Permutations with 0 or 1 fixed point in hyperoctahedral groups

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(1)國立臺灣師範大學數學系教學碩士班碩士論文. 指導教授：林延輯博士. Permutations with 0 or 1 fixed point in hyperoctahedral groups. 研究生：周鈺偵. 中華民國 108 年 6 月.

(2) 致謝. 這三年來邊工作邊念書的時光很踏實，雖然選擇了暑期在職教學碩士班，但仍希望自己能完成做純數研究的夢想，於是選擇了做組合學研究，雖然這條路很辛苦，但也因為有太多人的幫忙得以走完，謝謝所有曾經幫助過我的人。我要特別感謝指導教授林延輯老師，在我就讀教碩班的第一個暑假結束後，我拜託老師擔任我的指導教授，然而念教學碩士班做純數的人少之又少，而我又沒有任何組合學的基礎，大學時期修的離散課程也是很久遠的事。很感謝林延輯老師願意耐心從頭教導我，兩年來念組合學課本、寫習題、旁聽組合學課程、學習寫程式、看一些文章討論想法到慢慢地有基礎後可以做點東西，在老師的身邊學習到的不只是專業的知識，更多的是身為「老師」在教導學生時的恩威並施、正向積極、站在學生的角度思考和不降低標準的細心陪伴你成長，希望我也能成為這樣的老師。另外，我也要特別感謝游森棚老師，除了讓我旁聽組合學的課程，並在我研究論文的期間給了很重要的想法，在論文口試時也給予實質意見與肯定讓我的論文更完整。感謝口試時徐祥峻老師對論文的建議，感謝溫雅婷學妹在論文前後和當天體貼的幫忙，感謝數學系女排的教練和學妹在我的碩班生涯陪伴我打球讓我更有動力念書。最後感謝我的家人總是無條件地陪伴我做任何事，無論遇到什麼挫折總是可以回到家充個電再出發，謝謝你們成就今天的我，希望自己能持續進步，讓自己能更好。鈺偵 108.07.03.

(3) Contents 1. Introduction. 1. 2. Derangements in hyperoctahedral groups. 5. 3. Alternating permutations with maximal number of fixed points in. 14. hyperoctahedral groups 4. Strictly decreasing permutations. 16. 5. Conclusion and Discussion. 20. 6. Acknowledgement. 22. References.

(4) Permutations with 0 or 1 fixed point in hyperoctahedral groups Chou Yu-Jen Abstract In this thesis, we extend the work of fixed points on the permutations of [n] in two directions: firstly, we discuss the fixed points problems of hyperoctahedral groups Bn ; secondly, elements in Bn can be thought the letters are painted by two colors, it can be generalized with r colors. Moreover, we discuss the fixed point problems in the subsets alternating permutations of Bn (r). and strictly decreasing permutations of Sn . After removing all fixed points and standardizing the remaining letters, we focus on colored permutations with 0 or 1 fixed point. We obtain combinatorial correspondence between derangements and elements with exactly one fixed point together with their recursions and generating functions. Keywords. Derangements, hyperoctahedral groups, alternating permutations, colored permutations. 1. Introduction For any positive integer n, denote the set {1, 2, . . . , n} by [n]. Given a permuta-. tion σ on [n], a label k ∈ [n] is called a fixed point of σ if and only if σ(k) = k. A 1.

(5) permutation of [n] is called a derangement if it has no fixed point. Let dn denote the number of derangements of [n]. It is well-known (cf. [Tuc80]) that the sequence dn satisfies an easier recursion dn = (n − 1)(dn−1 + dn−2 ),. n ≥ 2,. (1). and also a harder recursion dn = ndn−1 + (−1)n ,. n ≥ 1.. (2). Combinatorial proofs of the latter identity (2) were given by Remmel [Rem83] and Wilf [Wil84] independently. And the generating function of dn is also wellknown [WG16]: f (x) =. X. dn. n≥0. xn e−x = n! 1−x. (3). Ordinary permutations on [n] can be thought as painted by only one color. Bagno ([Bag04, BG06]) generalizes this notion that allows r colors to be painted on permutations, which results in the colored permutations. In this article, we would like to generalize the notion of derangements to the colored permutations. (r). We first introduce the groups in Bn and Sn . Let Bn be the hyperoctahedral group of order n, that is, elements in Bn are permutations σ on the set {±1, ±2, ±3, · · · , ±n} such that σ(−i) = −σ(i) for all i ∈ [n], we use the notation ı := −i in this article [FH05]. In particular, elements of Bn can be thought as permutations whose letters are painted by either of two colors. Therefore when r (r). colors are available to paint the letters, we get the wreath product Sn of the cyclic group Cr of order r with the permutation group Sn (cf. [Bag04]). We call an index (r). i ∈ [n] is called a fixed point of σ ∈ Bn or σ ∈ Sn if σ(i) = i and i is uncolored, (in the case of Bn , we will refer the first color as uncolored.) A permutation σ without 2.

