Enumeration of Dumont permutations avoiding certain four-letter patterns

In this paper, we enumerate Dumont permutations of the fourth kind avoiding or containing certain permutations of length 4. We also conjecture a Wilf-equivalence of two 4-letter patterns on Dumont permutations of the first kind.


Preliminaries
This paper is concerned with the enumeration of pattern-restricted Dumont permutations. We will begin by defining patterns and Dumont permutations of various kinds, the original two kinds defined by Dumont [13] and the more recent third and fourth kinds defined by Burstein et al. [11]. We will then discuss previous results on this topic and present several new avoidance results for Dumont permutations of the fourth kind, an enumeration result for Dumont permutations of the fourth kind containing a single occurrence of a pattern, as well as a Wilf-equivalence conjecture on restricted Dumont permutations of the first kind.

Patterns
We begin with an example of pattern containment. Suppose we are given a permutation, say p = 26483751, and another permutation usually of shorter or equal length, e.g. q = 132. We say that the string 265 is order-isomorphic to 132. In such a case, we say that p contains q as a pattern, and that 265 is an instance, or occurrence, of q in p. Notice also that 265 is not the only instance of q in p, so are the subsequences 287 and 475.
On the other hand, for the same p and a pattern q ′ = 1234, we see that there is no length 4 subsequence of p that is order-isomorphic to q ′ . In this case we say that p avoids q ′ . Let us now define pattern containment and pattern avoidance more formally.

Dumont permutations
Dumont permutations are a certain class of permutations shown by Dumont [13] to be counted by the (unsigned) Genocchi numbers [23, A110501], a multiple of the Bernoulli numbers. Let the Genocchi and Bernoulli numbers be denoted G 2n and B 2n , respectively, then we have G 2n = 2(1 − 2 2n )(−1) n B 2n .
The exponential generating functions for the unsigned and signed Genocchi numbers are given by The definition of Dumont permutations of the first and third kinds involves descents and ascents, and thus, the linear structure of permutations. The definition of Dumont permutations of the second and fourth kinds involves fixed points, excedances, and deficiencies, and thus the cyclic structure of permutations.
We only consider Dumont permutations of even size since it will be clear from the definitions to follow that Dumont permutations of each kind of size 2n + 1 are obtained by appending the fixed point 2n + 1 to the right of a Dumont permutation of the same kind of size 2n. Definition 2.3. A Dumont permutation of the first kind (or Dumont-1 permutation for short) is a permutation wherein each even entry must be immediately followed by a smaller entry, and each odd entry must be immediately followed by a larger entry, or ends the permutation (i.e. the last entry must be odd). In other words, for all i, 1 ≤ i ≤ 2n, and some k, 1 ≤ k < n, π(i) = 2k =⇒ i < 2n and π(i) > π(i + 1), π(i) = 2k − 1 =⇒ π(i) < π(i + 1) or i = 2n.

Dumont permutations of the first and second kinds
Example 2.4. Let π = 435621 ∈ S 6 . Then 2, 4, and 6, the descent tops of π, are all even, while the descent bottoms of π are 3 and 5, and its final entry is 1, all odd. Thus, π is a Dumont permutation of the first kind.
Definition 2.5. A Dumont permutation of the second kind (or Dumont-2 permutation for short) is a permutation wherein each entry at an even position is a deficiency, and each entry at an odd position is a fixed point or an excedance. In other words, for all i, 1 ≤ i ≤ n, Example 2.6. Let ρ = 614352 ∈ S 6 . We see that the deficiencies of ρ occur at positions 2, 4 and 6, all even while excedances occur at positions 1 and 3, and a fixed point is at position 5, all odd. Thus, ρ is a Dumont permutation of the second kind.
In particular, the first entry of π is always a left-to-right maximum π. The map f , called the Foata's fundamental transformation [15], is defined as follows: • Start with a Dumont permutation of the first kind, say π.
• Insert parentheses to display the permutation in cycle notation, so that each cycle starts with a left-to-right maximum.

