equidistributions in Table 1, and to many more results for other choices of X and Y, X ∩ Y = ∅.

Another thing to observe is that in the case of the same recurrence relations, we naturally get bijective proofs for the corresponding equidistributed statistics. Indeed, one can label positions in a permutation, say from left to right, in which we insert the largest element, n + 1, or do the other insertion procedure (see Subsection 2.3); then, it is enough to match insertions in the positions having the same labels. However, such straightforward approach is not necessarily the best one, as labeling positions differently, rather than just from left to right, one may preserve extra statistics in bijections (see Section 4 for conjectures, which should be possible to prove using our approach with different labeling).

2.1 Recurrences for D_{n,k}^{X,Y}

A recursion for D_{n,k}^{X,Y} is derived in [2]:

D_{n+1,k}^{X,Y} = \{ \begin{matrix}(k + 1)D_{n,k+1}^{X,Y} + (n + 1 − k)D_{n,k}^{X,Y} , & if n + 1 ∉ X; \\ (y_{n} − (k − 1))D_{n,k−1}^{X,Y} + (n + 1 − (y_{n} − k))D_{n,k}^{X,Y} , & if n + 1 ∈ X. \end{matrix}

An argument for deriving the recursion is as follows. We are thinking of inserting the element n + 1 in a permutation σ = σ_{1} ⋯ σ_{n}, and we consider which of the obtained permutations are counted by D_{n,k}^{X,Y}. If n + 1 ∉ X then one never increases the number of (X, Y)-descents by inserting n + 1. More precisely, the number of (X, Y)-descents is either unchanged, or it is decreased by 1, when n + 1 is inserted between σ_{i} ∈ X and σ_{i+1} ∈ Y where σ_{i} > σ_{i+1}. The corresponding recursion case follows.

For the second case, notice that if n + 1 ∈ X, then the number of (X, Y)-descents is unchanged if n + 1 is inserted at the end of the permutation, in front of σ_{j} ∉ Y, or between σ_{i} ∈ X and σ_{i+1} ∈ Y where σ_{i} > σ_{i+1}, and it is increased by 1 in other cases (that is, when n + 1 is inserted in front of σ_{j} ∈ Y not involved in an (X, Y)-descent). The second recursion case follows.

We use a similar approach to derive recurrence relations for A_{n,k}^{X,Y}. Our derivations for V_{n,k}^{X,Y} use a different insertion procedure.

2.2 Recurrences for A_{n,k}^{X,Y}

We consider 4 cases.

Case 1. n + 1 ∉ X ∪ Y. The number of (X, Y)-adjacent pairs is decreased by 1 when n + 1 is inserted between σ_{i} ∈ X and σ_{i+1} ∈ Y and it is unchanged otherwise. Thus, in this case

A_{n+1,k}^{X,Y} = (k + 1)A_{n,k+1}^{X,Y} + (n + 1 − k)A_{n,k}^{X,Y} .

Case 2. n + 1 ∈ X ∩ Y. Adding n + 1 after a σ_{i} ∈ X or before a σ_{j} ∈ Y increases adj_{X,Y} by 1, while it keeps adj_{X,Y} (σ) unchanged otherwise. However, we note that the place between σ_{i} ∈ X and σ_{i+1} ∈ Y is after a σ_{i} ∈ X and before a σ_{i+1} ∈ Y. Thus, in this case

A_{n+1,k}^{X,Y} = (x_{n} + y_{n} − (k − 1))A_{n,k−1}^{X,Y} + (n + 1 − (x_{n} + y_{n} − k))A_{n,k}^{X,Y} .

Case 3. n+1 ∈ X −Y. Inserting n+1 to the left of a σ_{i} ∉ Y does not change adj_{X,Y} (σ), which is also the case if n + 1 is inserted between σ_{i} ∈ X and σ_{i+1} ∈ Y, or n + 1 is inserted at the very end. On the other hand, if n + 1 is inserted between σ_{i} ∉ X and σ_{i+1} ∈ Y, the number of (X, Y)-adjacent pairs is increased by 1. Thus, in this case

A_{n+1,k}^{X,Y} = (y_{n} − (k − 1))A_{n,k−1}^{X,Y} + (n + 1 − (y_{n} − k))A_{n,k}^{X,Y} .

Case 4. n + 1 ∈ Y − X. Inserting n + 1 to the right of a σ_{i} ∉ X does not change adj_{X,Y} (σ), which is also the case if n + 1 is inserted between σ_{i} ∈ X and σ_{i+1} ∈ Y, or n + 1 is inserted at the very beginning. On the other hand, if n + 1 is inserted between σ_{i} ∈ X and σ_{i+1} ∉ Y, the number of (X, Y)-adjacent pairs is increased by 1. Thus, in this case

A_{n+1,k}^{X,Y} = (x_{n} − (k − 1))A_{n,k−1}^{X,Y} + (n + 1 − (x_{n} − k))A_{n,k}^{X,Y} .