(6) (r). fixed points in Bn or Sn is still called a derangement. For a permutation σ in Bn (r). or Sn , we denote the set of fixed points of σ by FIX(σ), and fix(σ) is its cardinality. (r). There are two ways to represent the elements in Bn and Sn in this paper. First is the window notation and the other is cycle notation. In the window notation, permutations in Bn can be written as σ = σ1 σ2 · · · σn where σi = σ(i); and we can also denote σ ∈ Bn with cycle notation introduced by Reiner [Rei93]: each cycle (l1 , l2 , . . . , lk , . . . , lj ) means σ(|li |) = l(i+1) , 1 ≤ i ≤ j − 1 and σ(|lj |) = l1 where l1 , l2 , . . . , lj ∈ {±1, ±2, . . . , ±n}, |lj | = 6 |lk | if j 6= k. For example, consider the element   1 2 3 4 5 σ=  ∈ B5 . 3 2 1 4 5 Its window notation is σ = 3 2 1 4 5 ∈ B5 , and its cycle notation is written by σ = (1 3)(2)(4)(5). Its only fixed point is 2. (r). As like as elements in Bn , we also can denote the elements in Sn with window (r). notation and cycle notation. Elements in Sn will be written in the window notation σ = σ1 σ2 . . . σn , where each σi is a pair (ci , ji ) with c ∈ Cr and j ∈ [n]. We can (r). also write σ ∈ Sn in the cycle notation: each cycle (l1 , l2 , · · · , lk , · · · , lj ) means σ(|li |) = l(i+1) , 1 ≤ i ≤ j − 1 and σ(|lj |) = l1 where l1 , l2 , · · · , lj on [n] with r colors, |lj | = 6 |lk | if j 6= k. For example, consider the element   1 2 3 4 5  (3) σ=  ∈ S5 . 3 2 1 4 5 (3). Its window notation is σ = 3 2 1 4 5 ∈ S5 , and its cycle notation is σ = (1 3)(2)(4)(5). Its only fixed point is 2. 3.

(7) (r). This article begins with focus on derangements in Bn and Sn . First, we count the numbers of elements with k fixed points in Bn . By removing all fixed points and standardizing [FH08] the remaining letters from a permutation in Bn , we can get a derangement on fewer letters. For instance, let us consider the permutation π = 5 2 3 4 1 ∈ B5 with FIX(σ) = {2,3}. By removing all fixed points, we have σ = 5 4 1, then we standardize the remaining letters σ, we obtain a derangement τ = 3 2 1 ∈ B3 . With this procedure, we realize that enumeration on derangements is essential. Formulas similar to Equations (1), (2) and (3) are obtained for hyperoctahedral groups. We prove the easier recursion of the derangement numbers in Bn by using the cycle notation. Similarly, it can be extended to an easier recurrence of the (r). derangement numbers in Sn . Next, we prove the harder recursion in two different ways, we use the easier recursion to prove and another way is the combinatorial proof by removing all fixed points and standardizing [HX09] the remaining letters to get a bijection. Finally, we get the correspondence between the derangements in Bn and elements with exactly one fixed point in Bn . Then, we extend the results to (r). Sn and get the generating function which is similar to the generating function of dn in the end of section 2. At the start of this project, we observed that the relationship between the number of derangements in Bn and the number of elements in Bn with fix(σ) = 1. Next (r). we consider two subsets of Bn and Sn . In Section 3, we denote the subset of alternating permutations (“snakes”) by Gn,k , that is, the elements σ ∈ Bn satisfying σ(1) > σ(2) < σ(3) > σ(4) < · · · and having k fixed points. We enumerate the permutations with maximal number of fixed points in Gn,k . We discover an interesting identity between the derangements in Bn and the elements with maximal number 4.

(8) of fixed points in G2n and G2n+1 . To conclude Section 3, we give a combinatorial bijection for that identity. (r). Another interesting subset of Sn is the subset of strictly decreasing permutations, which we discuss in Section 4. By strict decrease we employ an order on the set Cr × [n] is found in Bagno and Garber [BG06]. It is easy to see that any strictly decreasing permutation can have at most 1 fixed point, therefore we only (r). need to enumerate the set of all strictly decreasing derangements in Sn , which is denoted by Sn,0 , as well as the set of all strictly decreasing permutations with only (r). one fixed point in Sn , which is denoted by Sn,1 . We give some enumerative results like the number of elements in Sn,0 , the connection between Sn,0 and Sn+1,1 , and we give the recursion and the generating function. In the end, we obtain a formula (r). analogous to Equation (2) for the strictly decreasing permutations in Sn . The paper is organized as follows. In Section 2, we discuss the derangements in hyperoctahedral groups together with some generalization. In Section 3, we give bijections between alternating permutations in B2n with maximal number of fixed points and the derangements in Bn . In Section 4, we enumerate the strictly (r). decreasing permutations in the wreath product Sn and provide their generating functions. We end this paper with some comments and future work in Section 5.. 2. Derangements in hyperoctahedral groups Our task in this section is to extend the easier recursion like Equation (1) and (r). the harder recursion like Equation (2) in Bn and Sn , finally we have the generating (r). function like Equation (3) in Sn .. 5.

