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Parity Alternating Permutations Starting with an Odd Integer

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Parity Alternating Permutations Starting with an Odd Integer Parity alternating permutations starting with an odd integer Frether Getachew Kebede1a, Fanja Rakotondrajaob aDepartment of Mathematics, College of Natural and Computational Sciences, Addis Ababa University, P.O.Box 1176, Addis Ababa, Ethiopia; e-mail: [email protected] bD´epartement de Math´ematiques et Informatique, BP 907 Universit´ed’Antananarivo, 101 Antananarivo, Madagascar; e-mail: [email protected] Abstract A Parity Alternating Permutation of the set [n] = 1, 2,...,n is a permutation with { } even and odd entries alternatively. We deal with parity alternating permutations having an odd entry in the first position, PAPs. We study the numbers that count the PAPs with even as well as odd parity. We also study a subclass of PAPs being derangements as well, Parity Alternating Derangements (PADs). Moreover, by considering the parity of these PADs we look into their statistical property of excedance. Keywords: parity, parity alternating permutation, parity alternating derangement, excedance 2020 MSC: 05A05, 05A15, 05A19 arXiv:2101.09125v1 [math.CO] 22 Jan 2021 1. Introduction and preliminaries A permutation π is a bijection from the set [n]= 1, 2,...,n to itself and we will write it { } in standard representation as π = π(1) π(2) π(n), or as the product of disjoint cycles. ··· The parity of a permutation π is defined as the parity of the number of transpositions (cycles of length two) in any representation of π as a product of transpositions. One 1Corresponding author. n c way of determining the parity of π is by obtaining the sign of ( 1) − , where c is the − number of cycles in the cycle representation of π. That is, if the sign of π is -1, then π is called an odd permutation, and an even permutation otherwise. For example, the permutation 42178635= (1 4 7 3)(2)(5 8)(6), of length 8, is even since it has sign 1. All basic definitions and properties not explained here can be found in for example [8] and [4]. According to [9], a Parity Alternating Permutation over the set [n] is a permutation, in standard form, with even and odd entries alternatively (in this general sense). The set of all parity alternating permutations is a subgroup of the symmetric group S , the Pn n group of all permutations over [n]. The order of the set set has been studied lately Pn in relations to other number sequences such as Eulerian numbers (see [10, 9]). However, in this paper we will deal only with the parity alternating permutations which in addition have an odd entry in the first position; and we call them PAPs. It can be shown that the set Pn containing all PAPs over [n] is a subgroup of the symmetric group S and also of the group . We consider this kind of permutations because, n Pn for odd n there are no parity alternating permutations over [n] beginning with an even integer. Avi Peretz determined the number sequence that count the number of PAPs (see A010551). Unfortunately, we could not find any details of his work. In A010551, we can also find the exponential generating function of these numbers due to Paul D. Hanna. Since there is no published proof of this formula we prove it here, as Theorem 2.2. Moreover, the numbers that count the PAPs with even parity and with odd parity (which were not studied before) are determined. By pn we denote the cardinality of the set Pn of all PAPs over [n]. Let φn denote a map n n from Pn to S S that relates a PAP σ to a pair of permutations (σ1, σ2) in the set ⌈ 2 ⌉ × ⌊ 2 ⌋ n n σ(2i 1)+1 σ(2i) S S , in such a way that σ1(i)= − and σ2(i)= . It is easy to see that ⌈ 2 ⌉ × ⌊ 2 ⌋ 2 2 this map is a bijection. For example, the PAPs 5 2 1 4 3 6 7 and 7 4 5 6 3 2 1 over [7] are mapped to the pairs (3124, 123) and (4321, 2 3 1), respectively. If we consider a PAP 2 σ in cycle representations, then each cycle consists of integers of the same parity. Thus, we immediately get cycle representation of σ1 and σ2. For instance, the cycle form of the two PAPs above are (153)(7)(2)(4)(6) and (17)(35)(246) which correspond to the pairs ((132)(4), (1)(2)(3)) and ((14)(23), (1 2 3)), respectively. (Unless stated otherwise we will always use (disjoint) cycle representation of