2.3 Recurrences for V_{n,k}^{X,Y}

Instead of inserting the largest element, n+1, in all possible places, we use another insertion procedure I_{n}^{(i)}(σ) that generates S_{n+1} from S_{n}. For σ = σ_{1} ⋯ σ_{n}, let I_{n+1}^{(n+1)}(σ) = σ(n + 1) = σ_{1} ⋯ σ_{n}(n + 1), and for 1 ≤ i ≤ n, let I_{n+1}^{(i)}(σ) = σ_{1} ⋯ σ_{i−1}(n + 1)σ_{i+1} ⋯ σ_{n}σ_{i} (that is, in the last case we replace σ_{i} in σ by n + 1 and move σ_{i} to the very end).

We now consider 4 cases.

Case 1. n + 1 ∉ X ∪ Y. In this case, one can only decrease the number of (X, Y)-place-value pairs. This happens when n + 1 occupies position i ∈ X in I_{n+1}^{(i)}(σ) for some σ, such that σ_{i} ∈ Y (σ_{i} is in position n + 1 in I_{n+1}^{(i)}(σ)). Thus, in this case

V_{n+1,k}^{X,Y} = (k + 1)V_{n,k+1}^{X,Y} + (n + 1 − k)V_{n,k}^{X,Y} .

Case 2. n+1 ∈ X ∩Y. It is straightforward to see that the number of (X, Y)-place-value pairs is unchanged if i ∉ X and σ_{i} ∉ Y, and it increases by 1 in each of the following three cases: i ∈ X and σ_{i} ∈ Y, i ∈ X and σ_{i} ∉ Y, and i ∉ X and σ_{i} ∈ Y. Note that we add 1 for each i ∈ X and 1 for each σ_{i} ∈ Y, so we count i ∈ X and σ_{i} ∈ Y twice. Moreover, having n + 1 in position n + 1 gives one more (X, Y)-place-value pair. Thus, in this case

V_{n+1,k}^{X,Y} = (1 + x_{n} + y_{n} − (k − 1))V_{n,k−1}^{X,Y} + (n + 1 − (1 + (x_{n} + y_{n} − k)))V_{n,k}^{X,Y} .

Case 3. n + 1 ∈ X − Y. One can check that in this case, the number of (X, Y)-place-value pairs increases by 1 if i ∉ X and σ_{i} ∈ Y, and it is unchanged otherwise. Thus, in this case

V_{n+1,k}^{X,Y} = (y_{n} − (k − 1))V_{n,k−1}^{X,Y} + (n + 1 − (y_{n} − k))V_{n,k}^{X,Y} .

Case 4. n + 1 ∈ Y − X. One can check that in this case, the number of (X, Y)-place-value pairs increases by 1 if i ∈ X and σ_{i} ∉ Y, and it is unchanged otherwise. Thus, in this case

V_{n+1,k}^{X,Y} = (x_{n} − (k − 1))V_{n,k−1}^{X,Y} + (n + 1 − (x_{n} − k))V_{n,k}^{X,Y} .

There are a number of cases where the recursions for V_{n,k}^{A,B}, A_{n,k}^{C,D}, and D_{n,k}^{E,F} coincide so that we immediately have equality between the various pairs of statistics. For example, comparing the recursions for A_{n,k}^{X,Y} and V_{n,k}^{X,Y}, we immediately have the following theorem.

Theorem 6. For all X, Y ⊆ ℕ such that X ∩ Y = ∅, n ≥ 1, and 0 ≤ k ≤ n, V_{n,k}^{X,Y} = A_{n,k}^{X,Y}.

In fact, it is easy to see that our proofs of the recursions can be used to give an inductive proof that there exists a bijection from S_{n} onto S_{n} for all n that will witness this equality. That is, our proofs of the recursions immediately allow us to construct inductively bijections Θ_{n} : S_{n} → S_{n} for all n such that for all σ ∈ S_{n},

adj_{X,Y} (σ) = val_{X,Y} (Θ_{n}(σ)).

For example, suppose that we have constructed Θ_{n} and n + 1 ∉ X ∪ Y. First consider our insertion procedure to prove the recursions for A_{n,s}^{X,Y}. If σ ∈ S_{n}, then we consider the places where we can insert n + 1 to σ. We first label the spaces between the elements σ_{i} ∈ X and σ_{i+1} ∈ Y from left to right with 1, …, adj_{X,Y} (σ) and then label the rest of the spaces from left to right with adj_{X,Y} (σ) + 1, …, n + 1. For example, if X = 𝔼, Y = 𝕆, and σ = 1 4 3 2 5, the spaces would be labeled by

3^{−}1_{4^{−}}4_{1^{−}}3_{5^{−}}2_{2^{−}}5_{6^{−}}.