(9) We first classify the permutations in Bn by their numbers of fixed points. Define Fn,k := {σ ∈ Bn : fix(σ) = k},. fn,k := |Fn,k |,. k = 0, 1, . . . , n.. Elements in Fn,0 are also known as the derangements in Bn . Below is the table of the first few values of fn,k : Table 1: Statistics fix on hyperoctahedral groups Bn n\k. 0. 0. 1. 1. 1. 1. 2. 5. 2. 1. 3. 29. 15. 3. 1. 4. 233. 116. 30. 4. 1. 5. 2329. 1165. 290. 50. 5. 1. 6. 27949. 13974. 3495. 580. 75. 6. 1. 7. 391285. 195643. 48909. 8155. 1015. 105. 7. 1. 2. 3. 4. 5. 6. 7. 1. We are going to prove a few identities involving the sequence fn,k . By picking out all fixed points and standardizing the remaining letters, the following result follows immediately. Proposition 2.1. Let n, k be integers with 0 ≤ k ≤ n. Then we have fn,k. n = fn−k,0 . k. (4). Proof. Let σ ∈ Bn have k fixed points. The number of ways to choose k fixed points from n is nk . By removing all fixed points of σ and standardizing the remaining letters, we obtain a derangement in fn−k,0 . 6.

(10) The derangement numbers satisfy the harder recurrence dn = ndn−1 + (−1)n . T. Benjamin and Joel Ornstein [BO17] used the cycle notation to present a simple combinatorial proof, and we use the similar idea to prove the recurrence of the derangement numbers fn,0 as follows. Theorem 2.2. The sequence hfn,0 in satisfies the following recursion: f0,0 = f1,0 = 1;. fn,0 = (2n − 1)fn−1,0 + 2(n − 1)fn−2,0 ,. ∀ n ≥ 2.. (5). Proof. Let σ ∈ Fn,0 be written in the cycle notation. We consider how to generate σ from elements in Fn−1,0 and Fn−2,0 : 1. In Fn−1,0 , we can insert a cycle (n) and get fn−1,0 elements in Fn,0 . Or we can put n or n behind n − 1 letters within any cycle, so we can get (n − 1) · 2 · fn−1,0 elements in Fn,0 . 2. Elements in Fn−1,1 can be inserted n or n into the cycle (j), where j ∈ {1, 2, . . . , n − 1}, and we have n − 1 choice of fixed point, by removing the only fixed point and standardizing, we have element in Fn−2,0 , so we can get (n − 1) · 2 · fn−2,0 elements in Fn,0 . For n ≥ 2, we have fn,0 = fn−1,0 + (n − 1) · 2 · fn−1,0 + (n − 1) · 2 · fn−2,0 = (2n − 1)fn−1,0 + 2(n − 1)fn−2,0. The previous result is parallel to the recursive formula satisfied by dn like Equation (1). Next we prove the result for fn,0 which is analogous to Equation (2).. 7.

(11) Theorem 2.3. For n ≥ 1, we have fn,0 = 2 · fn,1 + (−1)n . Proof. We provide two proofs here. First proof. From the Equation (4) and (5), we have f0,0 = 1, f0,1 = 0, and fn,0 − 2 · fn,1 = fn,0 − 2n · fn−1,0 = (2n − 1) · fn−1,0 + 2(n − 1) · fn−1,0 − 2n · fn−2,0 = −fn−1,0 + 2(n − 1) · fn−2,0 = −fn−1,0 + 2 · fn−1,1 (6) = (−1) · (fn−1,0 − 2 · fn−1,1 ) = · · · = (−1)n · (f0,0 − 2 · f0,1 ) = (−1)n · (1 − 0) = (−1)n Second proof. We give a combinatorial proof here. It is obvious that the identity holds for n = 1. The statement will then follow recursively by the identity fn,0 − 2 · fn,1 = 2 · fn−1,1 − fn−1,0 ,. ∀ n ≥ 2.. (7). We now give a combinatorial proof for Equation (7). Let 2 · Fn,1 denote the set . 2 · Fn,1 := (ε, σ) : ε ∈ {+, −}, σ ∈ Fn,1 , and 2 · Hn,1 be the subset of 2 · Fn,1 consisting of the elements (ε, σ) with σ(n) 6= n. We now construct a map θ from 2 · Hn,1 into Fn,0 . For (ε, σ) ∈ 2 · Hn,1 and ` being the fixed point of σ, let θ(ε, σ) be the signed permutation on [n] with 2 colors given by    σ(n), if k = `;    θ(ε, σ) (k) := ε · `, if k = n;      σ(k), otherwise 8. 1 ≤ k ≤ n..