permutations.) Another way of looking at the mapping φn is that σ1 and σ2 correspond to the parts that contain the odd and even integers in σ, respectively. Therefore, studying PAPs is similar to studying the two permutations that correspond to the even and the odd integers in the PAP separately and then combining the properties. In Table 1, we give a short summary of properties that permutations and PAPs satisfy (for detailed discussions, see Section 2). Permutations PAPs Seq 1, 1, 2, 6, 24, 120,.... (A000142) 1, 1, 1, 2, 4, 12,.... (A010551) 2√4 x2+2 cos−1(1 x2/2) EGF 1 − − 1 x (2 x)√4 x2 − − − Even (seq) 1, 1, 1, 3, 12, 60,.... (A001710) 1, 1, 1, 1, 2, 6, 18, 72,... Odd (seq) 1, 1, 1, 3, 12, 60,.... (A001710) 0, 0, 0, 1, 2, 6, 18, 72,... x2 √4 x2+cos−1 1 2 x2 − − 2 x2 x 1 Even (EGF) − + + + 2 2x (2 x)√4 x2 4 2 2 − − − 2 2 −1 x 2 √4 x +cos 1 2 Odd (EGF) x − − 2 x x 1 2 2x (2 x)√4 x2 4 2 2 − − − − − − Table 1: A comparison table of permutations and PAPs (EGF mean exponential generating function). One interesting subset of S is the set D of derangements. For d = D , we have a n n n | n| well known relation dn =(n 1)[dn 1 + dn 2], d0 = 1 and d1 =0 (1) − − − for n 2. A proof of this relation may be found in any textbook on combinatorics, but ≥ we will have later use of the following bijection due to Mantaci and Rakotondrajao ([6]). They define ψn to be the bijection between Dn and [n 1] (Dn 1 Dn 2) as follows: − × − ∪ − 3 (1) let Dn denote the set of derangements over [n] having the integer n in a cycle of length (2) greater than 2, and Dn be the set of derangements over [n] having n in a transposition. These two sets are disjoint and their union is D . Then for δ D define ψ (δ)=(i, δ′), n ∈ n n 1 where i = δ− (n) and δ′ is the derangement obtained from (1) δ Dn by removing n or • ∈ (2) δ Dn by removing the transposition (i n) and then decreasing all integers • ∈ greater than i by 1. For instance, the pairs (2, (152)(34)) and (2, (1 2)(3 4)) correspond to the derangements (1526)(34) and (13)(45)(26), respectively, for n = 6. We denote the restricted bijec- (1) (2) tions ψn (1) and ψn (2) by ψn and ψn , respectively. |Dn |Dn Another important, and more difficult to prove, recurrence relation that the numbers dn satisfy is n dn = ndn 1 +( 1) , d0 =0 (2) − − for n 1. We will later make a use of the bijection τn : ([n] Dn 1) Fn Dn En ≥ × − \ −→ \ given by the second author ([7]) proving the recurrence. Where En is the set containing the derangement ∆ = (12)(34) (n 1 n) for even n, and is empty for odd n. F n ··· − n is the set containing the pair (n, ∆n 1) when n is odd, and is empty when n is even. − Thus, the inverse ζn of τn relates an element of [n 1] Dn 1 with every derangement − × − over [n] that has the integer n in a cycle of length greater than 2, and an element of n Dn 1 Fn with every derangement over [n] in which n lies in a transposition. { } × − \ Classifying derangements by their parity, we denote the number of even and odd de- e o e o rangements over [n] by dn and dn, respectively. Clearly dn = dn + dn. Moreover, the e o numbers dn and dn satisfy the relations e o o o e e dn =(n 1)[dn 1 + dn 2] and dn =(n 1)[dn 1 + dn 2], (3) − − − − − − for n 2 with initial conditions de = 1, de = 0, do = 0, and do = 0 ([6], Proposition ≥ 0 1 0 1 4.1). 4 We will put a major interest on Parity Alternating Derangements (PADs) which are the derangements which also are parity alternating permutations starting with odd integers. Let d denote cardinality of the set of PADs D = D P . The restricted bijection n n n ∩ n n n Φn = φn Dn : Dn D D will let us consider the odd parts and the even parts of | −→ ⌈ 2 ⌉ × ⌊ 2 ⌋ any given PAD regarded as ordinary derangements with smaller length than the length of the PAD. The mapping Φn plays the central role in our investigations. In Table 2, we display the connection of ordinary derangements and PADs (for detailed discussions, see Section 3). Finding explicit expressions for some of the generating functions are still open questions. On the other hand the EGF for the PADs for example is the solution to an eighth order differential equation with polynomial coefficients, and also is expressible in terms of Hadamard products of some known generating functions. Derangements PADs Seq 1, 0, 1, 2, 9, 44,... (A000166) 1, 0, 0, 0, 1, 2, 4, 18, 81, 396,..
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