We then let σ^{(i)} be the permutation that results by inserting n + 1 into the space labeled i. For example, in our example, σ^{(4)} = 1 6 4 3 2 5. Next we consider our insertion procedure for proving the recursions for V_{n,s}^{X,Y}. Now if τ ∈ S_{n}, then we label the positions of τ by first labeling the positions i such that i ∈ X and τ_{i} ∈ Y from left to right with 1, …, val_{X,Y} (τ) and then label the remaining positions from left to right with val_{X,Y} (τ) + 1, …, n. For example, if X = 𝕆 and Y = 𝔼 and τ = 1 4 2 5 3, then we would label the positions

\frac{1}{2} \frac{4}{3} \frac{2}{1} \frac{5}{4} \frac{3}{5}

where we have indicated the labels in boldface. If label j is in position i, then we let τ^{(j)} = I_{n}^{(i)}(τ) and we let τ^{(n+1)} = I_{n}^{(n+1)}(τ). For example, in our case, τ^{(2)} = 6 4 2 5 3 1. Then for any σ ∈ S_{n} and i ∈ {1, …, n + 1}, we can define

Θ_{n+1}(σ^{(i)}) = Θ_{n}(σ)^{(i)}.

We can extend Θ_{n} to Θ_{n+1} in the other cases of the recursions in a similar manner.

Similarly, comparing the recursions for the V_{n,k}^{A,B}, A_{n,k}^{C,D}, and D_{n,k}^{E,F}, we can also derive bijective proofs of the following theorems.

Theorem 7. If X and Y are subsets of ℕ, A = X ∪ Y and there exists a B ⊆ ℕ such that b_{n} = |B ∩ [n]| satisfies

b_{n} = \begin{cases}x_{n} + y_{n} = |X ∩ [n]| + |Y ∩ [n]|, & if n + 1 ∈ X ∩ Y; \\ y_{n} = |Y ∩ [n]|, & if n + 1 ∈ X − Y; \\ x_{n} = |X ∩ [n]|, & if n + 1 ∈ Y − X, \end{cases}

then D_{n,k}^{A,B} = A_{n,k}^{X,Y}.

Theorem 8. If X and Y are subsets of ℕ, A = X ∪ Y and there exists a B ⊆ ℕ such that b_{n} = |B ∩ [n]| satisfies

b_{n} = \begin{cases}1 + x_{n} + y_{n} = 1 + |X ∩ [n]| + |Y ∩ [n]|, & if n + 1 ∈ X ∩ Y; \\ y_{n} = |Y ∩ [n]|, & if n + 1 ∈ X − Y; \\ x_{n} = |X ∩ [n]|, & if n + 1 ∈ Y − X, \end{cases}

then D_{n,k}^{A,B} = V_{n,k}^{X,Y}.

2.4 Explanation of Table 1 using our general results

1. The first group of statistics. D_{n,k}^{𝔼,ℕ} = E_{n,k}^{ℕ,𝔼} by Foata’s first transformation. Also, D_{n,k}^{𝔼,ℕ} = V_{n,k}^{𝔼,𝔼} by Theorem 8. Indeed, in this case A = X = Y = 𝔼 and B = ℕ leading to A = X ∪ Y, X − Y = Y − X = ∅, and b_{n} = n = 1 + 2|𝔼 ∩ [n]| if n + 1 ∈ 𝔼. As for the formulas, we can apply Theorem 2 with X = Y = 𝔼:

a_{2n,k} = V_{2n,k}^{X,Y} = k!(n − k)!n!\binom{n}{k}\binom{n}{k}\binom{n}{n − k} = [n!\binom{n}{k}]^{2} ;

a_{2n+1,k} = V_{2n+1,k}^{X,Y} = k!(n − k)!(n + 1)!\binom{n}{k}\binom{n}{k}\binom{n + 1}{n − k} = n!(n + 1)!\binom{n}{k}\binom{n + 1}{k + 1}.

2. The second group. D_{n,k}^{ℕ,𝕆} = E_{n,k}^{𝕆,ℕ} by Foata’s first transformation. Applying the reverse operation to each permutation, one sees that A_{n,k}^{𝕆,𝔼} = A_{n,k}^{𝔼,𝕆}. Applying the inverse operation to each permutation, one gets V_{n,k}^{𝕆,𝔼} = V_{n,k}^{𝔼,𝕆}. By Theorem 6, V_{n,k}^{𝕆,𝔼} = A_{n,k}^{𝕆,𝔼} as 𝕆 ∩ 𝔼 = ∅. Finally, by Theorem 7, D_{n,k}^{ℕ,𝕆} = A_{n,k}^{𝕆,𝔼}. Indeed, in this case A = ℕ, B = X = 𝕆, and Y = 𝔼 leading to A = X ∪ Y, and

b_{n} = # of odd numbers in [n] = \{ \begin{matrix}𝔼 ∩ [n], & if n + 1 ∉ 𝕆; \\ 𝕆 ∩ [n], & if n + 1 ∈ 𝔼. \end{matrix}