Dumont permutations of the third and fourth kinds
With respect to the next two types of Dumont permutations, Kitaev and Remmel [17,18] first conjectured that sets of permutations where each descent is from an even value to an even value are also counted by the Genocchi numbers. Burstein and Stromquist [12] proved this conjecture and called those sets of permutations Dumont permutations of the third kind. As with Dumont-1 and Dumont-2 permutations, the Foata's fundamental transformation maps Dumont permutations of the third kind onto a related equinumerous set that Burstein and Stromquist [12] called Dumont permutations of the fourth kind. We now introduce these two sets.
Definition 2.10. A Dumont permutation of the third kind (or Dumont-3 permutation for short) of size 2n (n ∈ N) is a permutation where each descent is from an even value to an even value, that is all descent tops and all descent bottoms are even. In other words, for all i, 1 ≤ i ≤ 2n − 1, π(i) > π(i + 1) =⇒ π(i) = 2l and π(i + 1) = 2k for some k and l, 1 ≤ k < l ≤ n.
Example 2.11. Let π = 16238457 ∈ S 8 . Then the descent tops of π are 6 and 8 are and the descent bottoms are 2 and 4, all even entries. Thus, π is a Dumont permutation of the third kind.
Definition 2.12. A Dumont permutation of the fourth kind (or Dumont-4 permutation for short) of size 2n (n ∈ N) is a permutation where deficiencies must be even values at even positions. In other words, for Example 2.13. Let ρ = 13657284 ∈ S 8 . Then the deficiencies of ρ are π(6) = 2, π(8) = 4, where all the positions and entries involved are even. Thus, ρ is a Dumont permutation of the fourth kind.

Patterns of length three
For the six patterns of length three, all the corresponding pattern-restricted sets have been enumerated, both in Dumont-1 and Dumont-2 permutations. Theorem 2. 16 ([7]). For all n ≥ 1, It is important to note here that although reverses, complements, and inverses of patterns create symmetry classes and Wilf-equivalences in the set of all permutations, Dumont permutations are not closed under any symmetry operations, so that applying a symmetry operation to a pattern does not necessarily yield a pattern-avoiding set of the same cardinality.  This follows from the fact that every even position is a deficiency, every odd position is an excedance or a fixed point, and for any Dumont-2 permutation π(2) = 1 and π(2n − 1) = 2n or 2n − 1, which eventually leads to the conclusion that it is impossible to avoid 123, 132, and 213.

Patterns of length four
Now, for Dumont-1 and Dumont-2 permutations avoiding patterns of length four, there are several cases which are open, as well as several enumerations of permutations that avoid two patterns of length four simultaneously. We begin with Dumont-2 permutations avoiding a single pattern of length four.
Theorem 2.22 ([7]). For all n ≥ 0, |D 2 2n (3142)| = C n . For pattern 4132, note that 321 is a subsequence of 4132, therefore D 2 2n (321) ⊆ D 2 2n (4132). The following theorem shows that, in fact, the two sets are equal. To date, there are no other enumerations of Dumont-1 and Dumont-2 permutations avoiding a single pattern of length four. However, we conjecture a Wilf-equivalence on Dumont-1 permutations of patterns 2143 and 3421 (see Section 5.1). Now, we will consider simultaneous avoidance of a pair of patterns. We begin with Dumont-1 permutations. with the generating function Note that the sequence (b n ) = (1, 1, 3, 11, 39, 139, 495, . . .) is [23, A007482] shifted one position to the right. In other words, D 1 2n (2341, 1423) is equinumerous to the set of of subsets of [2n − 2] where each odd element m has an even neighbor (m − 1 or m + 1).