(12) For example, if σ = 2 4 3 5 1 with the fixed point 3, then θ(−, σ) = 2 4 1 5 3. Clearly θ is an injective map into Fn,0 . The derangements π ∈ Fn,0 which is not in the image of θ must assume one of the two following forms: • π(n) = |n|. By deleting the trailing |n| from the word form of π, we obtain a derangement in Fn−1,0 . • ` := |π(n)| < n, and π(`) = n. In this case, these derangements can be mapped bijectively to 2 · Fn−1,1 by setting π 7→ (ε, π 0 ), where ε := sgn(π(n)), and π 0 (k) :=.    π(k), if k 6= `;   `,. 1≤k ≤n−1. if k = `;. Besides, the elements in 2 · Fn,1 which are not mapped by θ are those (ε, σ) such that σ(n) = n; by deleting the trailing n, it is easy to see that the number of those elements are 2 · fn−1,0 . By interpreting θ being a bijection between subsets of Fn,0 and 2 · Fn,1 , we see that fn,0 − fn−1,0 + 2 · fn−1,1 = 2 · fn,1 − 2 · fn−1,0 which is clearly equivalent to Equation (7), and the theorem is now proved. From the second proof of Theorem 2.3, we construct an almost bijection θn from 2Fn,1 to Fn,0 . Let (ε, σ) ∈ 2Fn,1 , where ε ∈ {+, −} and σ ∈ Fn,1 . The action of θn on (ε, σ) depends on the fixed point σ as follows. • FIX(σ) = {j}, j < n. We simply define    σ(n), if k = j,    θn (ε, σ) (k) := σ(k), if k = 6 j, n,      ε · j, if k = n, 9. 1 ≤ k ≤ n..

(13) Figure 1: Bijection between sets Fn,0 and 2 · Fn,1 For example, if σ = 2 4 3 5 1 with the fixed point j = 3, by exchanging j with σ(n), then we have θ5 (+, σ) = 2 4 1 5 3, and θ5 (−, σ) = 2 4 1 5 3. Clearly θ is (n). an injective map into F0 . • FIX(σ) = {n}. The sign ε will be significant in defining θn as shown below.    σ(k), if k 6= n, (a) θn (−, σ) (k) := 1 ≤ k ≤ n.   −n, if k = n, (b) When ε = +, we first purge the fixed point n from σ to obtain the derange−1 ment σ 0 = σ1 σ2 . . . σn−1 , then we look for the inverse image θn−1 (σ 0 ) = (ε0 , π),. where π ∈ Fn−1,1 (whenever it is possible). Let j be the fixed point of π. Then we set.    π(k), if k 6= j, n,    θn (+, σ) (k) := n, if k = j,     0  ε · j, if k = n,. 1 ≤ k ≤ n.. For example, if σ = 2 3 1 4 with the fixed point 4, then θ4 (−, σ) = 2 3 1 4. And by deleting the fixed point 4, we have θ4 (+, σ) from the word form of σ, we 10.

(14) obtain a derangement π = 2 3 1 ∈ F3,0 , then we know θ3−1 (π) = (−, 1 3 2) in F3,1 , so we can construct the map θ4 (+, 1 3 2 4) = 4 3 2 1. The correspondences F2,0 ↔ F2,1 and F3,0 ↔ F3,1 are shown below for the cases n = 2 and n = 3, where an element σ ∈ B2 and an element π ∈ B3 is represented by the sequence σ = σ1 σ2 and π = π1 π2 π3 , respectively: F2,0. ↔ (±, F2,1 ). 21. (+, 12). 21. (−, 12). 21. (+, 12). 12. (−, 12). 21. **. F3,0. ↔ (±, F3,1 ). F3,0. ↔ (±, F3,1 ). F3,0. ↔ (±, F3,1 ). 321. (+, 1 2 3). 231. (+, 1 3 2). 321. (+, 2 1 3). 321. (−, 1 2 3) 2 3 1. (−, 1 3 2) 2 1 3. (−, 2 1 3). 132. (+, 1 2 3). (+, 1 3 2). 312. (+, 3 2 1). 132. (−, 1 2 3) 2 3 1. (−, 1 3 2) 3 1 2. (−, 3 2 1). 132. (+, 1 2 3). (+, 2 1 3). 312. (+, 3 2 1). 123. (−, 1 2 3) 2 1 3. (−, 2 1 3). 312. (−, 3 2 1). 231. (+, 1 3 2). 132. (+, 2 1 3). 312. (+, 3 2 1). 231. (−, 1 3 2). 213. (−, 2 1 3) 3 1 2. (−, 3 2 1). 231. (+, 1 3 2). 321. (+, 2 1 3). 312. (+, 3 2 1). 231. (−, 1 3 2) 2 1 3. (−, 2 1 3). 312. (−, 3 2 1). 231. ***. Elements of Bn can be thought as permutations whose letters are painted by (r). either of two colors, next, we consider the elements in Sn which is painted by r 11.