As for the formulas, we can apply Theorem 2 with X = 𝔼 and Y = 𝕆:

a_{2n,k} = V_{2n,k}^{𝕆,𝔼} = k!(n − k)!n!\binom{n}{k}\binom{n}{k}\binom{n}{n − k} = [n!\binom{n}{k}]^{2} ;

a_{2n+1,k} = V_{2n+1,k}^{𝕆,𝔼} = k!(n − k)!(n + 1)!\binom{n}{k}\binom{n + 1}{k}\binom{n}{n − k} = n!(n + 1)!\binom{n}{k}\binom{n + 1}{k}.

3. The third group. Again, D_{n,k}^{𝕆,ℕ} = E_{n,k}^{ℕ,𝕆} by Foata’s first transformation. Moreover, by Theorem 7, D_{n,k}^{𝕆,ℕ} = A_{n,k}^{𝕆,𝕆}. Indeed, in this case A = X = Y = 𝕆 and B = ℕ leading to A = X ∪ Y, X − Y = Y − X = ∅, and b_{n} = n = 2|𝕆 ∩ [n]| if n + 1 ∈ 𝕆. As for the formulas, we can apply Theorem 3 with X = 𝕆:

a_{2n,k} = A_{2n,k}^{𝕆,𝕆} = (n!)^{2}\binom{n − 1}{k}\binom{n + 1}{k + 1};

a_{2n+1,k} = A_{2n+1,k}^{𝕆,𝕆} = n!(n + 1)!\binom{n}{k}\binom{n + 1}{k}.

4. The fourth group. D_{n,k}^{ℕ,𝔼} = E_{n,k}^{𝔼,ℕ} by Foata’s first transformation. The formulas for D_{n,k}^{ℕ,𝔼} are proved in [4, Section 4].

5. The fifth group. We use Theorem 2 with X = Y = 𝕆, to get

a_{2n,k} = V_{2n,k}^{𝕆,𝕆} = k!n!n!\binom{n}{k}\binom{n}{k}\binom{n}{n − k} = [n!\binom{n}{k}]^{2} ;

a_{2n+1,k} = V_{2n+1,k}^{𝕆,𝕆} = k!(n+1−k)!n!\binom{n + 1}{k}\binom{n + 1}{k}\binom{n}{k − 1} = n!(n+1)!\binom{n}{k − 1}\binom{n + 1}{k}.

6. The sixth group. We use Theorem 3 with X = 𝔼, to get

a_{2n,k} = A_{2n,k}^{𝔼,𝔼} = (n!)^{2}\binom{n − 1}{k}\binom{n + 1}{k + 1};

a_{2n+1,k} = A_{2n+1,k}^{𝔼,𝔼} = n!(n + 1)!\binom{n − 1}{k}\binom{n + 2}{n − k}.

3 Applications

In this section, we shall generalize several of the results that appear in Table 1. That is, in Table 1, we consider the parity of the elements in a descent, adjacency, or place-value pair. We shall show that we can get similar formulas when we consider the equivalence class modulo k of the elements in a descent, adjacency, or place-value pair. See [3] for related research on descents generalizing results of [4]. For any k ≥ 2 and 0 ≤ i ≤ k − 1, we let i + kℕ = {i + kn : n ≥ 0}.

First, we shall consider V_{n,k}^{X,Y} and A_{n,k}^{X,Y} where X = i + kℕ and Y = j + kℕ and 0 ≤ i < j ≤ k − 1. It follows from Theorem 6 that V_{n,k}^{X,Y} = A_{n,k}^{X,Y} in this case. Suppose that A = i + kℕ ∪ j + kℕ and B = i + kℕ. Note that when m + 1 = kn + i ∈ X − Y, then y_{m} = n = b_{m} = |B ∩ [m]| and when m + 1 = kn + j ∈ Y − X, then x_{m} = n + 1 = b_{m}. Thus it follows from Theorems 7 and 8 that D_{n,k}^{A,B} = V_{n,s}^{X,Y} = A_{n,s}^{X,Y} for all n and s. We then have three cases.

