Descent polynoials arxiv:1710.11033v2 [ath.co] 13 Nov 2017 Alexander Diaz-Lopez Departent of Matheatics and Statistics, Villanova University, 800 Lancaster Avenue, Villanova, PA 19085, USA, alexander.diaz-lopez@villanova.edu Paela E. Harris Matheatics and Statistics Departent, Willias College, 18 Hoxsey Street, Williastown, MA 01267, USA, paela.e.harris@willias.edu Erik Insko Departent of Matheatics, Florida Gulf Coast University, 10501 FGCU Blvd. South, Fort Myers, FL 33965-6565, USA, einsko@fgcu.edu Mohaed Oar Departent of Matheatics, Harvey Mudd College, 301 Platt Boulevard, Clareont, CA 91711-5901, USA, oar@g.hc.edu Bruce E. Sagan Departent of Matheatics, Michigan State University, East Lansing, MI 48824-1027, USA, sagan@ath.su.edu Noveber 15, 2017 Key Words: coefficients, consecutive pattern avoidance, Coxeter group, descent polynoial, descent set, peak polynoial, peak set, roots AMS subject classification (2010): 05A05 (Priary) 05E15, 20F55 (Secondary) Abstract Let n be a nonnegative integer and I be a finite set of positive integers. In 1915, MacMahon proved that the nuber of perutations in the syetric group S n with descent set I is a polynoial in n. We call this the descent polynoial. However, basic properties of these polynoials such as a description of their coefficients and roots do not see to have been studied in the literature. Much ore recently, in 2013, Billey, Burdzy, and Sagan showed that the nuber of eleents of S n with peak set I is a polynoial in n ties a certain power of two. Since then, there have been a flurry of papers investigating properties of this peak polynoial. The purpose of the present paper is to study the descent polynoial. We will see that it displays soe interesting parallels with its peak relative. Conjectures and questions for future research are scattered throughout. 1 Introduction For the rest of this paper, we let n be a nonnegative integer and I be a finite set of positive integers. (In Section 5 we will perit I to contain 0.) We will also use the notation = ax(i {0}), (1) 1
where the presence of zero ensures that is well defined even when I is epty. We also use the standard notation [n] = {1, 2,..., n}. More generally, given integers l, n we set [l, n] = {l, l + 1...., n}, and siilarly for other interval notations. Denote by S n the syetric group of perutations π = π 1 π 2... π n of [n] written in oneline notation. Note that we will soeties insert coas into such sequences for clarity in distinguishing adjacent eleents. The descent set of π is Des π = {i π i > π i+1 } [n 1]. Note that a siilar definition can be given for any sequence π of integers and we will have occasion to use that level of generality. Given I and n >, where is defined by (1), we wish to study the set D(I; n) = {π S n Des π = I}, and its cardinality As an exaple, if I = {1, 2} then d(i; n) = #D(I; n). D({1, 2}; n) = {π S n π 1 > π 2 > π 3 < π 4 < < π n }. (2) It follows that π 3 = 1. Furtherore, one can pick any two integers fro [2, n] to be to the left of π 3. Placing the integers to the left of π 3 in decreasing order and the reaining ones to the right of π 3 in increasing order copletely deterines π. Thus ( ) n 1 (n 1)(n 2) d({1, 2}; n) = =, (3) 2 2 which is a polynoial in n. Using the Principle of Inclusion and Exclusion, MacMahon [Mac04, Art. 157] proved that this is always the case. Theore 1.1 ([Mac04]). For any I and all n > we have that d(i; n) is a polynoial in n. We call d(i; n) the descent polynoial of I. Although this result was proved in 1915, very little work has been done in the intervening years to study these polynoials in ore detail. The purpose of this work is to rectify this oversight. We also note that since d(i; n) is a polynoial, we can extend its doain of definition to all coplex n, which will be a useful viewpoint in the sequel. Another well-studied statistic on perutations is the peak set defined by Peak π = {i π i 1 < π i > π i+1 } [2, n 1]. It is not true that any set of integers I [2, ) is the peak set of soe perutation. For exaple, clearly I can not contain two consecutive indices. Say that I is adissible if there is soe perutation π with Peak π = I. For I adissible and n >, consider the set P (I; n) = {π S n Peak π = I}. 2