(15) colored in Sn . Define (r). (r). Fn,k := {σ ∈ S(r) n : fix(σ) = k},. (r). fn,k := |Fn,k |,. (r). k = 0, 1, . . . , n. (r). Elements in Fn,0 are also known as the derangements in Sn . By the same reasoning (r) as in Propsition 2.1, we see that fn,k = nk f (r) n − k, 0. An argument similar to Theorem 2.3 can be used to prove the recurrence of the (r). derangement numbers fn,0 as follows: (r). Theorem 2.4. For r ∈ N. The sequence hfn,0 in satisfies the following recursion: (r). (r). (r). fn,0 = (rn − 1) · fn−1,0 + r · (n − 1) · fn−2,0 ,. ∀ n ≥ 2.. (8). (r). Proof. Let σ ∈ Sn be the cycle notation. As before, we consider to generate σ (r). (r). from elements in Fn−1,0 and Fn−2,0 , (r). 1. If elements in Fn−1,0 , we can insert a cycle (n), the letter n can be painted by (r). (r). (r − 1) colors to get (r − 1) · fn−1,0 elements in Fn,0 . Or we can put the letter n which is painted by r colors behind the letter j, j = 1, 2, ..., n − 1, so we can (r). (r). get r · (n − 1) · fn−1,0 elements in Fn,0 . (r). 2. If elements in Fn−2,0 , each element can be inserted a cycle which is (j n), j = 1, 2, ..., n − 1, and the letter n can be painted by r colors. Standardize, we (r). (r). can get r · (n − 1) · fn−2,0 elements in Fn,0 . For n ≥ 2 we have (r). (r). (r). (r). fn,0 = (r − 1) · fn−1,0 + r · (n − 1) · fn−1,0 + r · (n − 1) · fn−2,0 (r). (r). = (rn − 1) · fn−1,0 + r · (n − 1) · fn−2,0. 12.

(16) Theorem 2.5. For n ∈ N, we have: (r). (r). fn,0 = rfn,1 + (−1)n .. (9). Proof. The proof is analogous to that of Theorem 2.3. Using Theorem 2.4, we get (r). (r). (r). (r). fn,0 − rfn,1 = fn,0 − rnfn−1,0 (r) (r) (r) = (rn − 1)fn−1,0 + r(n − 1)fn−2,0 − rnfn−1,0 (r) (r) = − fn−1,0 − r(n − 1)fn−2,0 (r) (r) = − fn−1,0 − rfn−1,1 (r) (r) = · · · = (−1)n f0,0 − rf0,1 = (−1)n (1 − 0) = (−1)n .. We know the generating function of dn is given by f (x) =. P n≥0. n. dn xn! =. e−x . 1−x. (10). In the. end of this section, we prove the analogous closed form for the generating function (r). of fn,0 . (r). Theorem 2.6. The generating function of fn,0 is F (x) =. X. (r) x. fn,0. n≥0. n. n!. =. e−x 1 − rx. (11). (r). Proof. Let an = fn,0 , and we have the recurrences for the number of an is a0 = 1 and an = r · n · an−1 + (−1)n for n > 0. A routine computation shows that X n≥1. an. xn X xn X xn = n · r · an−1 + (−1)n . n! n! n≥1 n! n≥1. F (x) − 1 = r · x · F (x) + (e−x − 1). F (x)(1 − rx) = e−x . F (x) =. e−x . 1 − rx. 13. (12).

(17) Notice that the special case r = 1 reduces to the generating function of the number of derangements of the ordinary permutations.. 3. Alternating permutations with maximal number of fixed points in hyperoctahedral groups Han and Xin have extended alternating permutations with maximal number. of fixed points [HX09] in Sn . A permutation π = π1 π2 · · · πn ∈ Sn is said to be alternating (respectively reverse alternating) if π1 > π2 < π3 > π4 < · · · (respectively π1 < π2 > π3 < π4 > · · · ). In this section, we consider the subset Gn of Bn which consists of “snakes” (a.k.a. alternating) in Bn , i.e., σ ∈ Gn when σ(1) > σ(2) < σ(3) > σ(4) < · · · . We know that |Gn | = 2n · En [KPP94] , where En is the nth Euler number (1, 1, 1, 2, 5, 16, 61, . . . ). Again we define Gn,k := {σ ∈ Gn : fix(σ) = k},. gn,k := |Gn,k |,. k = 0, 1, . . . , b. n+1 c. 2. Theorem 3.1. We have g2n+1,n+1 = g2n,n = fn,0 . Proof. If σ ∈ Gn has maximal fixed points, then exactly one of 2i − 1 and 2i is fixed by σ for each 1 ≤ i ≤ b N2+1 c; in particular σ(2n + 1) = 2n + 1 if σ ∈ G2n+1,n+1 . By shrinking 2n + 1 from σ, we obtain a permutation σ 0 ∈ G2n,n . Hence the equality g2n,n = g2n+1,n+1 is already established. Now a map G2n,n → Fn,0 is easily constructed by removing the fixed points of σ and standardizing the remaining letters, as seen in the following example: 1 3 (10) 4 8 6 7 2 9 5. 7→. 3 (10) 8 2 5 14. 7→. 2 5 4 1 3..