Case 1. m = kn + t where 0 ≤ t < i. In this case, x_{m} = |X ∩ [m]| = y_{m} = |Y ∩ [m]| = n and x_{m}^{c} = |[m] − X| = y_{m}^{c} = |[m] − Y | = (k − 1)n + t. Thus it follows from Theorem 2 that

V_{m,s}^{X,Y} = \binom{n}{s}^{2}\binom{(k − 1)n + t}{n − s}s!(n − s)!((k − 1)n + t)!.  (5)

On the other hand, it follows from Theorem 4 that

D_{m,s}^{A,B} = |A_{m}^{c}|! ∑_{r=0}^{s}(−1)^{s−r}\binom{|A_{m}^{c}| + r}{r}\binom{m + 1}{s − r} ∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}). (6)

In this case |A_{m}^{c}| = kn + t − 2n = (k − 2)n + t. For any x, it is easy to see that

α_{A,m,x} + β_{B,m,x} = kn + t − 1 − |A ∩ [x + 1, kn + t] − |B ∩ [x − 1]|

where [x + 1, kn + t] = {r : x + 1 ≤ r ≤ kn + t}. Thus for any 0 ≤ a ≤ n − 1,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (2n − (2a + 1)) − a
= (k − 2)n + t + a  (7)

and

α_{A,m,ak+j} + β_{B,m,ak+j} = kn + t − 1 − (2n − (2a + 2)) − (a + 1)
= (k − 2)n + t + a.  (8)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

(∏_{a=0}^{n−1}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i})) (∏_{a=0}^{n−1}(1 + r + α_{A,m,ak+j} + β_{B,m,ak+j})) =

(∏_{a=0}^{n−1}(1 + r + (k − 2)n + t + a))^{2} =

(1 + r + (k − 2)n + t)_{n}(1 + r + (k − 2)n + t)_{n}

where we define (a)_{n} by (a)_{0} = 1 and (a)_{n} = a(a+1) ⋯ (a+n−1) for n ≥ 1. Since we have a combinatorial proof of the fact that V_{m,s}^{X,Y} = D_{m,s}^{A,B} in this case and the proof of Theorem 4 is also completely combinatorial, it follows that we have a combinatorial proof of the following identity:

\binom{n}{s}^{2}\binom{(k − 1)n + t}{n − s}s!(n − s)!((k − 1)n + t)! =  (9)

((k − 2)n + t)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 2)n + t + r}{r}\binom{kn + t + 1}{s − r} ×

(1 + r + (k − 2)n + t)_{n}(1 + r + (k − 2)n + t)_{n}.

Case 2. m = kn + t where i ≤ t < j. In this case, x_{m} = |X ∩ [m]| = n + 1 and y_{m} = |Y ∩ [m]| = n and x_{m}^{c} = |[m] − X| = (k − 1)n + t − 1 and y_{m}^{c} = |[m] − Y | = (k − 1)n + t. Thus it follows from Theorem 2 that

V_{m,s}^{X,Y} = \binom{n + 1}{s}\binom{n}{s}\binom{(k − 1)n + t}{n + 1 − s}s!(n + 1 − s)!((k − 1)n + t − 1)!.  (10)

On the other hand, we can obtain a formula for V_{m,s}^{X,Y} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = kn + t − (2n + 1) = (k − 2)n + t − 1. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (2n + 1 − (2a + 1)) − a
= (k − 2)n + t − 1 + a  (11)

and, for any 0 ≤ a ≤ n − 1

α_{A,m,ak+j} + β_{B,m,ak+j} = kn + t − 1 − (2n + 1 − (2a + 2)) − (a + 1)
= (k − 2)n + t − 1 + a.  (12)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

(∏_{a=0}^{n}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i})) (∏_{a=0}^{n−1}(1 + r + α_{A,m,ak+j} + β_{B,m,ak+j})) =

(1 + r + (k − 2)n + t − 1)_{n+1}(1 + r + (k − 2)n + t − 1)_{n} =

(r + (k − 2)n + t)_{n+1}(r + (k − 2)n + t)_{n}.

As in Case 1, it follows that we have a combinatorial proof of the following identity:

\binom{n + 1}{s}\binom{n}{s}\binom{(k − 1)n + t}{n + 1 − s}s!(n − s)!((k − 1)n + t − 1)! =  (13)

((k − 2)n + t − 1)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 2)n + t − 1 + r}{r}\binom{kn + t + 1}{s − r} ×

(r + (k − 2)n + t)_{n+1}(r + (k − 2)n + t)_{n}.

Case 3. m = kn+t where j ≤ t ≤ k−1. In this case, x_{m} = |X∩[m]| = y_{m} = |Y ∩[m]| = n+1 and x_{m}^{c} = |[m] − X| = y_{m}^{c} = |[m] − Y | = (k − 1)n + t − 1. Thus it follows from Theorem 2 that

V_{m,s}^{X,Y} = \binom{n + 1}{s}^{2}\binom{(k − 1)n + t − 1}{n + 1 − s}s!(n + 1 − s)!((k − 1)n + t − 1)!.  (14)

On the other hand, we can obtain a formula for V_{m,s}^{X,Y} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = kn + t − (2n + 2) = (k − 2)n + t − 2. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (2n + 2 − (2a + 1)) − a
= (k − 2)n + t − 2 + a  (15)

and, for any 0 ≤ a ≤ n

α_{A,m,ak+j} + β_{B,m,ak+j} = kn + t − 1 − (2n + 2 − (2a + 2)) − (a + 1)
= (k − 2)n + t − 2 + a.  (16)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

(∏_{a=0}^{n}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i})) (∏_{a=0}^{n}(1 + r + α_{A,m,ak+j} + β_{B,m,ak+j})) =

(1 + r + (k − 2)n + t − 2)_{n+1}(1 + r + (k − 2)n + t − 2)_{n+1} =

(r + (k − 2)n + t − 1)_{n+1}(r + (k − 2)n + t − 1)_{n+1}.