Avoidance on Dumont permutations of the fourth kind
In this section, we will consider pattern avoidance on Dumont-4 permutations. In previous patternavoidance literature, the first nontrivial cases to be analyzed were patterns of length 3. Since 1 cannot be a deficiency in a Dumont-4 permutation, it must be a fixed point, so all Dumont-4 permutations start with 1. Thus, we will also consider avoiding patterns π = (1, π ′ + 1), where π ′ is a permutation in S 3 , i.e. π ∈ {1234, 1243, 1324, 1342, 1423, 1432}. Note that in all but the first two cases, i.e. if π ′ does not start with 1, we have D 4 2n (π) = D 4 2n (π ′ ).
Since π(1) = 1, the three conditions above mean that, of the 2n − 5 entries in [4, 2n − 2], at most one is to the left of 3 and at most one is to the right of π(2n − 1). This leaves at least 2n − 7 ≥ 1 entries in [4, 2n − 2] that are to the right of 3 and to the left of π(2n − 1). Any such entry, together with 1, 3, and π(2n − 1) would form an occurrence of pattern 1234.
Proof: By definition of a Dumont-4 permutation π(1) = 1, in other words, 1 is a fixed point. Now assume that all odd entries from 1 through 2j − 1 are fixed points. Consider the entry in the next odd position, π(2j + 1), and as before, suppose π(2j + 1) = 2j + 1. It follows that • the entry 2j + 1 must be to the left of π(2j + 1), • π(2j + 1) > 2j + 1 as an odd position cannot be a deficiency, and • at least one even entry 2l ≤ 2j must be to the right of π(2j + 1). This is because all odd entries at most 2j + 1 are to the left of π(2j + 1), so if all even entries less than 2j + 1 are also to the left of 2j + 1, then there will be 2j + 1 entries to the left of position 2j + 1, which is impossible.
By induction, the lemma is proved.
Next, we will prove that if we have a deficiency at position 2k, then it must be the entry 2k − 2.
Proof of Theorem 3.2: Given Lemma 3.3, all odd entries are fixed. Thus the even values are the only entries of interest, and moreover, all even values occur in even positions. Furthermore, due to Lemma 3.4, a deficiency at position 2k must be the entry 2k − 2.
Note also that the only nondeficiency among the first k letters is the value k. Thus, we see inductively that the entire structure of π e is determined by the values of its nondeficiencies, one of which must be the entry n. Combining π e as above with the odd entries of π (all fixed points), we see that π also avoids 231. Thus, π e (and hence, π) is determined by the choice of a subset S ⊆ [1, n−1] such that if k ∈ S, then 2k is a nondeficiency of π. All of these choices are unrestricted, therefore the number of such choices is 2 n−1 . Moreover, the semilengths of the resulting blocks of the same form as (1, 2k, π ′ ) yield a composition of n into nonzero parts. For example, π = 16325478 ∈ D 4 8 (1342) corresponds to the composition 4 = 3 + 1 (see Figure 1).
Consider a 2n × 2n board with the dots in the i-th column from the left being in π(i)-th row from the bottom, for 1 ≤ i ≤ 2n. Now consider the dots which represent even right-to-left minima (solid red dots in Figure 2), and lower them one cell down (hollow red dots in Figure 2). Notice that, in particular, this will place a hollow red dot in each row with an odd right-to-left minimum.
Now travel along the cell boundaries from (0, 0) to (2n, 2n) in an East-North fashion, using steps (1, 0) (east) and (0, 1) (north) and keeping all dots (both filled and hollow) to the left of the path, staying as close to the diagonal as possible. Equivalently, this is the path P where the peaks (instances where an east step is followed directly by a north step) are exactly the bottom and right boundaries of the cells with hollow red dots.
Suppose that π has k even non-excedances with (even) values 2 = b 1 < b 2 < · · · < b k at (even) positions a 1 < a 2 < · · · < a k = 2n, respectively. Also, let b k+1 = 2n + 2 and a 0 = 0. Then the runs (maximal blocks) of east steps have lengths a i − a i−1 for 1 ≤ i ≤ k, and the runs of north steps have Therefore, all runs of east and north steps in path P are of even length (see Figure 2, with n = 6 and even non-excedance values of 2, 4, 8, 10 at positions 4, 6,8,12). Dividing the lengths of these runs in half, we obtain a Dyck path of semilength n from (0, 0) to (n, n).

Enumerating Dumont-4 permutations avoiding certain permutations of length four
Now that all three Dumont-4 permutations of length four have been avoided by Dumont-4 permutations of length 2n, we will look at three other permutations of length four starting with 1, namely 1324, 1243, and 1423. Also note, excluding the entry "1", the three permutations in the previous section coupled with the three permutations in this section constitute all of S 3 .
When π(2n) = 2k < 2n, the number of possible choices for the position l of 2n is the number of elements in [2k, 2n − 1], i.e. 2n − 2k. Since 1 ≤ k ≤ n − 1, the total number of Dumont-4 permutations avoiding 1324 is  Proof: This enumeration is the same as that of D 4 2n (1324) since the removal of the entry 1 from patterns 1324 and 1243 yields the patterns 213 and 132 that are each other's reflections with respect to the antidiagonal. That is, 213 and 132 are inverses of reversals of complements of each other, and thus their respective avoidance classes are enumerated by the same sequence.
To obtain a permutation in D 4 2n (1243) from a permutation in D 4 2n (1324), remove the entry "1" and reflect the remaining entries from about the antidiagonal. This maps blocks onto blocks, with their diagonals mapping onto the diagonals of the images of those blocks. Lastly, add 1 to every entry and prepend the value 1 that was removed. See Figure 4 for an example. Note that reflection about the antidiagonal applied above implies that any permutation π ∈ D 4 2n (1243) is uniquely given by π −1 (2) and π(2) ∈ [2, π −1 (2)].