(18) Table 2: Numbers of alternating permutations with k fixed points in Bn n\k. 0. 0. 1. 1. 1. 2. 3. 1. 3. 10. 5. 1. 4. 50. 25. 5. 5. 312. 156. 39. 5. 6. 2400. 1200. 275. 29. 7. 21168. 10584. 2646. 389. 1. 2. 3. 4. 29. For the element σ = σ1 σ2 σ3 σ4 · · · σ2n in G2n , each pair of letters σ2k−1 σ2k , k = 1, 2, · · · , n is a decent set and have only one fixed points, then the resulting permutation cannot have any fixed point. Next, we need to establish the inverse for the map above, i.e., from Fn,0 back to G2n,n . Firstly 2n − 1 must be a fixed point, hence n or n must correspond to 2n or 2n, respectively. If 2n is present, then it must be filled at the left position of its block, hence the right position of that block is a fixed point; if 2n is present, then it must be filled at the right position of its block, hence the left position of that block is a fixed point; hence we know what the next (unfixed) number (and its sign) would be. In general, once we know the unfixed number of this current block, then we know it should be put at the left or right position of the next block if each block must be [big]-[small]. And for the element π = π1 π2 · · · π2i−1 π2i π2i+1 π2i+2 · · · π2n , if π2i−1 and π2i+1 are fixed points, then π2i+1 > π2i−1 > π2i ; if π2i−1 and π2i+2 are fixed points, then π2i+1 > π2i+2 > π2i−1 > π2i ; if π2i and π2i+1 are fixed points, then 15.

(19) π2i+1 > π2i ; if π2i and π2i+2 are fixed points, then π2i+1 > π2i+2 > π2i , so that the resulting permutation π is a snake (i.e. alternating). Just keep going until the cycle is finished. Now find the rightmost unfilled block, its left position must be a fixed point, and trace the cycle, so on and so forth, so we are done. Example 3.2. Consider 2 1 4 5 3 ∈ F5,0 . The recovery is made through the following procedure: ∗ ∗ ∗ ∗ ∗ ∗. ∗. ∗ 9 ∗ 5 7→ (10). ∗ ∗ ∗ ∗ ∗ ∗ (10) 8 9 ∗ 4 7→ 7 ∗ ∗ ∗ ∗ 5 7 (10) 8 9 ∗ 3 7→ 6 ∗ ∗ ∗ ∗ 5 7 (10) 8 9 6. ···. ∗ ∗ 3 ∗ 5 7 (10) 8 9 6. 2 7→ 4. 1 4 3 ∗ 5 7 (10) 8 9 6. 1 7→ 2. 1 4 3 2 5 7 (10) 8 9 6 So, we have the bijection between the alternating permutations in B2n (and also B2n+1 ) with maximal numbers of fixed points and the derangements in Bn . (r). (r). (r). This sequel also can extend to Fn,k . Consider the subset Gn,k of Fn,k which con(r). (r). (r). (r). sists alternating permutations in Fn,k , define gn,k = |Gn,k |, we also have g2n+1,n+1 = (r). (r). g2n,n = fn,0 , which can be verified analogously as in Theorem 3.1.. 4. (r). Strictly decreasing permutations in Sn (r). In this section, we discuss another subset in Sn . We regard Cr as the set (r). {0, 1, . . . , r − 1} with orders inherited from Z. An element σ ∈ Sn can be represented in the window notation: σ = σ1 σ2 . . . σn , where each σi is a pair (ci , ji ) with c ∈ Cr and j ∈ [n]. We use an order on the set Cr × [n] found in Bagno and Gar16.