It follows that we have a combinatorial proof of the following identity:

\binom{n + 1}{s}^{2}\binom{(k − 1)n + t − 1}{n + 1 − s}s!(n + 1 − s)!((k − 1)n + t − 1)! =  (17)

((k − 2)n + t − 1)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 2)n + t − 1 + r}{r}\binom{kn + t + 1}{s − r} ×

(r + (k − 2)n + t − 1)_{n+1}(r + (k − 2)n + t − 1)_{n+1}.

Next we shall consider V_{n,k}^{X,Y} and A_{n,k}^{X,Y} where X = Y = i + kℕ for 0 ≤ i ≤ k − 1 and k ≥ 2. In this case, it is no longer the case that A_{m,s}^{X,X} = V_{m,s}^{X,X} so we will handle the cases of A_{m,s}^{X,X} and V_{m,s}^{X,X} separately.

First we shall consider V_{n,s}^{X,Y}. Note that if A = i+kℕ and B = i+kℕ∪i+1+kℕ, then for m + 1 = kn + i ∈ X ∩ Y, then x_{n} = |X ∩ [m]| = n = y_{m} = |Y ∩ [m]| and b_{m} = |B ∩ [m]| = 2n. Thus it follows from Theorem 8 that V_{n,s}^{X,X} = D_{n,s}^{A,B} for all n and s. We then have two cases.

Case I. m = kn + t where 0 ≤ t < i. In this case, x_{m} = |X ∩ [m]| = n and x_{n}^{c} = |[m] − X| = (k − 1)n + t. Then it follows from Theorem 2 that

V_{m,s}^{X,X} = \binom{n}{s}^{2}\binom{(k − 1)n + t}{n − s}s!(n − s)!((k − 1)n + t)!.  (18)

On the other hand, we can obtain a formula for V_{m,s}^{X,X} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = kn + t − n = (k − 1)n + t. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (n − (a + 1)) − 2a
= (k − 1)n + t − a.  (19)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

∏_{a=0}^{n−1}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i}) =

∏_{a=0}^{n−1}(1 + r + (k − 1)n + t − a) =

(1 + r + (k − 1)n + t) ↓_{n}

where (a) ↓_{n} is defined by (a) ↓_{0}= 1 and (a) ↓_{n}= a(a − 1) ⋯ (a − n + 1) for n ≥ 1. Thus it follows that

\binom{n}{s}^{2}\binom{(k − 1)n + t}{n − s}s!(n − s)!((k − 1)n + t)! =  (20)

((k − 1)n + t)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 1)n + t + r}{r}\binom{kn + t + 1}{s − r}(1 + r + (k − 1)n + t) ↓_{n} .

Case II. m = kn + t where i ≤ t ≤ k − 1. In this case, x_{m} = |X ∩ [m]| = n + 1 and x_{n}^{c} = |[m] − X| = (k − 1)n + t − 1. Then it follows from Theorem 2 that

V_{m,s}^{X,X} = \binom{n + 1}{s}^{2}\binom{(k − 1)n + t − 1}{n + 1 − s}s!(n + 1 − s)!((k − 1)n + t − 1)!.  (21)

On the other hand, we can obtain a formula for V_{m,s}^{X,X} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = kn + t − (n + 1) = (k − 1)n + t − 1. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (n + 1 − (a + 1)) − (2a + 1)
= (k − 1)n + t − 1 − a.  (22)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

∏_{a=0}^{n}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i}) =

∏_{a=0}^{n−1}(1 + r + (k − 1)n + t − 1 − a) =

(r + (k − 1)n + t) ↓_{n+1} .

Thus it follows that

\binom{n + 1}{s}^{2}\binom{(k − 1)n + t − 1}{n + 1 − s}s!(n + 1 − s)!((k − 1)n + t − 1)! =  (23)

((k − 1)n + t − 1)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 1)n + t − 1 + r}{r}\binom{kn + t + 1}{s − r}(r + (k − 1)n + t) ↓_{n+1} .

Next we consider the case of computing A_{n,s}^{X,Y} where X = Y = i + kℕ where k ≥ 2 and 0 ≤ i ≤ k − 1. Let A = i + kℕ and B = i − 1 + kℕ. Then it is easy to see that if m + 1 = kn + i ∈ X ∩ Y = X, then x_{m} = y_{m} = n and b_{m} = 2n + 1. Thus it follows from Theorem 7 that A_{n,k}^{X,Y} = D_{n,s}^{A,B} for all n and s in this case. We then have two cases.