To illustrate, if I =, then P ( ; n) = {π S n π 1 > > π i < π i+1 < < π n for soe 1 i n}. Noting that π i ust be 1, such a perutation is deterined by picking soe subset of [2, n] to be to the left of π i, then arranging those eleents in decreasing order, and finally aking the rest an increasing sequence to the right of π i. It follows that #P ( ; n) = 2 n 1, which is certainly not a polynoial in n. But nearly one hundred years after MacMahon s theore, Billey, Burdzy, and Sagan [BBS13] proved the following result. Theore 1.2 ([BBS13]). For any adissible I and all n > we have that #P (I; n) = p(i; n)2 n #I 1, where p(i; n) is a polynoial in n taking on integer values in the range (, ). As ight be expected, p(i; n) is called the peak polynoial of I. Inspired by this theore, a nuber of papers have been written about properties of peak and related polynoials [BBPS15, BFT16, CVDLO + 17, DLHIO17, DLHIPL17, DNPT, Kas]. It turns out that any of our results about descent polynoials have analogues for peak polynoials. The rest of this paper is organized as follows. In the next section we derive two recursions for d(i; n) that prove useful in the sequel. Section 3 is devoted to the study of the coefficients of d(i; n) when expanded in an appropriately centered binoial coefficient basis for the polynoial ring Q[n]. In particular, we give a cobinatorial interpretation for these constants which perits us to prove a log-concavity result. We also explore a conjecture that the coefficients of d(i; n) when expanded in a differently centered basis alternate in sign. In Section 4, we study the roots of the descent polynoial, including those which are coplex. It will be shown that the eleents of I are always integral zeros, and progress will be ade towards a conjecture about the location of the full set of roots in the coplex plane. Analogues of d(i; n) in Coxeter groups of type B and D are considered in Section 5. We end with a section containing coents and open questions. There we present a result that unifies Theores 1.1 and 1.2 using the concept of consecutive pattern avoidance. 2 Two recursions In this section we derive two recursions for d(i; n). The first will be useful in a nuber of ways, for exaple in deterining the degree of d(i; n) and in finding soe of its roots. If I, then we let I = I {}. We first express d(i; n) in ters of d(i ; n) which will perit latter proofs by induction on or on #I. 3
Proposition 2.1. If I, then d(i; n) = ( ) n d(i ; ) d(i ; n). (4) Proof. Consider the set P of perutations π S n that can be written as a concatenation π = π π satisfying 1. #π = and #π = n, and 2. Des π = I and π is increasing. We can write P as the disjoint union of those π where π > π 1 and those where the reverse inequality holds. So #P = d(i; n) + d(i ; n). On the other hand, the eleents of P can be constructed as follows. Pick eleents of [n] to be in π which can be done in ( n ) ways. Arrange those eleents to have descent set I which can be done in d(i ; ) ways. Finally, put the reaining eleents in π in increasing order which can only be done in one way. If follows that #P = ( n ) d(i ; ). Coparing this with the expression for #P at the end of the previous paragraph copletes the proof. We can use the previous result to provide a new proof of MacMahon s theore and to also obtain the degree of d(i; n). Theore 2.2. For all I we have that d(i; n) is a polynoial in n with deg d(i; n) =. Proof. We