D 4 2n (1423)
As noted earlier, since the entry following 1 is not 2, it follows that D 4 2n (1423) = D 4 2n (312). Since our result for this pattern involves a continued fraction, we need a remark on notation.
Notation 3.11. For convenience, the following notation will be used for continued fractions: Additionally, we will need to use the truncations of the even and odd parts of the Catalan generating function C(z) = ∞ n=0 C n z n . Define where we set these functions equal to 0 when m < 0.
where the sequence of functions zR 2k+1 , k ≥ 0, satisfies the recurrence relation This yields the generating function for |D 4 2n (1423)| in the form of a continued fraction.
We consider a Dumont-4 permutation avoiding 1423, and we are interested in the parity of the smallest entry in each block using block decomposition after the "1". Note, since every Dumont-4 permutation begins with the entry "1", avoiding 1423 is the same as avoiding 312 by the Dumont-4 permutation with the "1" removed from the beginning of the permutation. Therefore we analyze the block decomposition of the resulting permutation diagram, ignoring the initial block of "1".
To analyze the blocks resulting from the iterations of the block decompositon, we will need an auxiliary parameter, namely, the length m of the maximal contiguous segment of allowed cells in the bottom row. We denote the blocks mnemonically using "E" for even and "N" for "not even," i.e. odd. However, for convenience in working with generating functions, we also let the EE (resp. N E, EN , and N N ) block with parity-restricted positions and values of m-deficiencies be represented by the generating function P m = P m (z) (resp. R m = R m (z), S m = S m (z), and T m = T m (z)), and refer to that block as a P -board (resp. R-board, S-board, and T -board). See Figure 5, where blue dots are odd positions and red dots are even positions in the starting D 4 -permutation diagram.
Proof: We now produce the generating function for the enumeration sequence of D 4 2n (1423) = D 4 2n (312). To begin, note that a P -board and T -board may be empty, while an R-board and S-board cannot be empty. This is due to the fact that P -boards and T -boards have even dimensions whereas R-boards and S-boards have odd dimensions.
Consider the recurrence relations for the generating functions corresponding to each of the four blocks. We use the position of the entry 1 in the block to produce those. Since a permutation in each block avoids 312, any value to the left of 1 must be less that any value to the right of 1. For the P -board with m = 2k, k ≥ 1, we have the generating function P 2k (z), the recurrence relation for which has three summands: • The P -board may be empty, which corresponds to the summand "1". the factor of z corresponds to the bottom row entry, in this case a blue dot on the P -board in Figure 5. The block to the left of the blue dot must be a square of even dimension ≤ 2k − 2 with all cells allowed, and the block to the right of the blue dot is an S-board with m = 2k, correponding to the generating function S 2k . This yields the summand zC e,2k−2 S 2k .
• Lastly, consider the entry in the bottom row that is in an even position ≤ 2k. Again, the factor z corresponds to the bottom row entry, in this case is a red dot on the P -board in Figure 5. The block to the left of the red dot must be a square of odd dimension ≤ 2k − 1 with all cells allowed, and the block to the right of the red dot is a P -board with m = 2k, corresponding to the generating function P 2k . This yields the summand zC o,2k−1 P 2k .
Thus, the recurrence formula for our generating function P 2k is given by We find the remaining recurrence formulas for R 2k+1 , S 2k , T 2k+1 in the same fashion, which results in the following system of equations: Now, consider the sequence {|D 4 2n (1423)|} n≥0 = {|D 4 2n (312)|} n≥0 . We claim that its ordinary generating function is R 1 /z. Indeed, removing the top row and the rightmost column of the R-board with m = 1 yields exactly the board of allowed cells in a Dumont-4 permutation. Now consider the function R 1 /z. Solving the system of equations (2) for R 2k+1 , k ≥ 0, yields the following recursive formula: Multiplying both sides by z, we obtain for all k ≥ 0: Note, the term on the left and the last term on the right are of the same form with k increasing by 1. This yields the continued fraction representation for the generating function R 1 /z = zR 1 /z 2 = ∞ n=0 |D 4 2n (312)|z 2n . Substituting √ z for z, we get the generating function for {|D 4 2n (312)|} n≥0 . Also note that the numerators of first three terms in the recurrence formula contain only the truncations of the even part of C(z), and the denominators contain only the truncations of the odd part of C(z). Moreover, all of those numerators and denominators are even functions. Furthermore, it is not difficult to see that the term zR 2k+3 in the recursive formula (1) does not contribute to the terms of degree at most 6 in zR 2k+1 (due to the factors of z 2 in the three numerators). Applying this observation iteratively, we see that deleting the term zR 2k+1 in the resulting recursive formula for R 1 /z does not affect the terms of degrees 2n ∈ [0, 6k − 2], i.e. 2n for 0 ≤ n ≤ 3k − 1.
Note that these are the coefficients for 0 ≤ n ≤ 11, so 3k − 1 = 11 yields k = 4, and therefore these terms can be found by expanding R 1 /z iteratively until we reach zR 9 , then removing the term zR 9 .