Case A. m = kn + t where 0 ≤ t < i − 1. In this case, x_{m} = n and x_{m}^{c} = (k − 1)n + t. Thus if follows from Theorem 3 that

A_{m,s}^{X,X} = n!((k − 1)n + t)!\binom{n − 1}{s}\binom{(k − 1)n + t + 1}{n − s}.  (24)

On the other hand, we can obtain a formula for A_{m,s}^{X,X} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = (k − 1)n + t. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (n − (a + 1)) − (2a + 1)
= (k − 1)n + t − 1 − a.  (25)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

∏_{a=0}^{n−1}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i}) =

∏_{a=0}^{n−1}(1 + r + (k − 1)n + t − 1 − a) =

(r + (k − 1)n + t) ↓_{n} .

Thus it follows that

n!((k − 1)n + t)!\binom{n − 1}{s}\binom{(k − 1)n + t + 1}{n − s} =  (26)

((k − 1)n + t)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 1)n + t + r}{r}\binom{kn + t + 1}{s − r}(r + (k − 1)n + t) ↓_{n} .

Case B. m = kn + t where i ≤ t < k − 1. In this case, x_{m} = n + 1 and x_{m}^{c} = (k − 1)n + t − 1. Thus if follows from Theorem 3 that

A_{m,s}^{X,X} = (n + 1)!((k − 1)n + t − 1)!\binom{n}{s}\binom{(k − 1)n + t}{n + 1 − s}.  (27)

On the other hand, we can obtain a formula for A_{m,s}^{X,X} = D_{m,s}^{A,B} from equation (6). In this case |A_{m}^{c}| = (k − 1)n + t − 1. For any 0 ≤ a ≤ n,

α_{A,m,ak+i} + β_{B,m,ak+i} = kn + t − 1 − (n + 1 − (a + 1)) − (2a + 1)
= (k − 1)n + t − 2 − a.  (28)

Thus

∏_{x∈A_{m}}(1 + r + α_{A,m,x} + β_{B,m,x}) =

∏_{a=0}^{n+1}(1 + r + α_{A,m,ak+i} + β_{B,m,ak+i}) =

∏_{a=0}^{n+1}(1 + r + (k − 1)n + t − 2 − a) =

(r + (k − 1)n + t − 1) ↓_{n+1} .

Thus it follows that

n!((k − 1)n + t − 1)!\binom{n + 1}{s}\binom{(k − 1)n + t}{n + 1 − s} =  (29)

((k − 1)n + t − 1)! ∑_{r=0}^{s}(−1)^{s−r}\binom{(k − 1)n + t − 1 + r}{r}\binom{kn + t + 1}{s − r}(r + (k − 1)n + t − 1) ↓_{n+1} .

4 Direction for future research

In this section, we shall describe some problems for further research that naturally arise from the work in this paper.

There are other statistics which are closely related to the statistics that we consider in this paper. For example, suppose that X, Y ⊆ ℕ and define

γ_{X,Y} (σ) = |{i ∈ X : σ_{i} ∈ X} ∪ {i ∈ Y : σ_{i} ∈ Y }|.

Let Γ_{n,s}^{X,Y} = |{σ ∈ S_{n} : γ_{X,Y} (σ) = s}|. Then we have the following theorem.

Theorem 9. For any X and Y such that X ∪ Y = ℕ and X ∩ Y = ∅, we have Γ_{n,s}^{X,Y} = 0 unless s = 2k + y_{n} − x_{n} for some k, in which case

Γ_{n,2k+y_{n}−x_{n}}^{X,Y} = (x_{n})!(y_{n})!\binom{x_{n}}{k}\binom{y_{n}}{x_{n} − k}.

Proof. Suppose we pick k positions in X_{n} to contain elements in X_{n} in \binom{x_{n}}{k} ways and we pick x_{n} − k positions Y_{n} to put the other elements of X_{n} in \binom{y_{n}}{x_{n}−k} ways. The remaining positions in the permutation must be filled with elements of Y_{n}. Next arrange elements in X_{n} in (x_{n})! ways and we arrange elements in Y_{n} in (y_{n})!. Clearly the number of permutations σ that can be constructed in this way is (x_{n})!(y_{n})!\binom{x_{n}}{k}\binom{y_{n}}{x_{n}−k}. Note that our construction forces y_{n} − (x_{n} − k) elements of Y_{n} to be in positions in Y_{n} so that for any σ constructed in this way γ_{X,Y} (σ) = 2k + y_{n} − x_{n}.