prove this by induction on #I. If I =, then d(i; n) = 1 and the result clearly holds. For nonepty I, we exaine (4). We have that ( n ) is a polynoial in n of degree. Multiplying by the nonzero constant d(i ; ) does not change this. And, by induction, d(i ; n) is a polynoial of lesser degree so that the first ter in the difference is doinant. MacMahon also gave an explicit forula for d(i; n) using the Principle of Inclusion and Exclusion. As a further application of (4), we will now rederive this expression. Before doing so, we set the following notation. Recall that a coposition of n is a sequence of positive integers suing to n. Given a set of positive integers I = {i 1 < < i k } and n > i k it will be convenient to let i 0 = 0 and i k+1 = n. Now we can for the difference coposition δ(i) = (i 1 i 0, i 2 i 1,..., i k+1 i k ). (5) To any coposition δ = (δ 1,..., δ k ) of n we associate the ultinoial coefficient ( ) n n! = δ δ 1!... δ k!. Finally, we let ( I i) be the set of all i-eleent subsets of I. Theore 2.3 ([Mac04]). If I is a set of positive integers with #I = k, then d(i; n) = ( ) n ( 1) k i. (6) δ(j) i 0 4 J ( I i)
Proof. We proceed by induction on #I. If I =, then d(i; n) = 1. In this case the right-hand side of (6) is ( n δ( )) = 1. We assue that the result holds for all sets I with #I k. Consider #I = k + 1 and = ax(i). Note that if δ is a coposition of then ( n ) ( )( δ = n δ) where δ is δ with n appended. Now using this fact, equation (4), and the induction hypothesis we have as desired. ( ) n d(i; n) = ( 1) k i i 0 = ( 1) k+1 i i 0 = i 0 ( 1) k+1 i J ( I i ) J ( I i), J J ( I i) ( n δ(j) ( ) ( 1) k i δ(j) i 0 ( ) n + δ(j) ), J ( I i), / J J ( I i ) ( ) n δ(j) ( ) n δ(j) It will be useful to have a recursion that does not contain any negative ters. We will see an application of this recursion when we investigate the expansion of d(i; n) in a certain binoial basis. A siilar recursion was used by Diaz-Lopez, Harris, Insko and Oar [DLHIO17] when they proved the peak polynoial positivity conjecture of Billey, Burdzy, and Sagan [BBS13]. To state our recursion, we need soe notation. Suppose I = {i 1,..., i l } where the integers are listed in increasing order. We define two related sets of positive integers. Specifically, for 1 k l we let and I k = {i 1,..., i k 1, i k 1,..., i l 1} {0}, Î k = {i 1,..., i k 1, i k+1 1,..., i l 1}. Note that subtracting {0} in I k is only necessary when k = 1 and i 1 = 1 so that I k is still a set of positive integers. The reason these sets are interesting is that if one reoves n + 1 fro a π D(I; n + 1) then the resulting π has Des π = I k or Des π = Îk for soe k. Also note that n + 1 can only appear at the end of π or at a position i k where i k 1 I. So define I = {i k i k 1 I} and I = I {1}. Note I and I are only different if 1 I. Theore 2.4. If I, then d(i; n + 1) = d(i; n) + d(i k ; n) + d(îk; n). i k I i k I 5
Proof. We partition D(I; n + 1) according to the position of n + 1. If π D(I; n + 1) then we let π be the perutation obtained fro π by deleting n + 1. If π n+1 = n + 1 then the corresponding π are the eleents of D(I; n) which gives the first ter in the su for d(i; n + 1). Now suppose π ik = n + 1 where i k > 1 and π ik 1 > π ik +1. Then the possible i k where this could occur are exactly the eleents of I, and the π which result for the set D(I k ; n). This explains the first suation. Siilarly, suppose π ik = n + 1 where either i k = 1, or i k > 1 and π ik 1 < π ik +1. Then the corresponding π are counted by the second su and we are done. 3 Coefficients In this section we show that the coefficients of descent polynoials, written in a certain polynoial basis, are nonnegative by providing a cobinatorial interpretation for the. Based on a partial result and coputer evidence, we then conjecture that these coefficients