(20) ber [BG06]: declaring (c1 , j1 ) (c2 , j2 ) when either c1 < c2 , or c1 = c2 but j1 > j2 . (r). A permutation σ ∈ Sn is called strictly decreasing if σ(1) σ(2) · · · σ(n). Notice that this order is different from that in Section 3, nonetheless an obvious isomorphism exists between these orders. Let Sn,k be the set of all r-colored strictly decreasing permutations on [n] with k fixed points and the number of elements in Sn,k denoted by sn,k . If a permutation has two or more fixed points, there must be an ascent between any pair of those (r). fixed points, therefore it is impossible for any strictly decreasing permutation in Sn. with two or more fixed points. Henceforth we know the number of fixed points of σ ∈ Sn,k is either 0 or 1, then a simple count gives sn,0 + sn,1 = rn . (r). Let Sn,0 be the set of all strictly decreasing derangements in Sn and Sn,1 be the (r). set of all strictly decreasing permutations with exactly one fixed point in Sn . Now we give the enumerative results on these strictly decreasing permutations subject to the numbers of fixed points. Theorem 4.1. We have: (i) For n ≥ 1, we have sn,0 = sn+1,1 . (ii) Let t ∈ [n]. The number of permutations in sn,1 with t being the fixed point is n X n−t n−t n−2t+1 t−1 (r − 1) r . Therefore sn,1 = (r − 1)n−2t+1 rt−1 . t−1 t−1 t=1 (iii) For n ≥ 3, we have sn,0 = (r − 1) · sn−1,0 + r · sn−2,0 . (iv) The generating function of sn,0 is g(z) =. ∞ X. sn,0 z n =. n=0. Proof.. 1 . 1 − (r − 1)z − rz 2. (i) Let σ ∈ Sn+1,1 , we can remove the only fixed point and standardize. the remaining letters, we have a derangement in Sn,0 . On the other hand, let 17.

(21) π ∈ Sn,0 , we can insert a fixed point k right before the earliest deficiency of π, before we insert the fixed point k, we need to destandardize the letters π(i) if |π(i)| > k, then we can get the element in Sn+1,1 . (ii) Let σ ∈ Sn,1 with the only fixed point t, t ∈ [n]. The letters σ(1), . . . , σ(t − 1) can be chosen from t + 1, . . . , n without any color; there are n−t possible t−1 choices. For those letters behind σ(t), it can be colored by all r colors if its absolute value is less than t, but only by r − 1 colors if otherwise. By the multiplication principle, the number of strictly permutations in Sn,1 with t n X n−t being the only fixed point is (r − 1)n−2t+1 rt−1 . t − 1 t=1 (iii) By (i), for n ≥ 3, we have. sn,0 + sn,1 = rn = r · (sn−1,0 + sn−1,1 ) sn,0 + sn−1,0 = r · (sn−1,0 + sn−2,0 ). (13). sn,0 = (r − 1) · sn−1,0 + r · sn−2,0 (iv) Let g(z) =. ∞ X. sn,0 z n =. ∞ X. n=0. sn+1,1 z n .. n=0. n. Using the result of sn,k = r , we have ∞ X. 1 1 − rz. sn,1 z n =. 1 1 − rz. g(z) + zg(z) =. 1 1 − rz. sn,0 z +. n=0 ∞ X n=0. ∞ X. sn,1 z n =. n. n=0. sn+1,1 z n +. ∞ X n=0. 1 (1 − rz)(1 + z) 1 = 1 − (r − 1)z − rz 2. g(z) =. 18. (14).

(22) As an example of (i), consider the permutation π = 2 1 3 ∈ S3,0 , we should put the fixed point 2 right before the element 1. Before we insert the fixed point 2, we need to add 1 to the absolute value of the letters π(i) which satisfy |π(i)| > 2. As a result we reach the element 3 2 1 4 ∈ S4,1 . On the other hand, for ρ = 3 2 1 4 ∈ S4,1 , by removing the fixed point 2 and standardizing the remaining letters, we have a derangement 2 1 3 ∈ S3,0 . We construct a bijection between elements with exactly one fixed point in Sn+1,1 and derangements in Sn,0 . By Theorem 4.1 (iv), we have the generating function of sn,0 , then we can get Theorem 4.2. The number of strictly decreasing derangements in Sn,0 is sn,0 =. 1 · (rn+1 + (−1)n ). r+1. (15). Proof. In Theorem 4.1 (iv), we have ∞ X. sn,0 z n =. n=0. 1 1 − (r − 1)z − rz 2. 1 (1 + z)(1 − rz) 1 1 r = ( + ) r + 1 1 + z 1 − rz ∞ ∞ X 1 X n n = ( (−1) z + rn+1 z n ) r + 1 n=0 n=0 =. =. ∞ X n=0. So we have sn,0 =. 1 r+1. (. (16). 1 (rn+1 + (−1)n ))z n r+1. · (rn+1 + (−1)n ).. The number of derangements in Sn , we have the harder recursion dn = ndn−1 + (−1)n , and in section 2, we already prove the recursion fn,r,0 = rfn,r,1 + (−1)n in (r). Sn . In the end of this section, we have 19.