Note that in the special case where X = 𝔼 and Y = 𝕆, we have that Γ_{2s,2n}^{𝔼,𝕆} = (n!)^{2}\binom{n}{s}^{2} and Γ_{2s+1,2n+1}^{𝔼,𝕆} = n!(n + 1)!\binom{n}{s}\binom{n+1}{s+1} which agrees with other formulas in our table in these special cases. However, for general X and Y, we get quite different recursions. For example, suppose that n + 1 ∈ X ∩ Y, σ ∈ S_{n}, i ∈ X − Y, and σ_{i} ∈ Y − X. Then it is easy to see that

γ_{X,Y} (I_{n+1}^{(i)}(σ)) = γ_{X,Y} (σ) + 2

so that the value of γ_{X,Y} can jump by 2 with a single insertion. This type of phenomenon does not happen with any of the other statistics studied in this paper. Thus it would be interesting to further study Γ_{n,s}^{X,Y} for arbitrary X, Y ⊆ ℕ to see if one can prove explicit formulas for Γ_{n,s}^{X,Y} which are similar to the formulas for D_{n,s}^{X,Y} given in Theorems 4 and 5.

Even though we found solutions to all the bijective questions related to the objects in our table, in some cases, one should be able to modify our bijections (find new ones) to preserve more than one statistic.

Recall that the statistic S_{10} is the number of odd descent-tops, and S_{12} is the number of (odd,odd) pairs. In this section, we will use the following statistics as well.

• S_{17} — the length of the maximal subsequence of the form 12 ⋯ i in a permutation. E.g., S_{17}(34152) = 2 while the increasing permutation of length n gives the maximum value of S_{17} in S_{n}. A modification of this statistic was studied by Zeilberger [5] in connection with 2-stack sortable permutations.

• T_{1} = S_{10} but not S_{12}.

• T_{2} = S_{12} but not S_{10}.

• T_{3} = S_{10} and S_{12}.

Our first conjecture is the following joint equidistribution:

Conjecture 10. The following should be true: (S_{10}, S_{12}, S_{17}) ∼ (S_{12}, S_{10}, S_{17}).

Notice, that Conjecture 10 suggests existence of an involution turning S_{10} to S_{12} and vice versa. This conjecture can be refined as follows.

Conjecture 11. (T_{1}, T_{2}, T_{3}, S_{17}) ∼ (T_{2}, T_{1}, T_{3}, S_{17}). That is, if the involution mentioned above exists, it is likely to leave pairs that are both S_{10} and S_{12} untouched.

Observe that to preserve statistic S_{17} in Conjectures 10 and 11, we need to require the increasing n-permutation to go to itself, and this is the only thing we need to worry about in our recursive construction of the bijection regarding S_{17} as otherwise it is not changed and thus preserved by induction no matter where we stick the largest element. So, it seems like we should be able to have the increasing permutation as a fixed point.

Here is how a proof of Conjecture 11 could be arranged for even n assuming the rest is constructed by induction. For odd n’s things seem to be much more complicated.

For n = 1, 1 is mapped to 1. Suppose we have constructed a bijection from S_{n−1} to S_{n−1} (n is even) such that it sends k (resp. ℓ, s) occurrences of T_{1} (resp. T_{2}, T_{3}) to k (resp. ℓ, s) occurrences of T_{2} (resp. T_{1}, T_{3}). Inserting n in T_{1} (resp. T_{2}, T_{3}) pair decreases the number of T_{1} (resp. T_{2}, T_{3}) by 1 keeping all other statistics unchanged. Clearly, we can manage the corresponding insertion on the other side that would decrease by 1 the number of occurrences of the corresponding statistic. Inserting n at any other position does not change a thing in either side and can be matched to each other. In particular, inserting n at the end corresponds to inserting n at the end and this guarantees that the statistic S_{17} is preserved (either it is unchanged in both cases, or assuming we deal with the increasing permutation 12 ⋯ (n − 1) going to itself, S_{17} is increased by 1 in both cases).

References

[1] D. C. Foata and M. P. Schützenberger, Théorie Géométrique des Polynômes Euleriens, Lecture Notes in Math. 138, Springer-Verlag, Berlin, 1970.

[2] J. Hall and J. Remmel, Counting descent pairs with prescribed tops and bottoms, J. Combin. Theory Ser. A 115 (2008), 693–725.

[3] S. Kitaev and J. Remmel, Classifying descents according to equivalence mod k, Electron. J. Combin. 13 (2006), #R64.

[4] S. Kitaev and J. Remmel, Classifying descents according to parity, Ann. Comb. 11 (2007), 173–193.

[5] D. Zeilberger, A proof of Julian West’s conjecture that the number of two-stack-sortable permutations of length n is 2(3n)!/((n+1)!(2n+1)!), Discrete Math. 102 (1992), 85–93.

2000 Mathematics Subject Classification: Primary 05A15, 05A19 Secondary 05A05.

Keywords: descent, adjacent pair, place-value-pair, equidistribution, binomial identity.

Received March 14 2009; revised version received June 21 2009. Published in Journal of Integer Sequences, June 21 2009.

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