for a log-concave sequence. We also ake a conjecture that the coefficients in another polynoial basis alternate in sign and prove it in a special case. The study of coefficients of polynoials has a rich history and any iportant exaples. For instance, Ehrhart polynoials [Sta80] and chroatic polynoials [Bre92] can be written in certain polynoial bases using nonnegative coefficients. In 2013 Billey, Burdzy, and Sagan conjectured that peak polynoials could be written with non-negative coefficients in a binoial basis [BBS13]. This conjecture was proved in 2017 by Diaz-Lopez et al. [DLHIO17]. We restate their result here and then prove a siilar, but stronger, result for descent polynoials in Theore 3.3. Theore 3.1 ([DLHIO17]). For any non-epty adissible set I we have ( ) ( ) ( ) n n n p(i; n) = b 0 (I) + b 1 (I) + + b 1 (I), 0 1 1 where b 0 (I) = 0 and for 1 k 1 the constant b k (I) is positive. Before proving our ain result of this section, we need a lea which is of interest in its own right. Recall the definition that for integers l, n [l, n] = {l, l + 1, l + 2,..., n}. We also use this notation for the sequence l, l + 1,..., n. Context should ake it clear which interpretation is eant. Lea 3.2. For any finite set of positive integers I and n > we have D(I; n). Proof. We induct on #I. If I = then the identity perutation is in D(I; n). If I then by induction there is a perutation π S with π D(I ; ) where, as usual, I = I {}. It follows that D(I; n) contains the concatenation σ = π 1[+2, n] where π is π with all its eleents increased by one. We can now state the ain result of this section for descent polynoials. 6
Theore 3.3. For any finite set of positive integers I we have that ( ) ( ) ( ) n n n d(i; n) = a 0 (I) + a 1 (I) + + a (I), (7) 0 1 where a 0 (I) = 0 and for k 1 the constant a k (I) is the nuber of π D(I; 2) such that Moreover, a k (I) > 0 for 1 k. {π 1,..., π } [ + 1, 2] = [ + 1, + k]. (8) Proof. By Theore 2.2, d(i; n) is a polynoial in n of degree, so we can write it uniquely as a linear cobination of the polynoial basis {( ) ( ) ( )} n n n,,...,. 0 1 For ease of notation, given π D(I; n) we let Now consider π[] = {π 1,..., π } [ + 1, n]. D k (I; n) = {π D(I; n) #π[] = k}. Clearly D(I; n) is the disjoint union of the sets D k (I; n) for k 0. So to prove the suation forula in (7), it suffices to deonstrate that #D k (I; n) = a k (I) ( ) n k. We also clai that D 0 (I; n) = which forces a 0 (I) = 0. Indeed, if there is an eleent π D 0 (I; n) then π[] =. This iplies that {π 1,..., π } = []. Thus π and π +1 > which contradicts the fact that is a descent. For the rest of the proof we will assue n 2. This assuption is without loss of generality since if we can show that the polynoials on both sides of equation (7) agree for an infinite nuber of values, then they ust agree everywhere. For k 1, consider the eleents π D k (I; n). There are ( ) n k ways to pick the k eleents of π[]. Furtherore, given any two k-eleent subsets X and Y of [ + 1, n], there is an order preserving bijection f : X Y. This induces a bijection fro the π D k (I; n) with π[] = X to the σ D k (I; n) with σ[] = Y by applying f to the eleents of π[], leaving the eleents in the first positions fro [] unchanged, and then listing the reaining eleents in increasing order. Note that all the eleents of [] reain unchanged as f is only applied to eleents of [ + 1, n]. This bijection clearly preserves the descent set everywhere except possibly at position. To see that the descent at is preserved, note that π +1 [] since the subsequence π +1 π n is increasing and there is at least one eleent of [] not in {π 1,..., π } because of the assuption k 1. But then in σ = f(π) we have σ +1 = π +1 since eleents of [] are unchanged. So if π [] then σ = π > π +1 = σ +1 and if π > then σ > σ +1 as desired. Letting X = [ + 1, + k] we have shown that ( ) n #D k (I; n) = #X. k 7