(23) Corollary 4.3. For n ∈ N, we have: sn,0 = r · sn,1 + (−1)n .. (17). Proof. By Theorem 4.1 (iv), we have ∞ X. sn,0 z n =. n=0. 1 . (1 − rz)(1 + z). And we also have ∞ X. n. (sn,0 + sn,1 )z =. n=0. ∞ X. rn z n =. n=0. 1 . (1 − rz). (18). Therefore, ∞ X. sn,1 z n =. n=0. 1 1 − (1 − rz) (1 − rz)(1 + z). 1+z−1 (1 − rz)(1 + z) z = (1 − rz)(1 + z) =. ∞ X. 1 rz (sn,0 − r · sn,1 )z = − (1 − rz)(1 + z) (1 − rz)(1 + z) n=0. (19). n. 1 1+z ∞ X = (−1)n · z n =. n=0. 5. Conclusion and Discussion In this study, we generalize the enumeration problems on derangements and. permutations with only one fixed point to colored permutations. In fact, the rela(r). tionship between the derangements in Sn and the elements with exactly only one 20.

(24) (r). fixed point in Sn can be classified with the descent sets. At first we observed the phenomenon on the enumeration on derangements for alternating permutations in Bn , i.e. the numbers in the first column of Table 2 are twice of those in the second column. It could be thought the special case of the following setting. For each subset J of [n − 1], define (r),J. Fn,k := {σ ∈ S(r) n : fix(σ) = k and DES = J},. (r),J. (r),J. fn,k := |Fn,k |,. k = 0, 1, . . . , n.. For example, the snakes in Bn have the descent sets {1, 3, 5, 7, 9, . . . } ∩ [n − 1]. (r),J. Using the mathematical software SAGE, we enumerate the cardinalities fn,k find the following relations:    r · f (r),J + (−1)n , if J = [n − 1]; n,1 (r),J fn,0 =  (r),J  r · fn,1 , if J 6= [n − 1].. and. (20). The case r = 1, which is the ordinary permutations, has been dealt with by Foata and Han [FH08]. They found a bijection on Sn that transforms the pair of statistics (fix, DES) to (pix, IDES) [DW93]. In section 4, we have already explained the situation with fn,0 = r · fn,1 + (−1)n , which corresponds to the case J = [n − 1]. We hope the we can find the correct notions for pixed points and IDES sets of colored permutations that enables us to establish the identity (20) for J 6= [n − 1] in the future.. 21.

(25) 6. Acknowledgement I am grateful to Professor Sen-Peng Eu gave the important ideas of permuata-. tions with r colors and also for his helpful comments of this paper.. References [Arn92] Vladimir I. Arnol’d, The calculus of snakes and the combinatorics of Bernoulli, Euler and Springer numbers of Coxeter groups, Russian Mathematical Surveys 47 (1992), no. 1, 1. [Bag04] Eli Bagno, Euler-Mahonian parameters on colored permutation groups, Séminaire Lotharingien de Combinatoire 51 (2004), B51f. [BG06]. Eli Bagno and David Garber, On the excedance number of colored permutation groups, Sém. Lothar. Combin 53 (2006), B53f.. [BO17]. Arthur T. Benjamin and Joel Ornstein, A bijective proof of a derangement recurrence, FIBONACCI QUARTERLY 55 (2017), no. 5, 28–29.. [DW93] Jacques Désarménien and Michelle L. Wachs, Descent classes of permutations with a given number of fixed points, Journal of Combinatorial Theory, Series A 64 (1993), no. 2, 311–328. [FH05]. Dominique Foata and Guo-Niu Han, Signed words and permutations II; the Euler-Mahonian polynomials, the electronic journal of combinatorics 11 (2005), no. 2, 22.. [FH08]. , Signed words and permutations, IV: Fixed and pixed points, Israel Journal of Mathematics 163 (2008), no. 1, 217–240. 22.

(26) [HX09]. Guo-Niu Han and Guoce Xin, Permutations with extremal number of fixed points, Journal of Combinatorial Theory, Series A 116 (2009), no. 2, 449– 459.. [KPP94] Alexander G. Kuznetsov, Igor M. Pak, and Alexandr E. Postnikov, Increasing trees and alternating permutations, Russian Mathematical Surveys 49 (1994), no. 6, 79. [Rei93]. Victor Reiner, Signed permutation statistics, European journal of combinatorics 14 (1993), no. 6, 553–567.. [Rem83] Jeffrey B. Remmel, A note on a recursion for the number of derangements, European Journal of Combinatorics 4 (1983), no. 4, 371–374. [Tuc80] Alan Tucker, Applied combinatorics, Wiley, 1980. [WG16] Walter D. Wallis and John C. George, Introduction to combinatorics, Chapman and Hall/CRC, 2016. [Wil84]. Herbert S. Wilf, A bijection in the theory of derangements, Mathematics Magazine 57 (1984), no. 1, 37–40.. 23.

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