A single occurrence of patterns in Dumont permutations
The results of this section focus on enumeration of Dumont permutations with a single occurrence of certain patterns.

One occurrence in Dumont-1 and Dumont-2 permutations
We first introduce notation that is useful for going back and forth between sequences and their generating functions. Then we will review results on single occurrences of patterns in Dumont-1 and Dumont-2 permutations. Finally, we will prove a related result on a single occurrence of a pattern in Dumont-4 permutations.

A single occurrence of patterns in Dumont-4 permutations
The results of this subsection focus on a single occurrence of patterns in Dumont-4 permutations. To begin, let us consider a theorem.  This result enumerates all permutations containing a single copy of pattern 321. The original proof by both Noonan [22] and Noonan and Zeilberger [25] uses a complicated induction and generating functions with multiple auxiliary parameters. A much shorter argument has been subsequently given by the first author in [6], which was then shortered further by Zeilberger [29] and recently generalized by Bóna and the first author in [4]. Given the methods used in the proofs in [6,29], as well as the corresponding permutation diagram of the decomposition of permutations in S n (321; 1) as in [6], we will now prove a theorem first presented in [9], enumerating a single occurrence of 321 in Dumont-4 permutations.
It follows from Theorems 4.9 and 4.10 that the generating function for the sequence |D 4 2n (321; 1)|, n ≥ 0, is Proof: Let π ∈ D 4 2n (321; 1). Suppose that the single occurrence of pattern 321 is formed by values c > b > a and π = τ 1 cτ 2 bτ 3 aτ 4 for some strings τ i , i = 1, 2, 3, 4. Then b is a fixed point. Let π ′ and π ′′ be the patterns of τ 1 cτ 2 a and cτ 3 aτ 4 , respectively, as in [29]. Then π ′ and π ′′ avoid 321, π ′ does not end on its top entry, and π ′′ does not start with its bottom entry. Moreover, the entry a, the "1" of the occurrence of 321, is a deficiency and hence an even value in an even position in π.
Consider two cases based on the parity of b, the entry "2" of the single 321-occurrence in π.
Case 1: Suppose b is even, say, b = 2k for some k ∈ [2, n − 1] (note that the entry b cannot be either 2 or 2n, since neither is involved as a "2" in an occurrence of 321: the former because 1 is a fixed point, the latter because 2n is the largest entry).
Note that the sizes of ρ ′ and ρ ′′ add up to (2k + 2) + (2n − 2k + 2) = 2n + 4 in this case. Now, given permutations ρ ′ and ρ ′′ , use the map of Theorem 3.6 to produce two permutations σ ′ and σ ′′ , respectively, that avoid 321. Note that ρ ′ does not end on its top entry, and therefore neither does σ ′ . Likewise, ρ ′′ does not start with 12 . . . , and therefore σ ′′ does not start with 1. Thus, the pair σ ′ and σ ′′ are as in the proof in [6] with the combined length of (2n + 2)/2 = n + 1 or (2n + 4)/2 = n + 2. Thus, we can combine them as in [6] to produce a single permutation σ of length n or n + 1 (depending on the parity of b) with a single occurrence of pattern 321.