Furtherore k = #X is less than or equal to, which eans that the largest interval we need to consider is [ + 1, 2] and this is contained in [ + 1, n] by our assuption that n 2. Thus #X = a k (I) which is clearly a constant independent of n. This copletes the proof of the suation forula (7). To prove the last stateent of the theore, suppose 1 k. It is enough to show that D k (I; 2). By Lea 3.2 there is π D(I ; ). Thus the concatenation σ = π [1, k][ + k + 1, 2] is in D k (I; 2) where π is π with every eleent increased by k. To illustrate this result, let I = {1, 2}. Then a 1 (I) is the nuber of π = π 1 π 2 π 3 π 4 D(I; 4) such that {π 1, π 2 } [3, 4] = [3]. Siilarly, a 2 (I) is the nuber of π D(I; 4) such that {π 1, π 2 } [3, 4] = [3, 4]. Out of the three eleents in D(I; 4) one can quickly check that only π = 3214 satisfies the condition for a 1 (I), thus a 1 (I) = 1. Siilarly, only π = 4312 satisfies the condition for a 2 (I), so a 2 (I) = 1. Theore 3.3 states that ( ) ( ) n 2 n 2 d(i; n) = +. 1 2 By the binoial recursion, this expression agrees with (3). Many coefficient sequences of cobinatorial polynoials have interesting properties, one of which we will investigate in the context of the previous theore. A sequence of real nubers (a k ) = (a k ) k 0 is log-concave if, for every k 1, we have a k 1 a k+1 a 2 k. Log-concave sequences appear naturally in cobinatorics, algebra, and geoetry; we refer the reader to [Sta89] and [Bre94] for iportant exaples and results. We ake the following conjecture about the sequence (a k (I)) which has been verified for any set I with 18. Conjecture 3.4. For any finite set of positive integers I, the sequence (a k (I)) is log-concave. We are able to prove this conjecture for certain I, but first we need a lea. In it, the sequence (a k ) is said to have a certain property, such as nonnegativity, if all the individual a k do. Also, the sequence has no internal zeros if the eleents between any two nonzero eleents of the sequence are also nonzero. Lea 3.5. 1. If (a k ) and (b k ) are log-concave sequences, then so is (a k b k ). 2. Let (a k ) be a nonnegative log-concave sequence with no internal zeros and let l be a positive integer. Then the sequence (a k + a k+1 + + a k+l ) is log-concave. Proof. Stateent 1 follows easily fro the definition of log-concavity. For stateent 2 note that if we can prove the case l = 1 then the general case will follow by induction since a k + + a k+l = (a k + + a k+l 1 ) + a k+l. A nonnegative log-concave sequence (a k ) with no internal zeros satisfies a k+1 /a k a k /a k 1 for all k. In particular, if j k then a k+1 /a k a j /a j 1 and thus a j 1 a k+1 a j a k. So as desired. (a k 1 + a k )(a k+1 + a k+2 ) = a k 1 a k+1 + a k 1 a k+2 + a k a k+1 + a k a k+2 a 2 k + a k a k+1 + a k a k+1 + a 2 k+1 = (a k + a k+1 ) 2, 8
( k 1 ) i 1 ( ) l k+i 1 ( l+k i ) k 1 1 l k + i l l + i 1 + 1 2 Figure 1: The diagra of a π D(I; 2). The binoial nubers correspond to the possible of ways of choosing each of the three highlighted segents. The next result shows that the sequence (a k (I)) is log-concave in a special case. Proposition 3.6. Let l be positive integers and let I = {l, l + 1,..., }. Then (a k (I)) is log-concave. Proof. We first use the cobinatorial description of a k (I) in Theore 3.3 to derive an explicit forula for this quantity. Let π D(I; 2) satisfy equation (8). In Figure 3 we create a diagra of the perutation π by plotting the points (i, π i ) and connecting the by, possibly dotted or dashed, segents. Note that the for of I iplies that π 1... π has a single local axiu at π l. Cobining this with (8) we see that π l = + k and the eleents of [ + 1, + k] are π l k+i, π l k+i+1,..., π l+i 1 for soe i with 1 i l+1. Now there are ( k 1 i 1) ways of selecting the