A conjecture on restricted Dumont-1 permutations
The table at the end of [7, Section 3.1] implies that no two Dumont-1 permutations of length 4 (recall: D 1 4 = {2143, 3421, 4213}) are Wilf-equivalent. However, we found that this is due to an incorrect calculation of |D 10 (3421)|, which is, in fact, equal to |D 10 (2143)|. Furthermore, we found that |D 10 (3421)| = |D 10 (2143)| for n ≤ 6; an additional computation by Albert [1] showed that the two sequences are the same for n ≤ 10, i.e. for Dumont-1 permutations of length up to 20. Therefore, we conjecture that the two sequences are equinumerous for all n.
2n , that is |D 1 2n (2143)| = |D 1 2n (3421)| for all n ≥ 0. Our conjecture is based on the following empirical observation. In all cases where sequences |S n (π 1 )| and |S n (π 2 )| are known to diverge (where π 1 and π 2 are patterns of the same length k), the first index at which the corresponding terms differ was found to be at most 2k − 1. Similarly, we expect Dumont permutations avoiding patterns of length k to be Wilf-equivalent if their avoidance sequences coincide for avoiding permutations of length up to 4k − 2. In our case, the patterns are of length k = 4, and the sequences coincide up to n = 10, i.e. up to length 20, greater than 4k − 2 = 14. Thus, we conjecture that they coincide for all n ≥ 0. See Table 1  This conjecture has turned out to be very difficult to prove, and only some partial inroads are made as of this writing. One approach to proving such a conjecture is to refine it by considering distributions of some combinatorial statistics on both D 1 2n (2143) and D 1 2n (3421). For this, we will first need to define a generalized version of a permutation pattern.
Definition 5.2 ([2]). A vincular (or generalized, or dashed) permutation pattern is a pattern that allows the additional requirement that two (or more) adjacent letters in a pattern be also adjacent in the containing permutation.
Note, in the pattern 2-31, the "2" and "3" need not be adjacent, however the "3" and the "1" must be adjacent. Now that we have the definition for vincular patterns, we can refine our conjecture slightly. After studying the behavior of distributions of occurrences of various vincular patterns on D 1 2n (2143) and D 1 2n (3421), we were able to form the following conjecture. Conjecture 5.3 ([9]). For all n ≥ 0, we conjecture the following. Let a n,k = |{π ∈ D 1 2n (2143)|(2-31)π = k}|, b n,k = |{π ∈ D 1 2n (3421)|(13-2)π = k}|, where (2-31)π (resp. (13-2)π) is the number of occurrences of 2-31 (resp. 13-2) in π. Then a n,k = b n,k = 1 for k = n 2 , a n,k = b n,k = 0 for k > n 2 , and m k=0 a n,k ≥ m k=0 b n,k for 0 ≤ m ≤ n 2 , with equality for m = n 2 . This is less optimal than statistic distributions that are equal to each other; however, of the statistics we studied, this relation is the only interesting one we found. Additionally, we conjecture that both sequences (a k ) and (b k ) are unimodal for each n ≥ 0, equal at the tails of the distributions, while in the middle of the distributions we first have a block of a k > b k followed by a block of a k < b k with the switch in direction of the inequality occurring at approximately k = 2n − 5. See Table 2 for n = 5, Table 3 for n = 6, and Table 4 on the following page for n = 7.

Conclusion
In this paper, we have enumerated Dumont-4 permutations avoiding certain patterns, as well as Dumont-4 permutations containing a single occurrence of certain patterns. The Catalan numbers and powers of 2 that occur in the enumeration of permutations avoiding three-letter patterns were also encountered in our results. However, for some patterns we could only find a generating function in the form of a continued fraction, as in [27].
We also gave an intriguing conjecture regarding the Wilf-equivalence of a pair of patterns on Dumont permutations of the first kind. This conjecture has proven quite unyielding, and despite being presented twice at the open problem sessions at Permutation Patterns 2016 and 2018, progress on it is yet to be made.
It would also be interesting to see if other Wilf-equivalences exist on Dumont permutations of the first kind avoiding a single four-letter pattern. The same question may be posed for the other kinds of Dumont permutations as well.