eleents π l+1,..., π l+i 1. Once these eleents are put in a decreasing sequence just after π l, the rest of the eleents of [+1, +k] ust for an increasing sequence just before π l. Next we choose the eleents of the increasing sequence π 1,..., π l k+i 1 fro [] in ( l k+i 1) ways. The reaining l + k i + 1 eleents of [] ust be arranged as the eleents π l+i,..., π +k with unique local iniu at π +1. So the nuber of ways to choose π +2,..., π +k is ( ) l+k i k 1. And once these eleents are chosen there is only one way to arrange the and the reaining eleents since they are all in increasing or decreasing order. So a k (I) = l+1 i=1 ( )( k 1 i 1 l k + i 1 )( l + k i k 1 Now for any fixed c, the binoial coefficient sequences (( k c ))k 0 and (( )) c are well known to k k 0 be log-concave. In cobination with Lea 3.5, this shows that the sequence (a k (I)) is logconcave. If we expand d(i; n) in the binoial basis centered at 1 then these coefficients also sees to be well behaved. The following conjecture has been verified for all I with 12. ). Conjecture 3.7. For any I we have d(i; n) = ( ) n + 1 ( 1) k c k (I), k k=0 9
where c k (I) is a nonnegative integer for all 0 k. We are able to prove this conjecture for c 0 (I). To do so, we need a couple of leas. Recall that since d(i; n) is a polynoial in n, it is defined for all coplex nubers. Lea 3.8. We have d(i; 0) = ( 1) #I. Proof. We proceed by induction on #I. The result is clear when I = as d( ; n) = 1. Consider any set I with #I 1, then by Proposition 2.1 and the inductive hypothesis ( ) 0 d(i; 0) = d(i ; ) d(i ; 0) = 0 ( 1) #I = ( 1) #I, which is what we wished to prove. Keeping the notation of Conjecture 3.7, we note that This is why our next result will be useful. Proposition 3.9. For any I and any n + 2 we have d(i; 1) = ( 1) c 0 (I). (9) d(i; n) d(i; 1). Proof. Note that d(i; n) is an increasing function of n for integral n > because any perutation π D(I; n) can be extended to one in D(I; n + 1) by erely appending n + 1. So it suffices to prove the result when n = + 2. We proceed by induction on. If = 0 then I = and d(i; n) = 1 and the result follows. For the induction step, we first note that by Leas 3.2 and 3.8 d(i; 0) = 1 d(i; + 1). We now apply Theore 2.4, keeping the notation therein, as well as induction and the previous displayed equation to obtain d(i; + 2) = d(i; + 1) + d(i k ; + 1) + d(îk; + 1) i k I i k I d(i; + 1) + d(i k ; 1) + d(îk; 1) i k I i k I d(i; 0) + d(i k ; 1) + d(îk; 1) i k I i k I d(i; 0) d(i k ; 1) d(îk; 1) i k I i k I as desired = d(i; 1), 10
Proposition 3.10. For any I we have c 0 (I) 0. Proof. By equation (9), it suffices to show that the sign of d(i; 1) is ( 1). We will proceed by induction on #I. As usual, the case I = is trivial. For I, applying recursion (4) yields ( ) 1 d(i; 1) = d(i ; ) d(i ; 1) = ( 1) d(i ; ) d(i ; 1). (10) By Lea 3.2 we have d(i ; ) > 0. And by induction, the sign of d(i ; 1) is ( 1) where = ax(i {0}). So if and have opposite parity, then the result follows fro (10). If they have the sae parity, then + 2. Applying Proposition 3.9 to I we get d(i ; ) d(i ; 1). So, using equation (10) again, the sign of d(i; 1) is ( 1) in this case as well. 4 Roots We defined d(i; n) only for n > because we wished to count a nonepty set of perutations. However, by Theore 1.1, d(i; n) is a polynoial in n so we can extend the definition to d(i; z) for any coplex nuber z. In this context, it akes sense to talk about the roots of d(i; z) and we study the in this section. We start by showing that eleents of I are roots of d(i; z), a result analogous to one for peak polynoials [BFT16]. Theore 4.1. If I is a set of positive integers and i I then d(i; i) = 0. Proof. We induct on #I using the recursion (4). The result is vacuously true when I is epty. If i I then, by the induction hypothesis, d(i ; i) = 0. Also ( i ) = 0 since i <. Substituting these values into (4) shows that d(i; i) = 0. The only other case is i =. But then, using equation (4) again, we have that ( ) d(i; ) = d(i ; ) d(i ; ) = 0, as desired. Now that we have established that the eleents of I are theselves roots of d(i; z), the reainder of this section focuses on understanding the reaining roots of this polynoial lying in the coplex plane. Throughout we denote by z 0, R(z 0 ) and I(z 0 ) the nor, real and iaginary parts, respectively, of the coplex nuber z 0. We begin by coenting on the analogous proble for peak polynoials. Billey, Fahrbach and Talage [BFT16] extensively studied the roots of peak polynoials. Their observations led to the following conjecture regarding the position of the roots in the coplex plane. Conjecture 4.2 ([BFT16]). For any adissible I and z 0 C which is a root of p(i; z), we have 1. z 0, and 2. R(z 0 ) 3. 11
In fact, in Section 2 of their paper, Billey, Fahrbach and Talage establish that Theore 3.1 for peak polynoials was iplied by this conjecture. They verified Conjecture 4.2 coputationally for all polynoials p(i; z) where 15. We have coputed the roots of descent polynoials d(i; z) for all sets I with 12 and arrived at a siilar, but ore restrictive, conjecture. Conjecture 4.3. For any I and z 0 C which is a root of d(i; z) we have 1. z 0, and 2. R(z 0 ) 1. We start by establishing that this conjecture holds for #I = 1 by ad hoc eans. Although this approach does not see to generalize, it gives soe intuition about why the two bounds hold. Theore 4.4. If I = {} and d(i; z 0 ) = 0 then 1. z 0, and 2. R(z 0 ) 1. Proof. Consider the equation 0 = d(i; z) = ( ) z 1. First suppose that z >. Then, by the triangle inequality, z k z k > k and it follows that ( ) z z z 1 z + 1 = > 1.! So such z can not be a root of d(i; z) and the first stateent in the theore is proved. Now suppose R(z) < 1. Then z k R(z k) > k + 1 and the previous displayed equation still holds. This finishes the proof of the second stateent. We note that one can use siilar techniques to show that if I = {1, } then the roots of d(i; z) satisfy the conjecture. But since we were not able to push this ethod further we will not present the proof. In order to establish further bounds for z 0, we introduce soe necessary background on bounding the oduli of roots of polynoials. Recall that given a nonconstant polynoial f(z) = d i=0 c iz i, the axiu odulus of a root of f(z) is bounded above by the Cauchy bound of f, denoted ρ(f), which is the unique positive real solution to the equation c 0 + c 1 z + + c d 1 z d 1 = c d z d, (11) when f is not a onoial, and zero otherwise [RS02, Theore 8.1.3]. Although the Cauchy bound of f(z) does not yield an explicit bound for the oduli of the roots of f(z) there are any results that provide such upper estiates for the Cauchy bound. For exaple [RS02, Corollary 8.1.8] gives various bounds for ρ(f) including ρ(f) < 1 + ax c i, (12) 12 0 i d 1 c d
which we will use in the proof of Theore 4.12. It is possible to obtain bounds for polynoials expressed in other bases, such as Newton bases, which we define now. Given a sequence of coplex nubers ξ 1, ξ 2,..., the polynoials P k (z) = k (z ξ i ), i=1 k 0, for a basis for the vector space of all real polynoials called the Newton basis with respect to the nodes ξ 1, ξ 2,.... Furtherore, since deg(p k (z)) = k then {P 0 (z), P 1 (z),..., P d (z)} fors a basis for the vector space of real polynoials of degree at ost d, for any d. Theore 4.5 (Theore 8.6.3 in [RS02]). Let f(z) = d k=0 c kp k (z) be a polynoial of degree d where the P k s for the Newton basis with respect to the nodes ξ 1,..., ξ d. Then f has all of its zeros in the union of the discs where k = 1,..., d and ρ is the Cauchy bound of d k=0 c kz k. D k := {z C z ξ k ρ}, (13) Theore 4.5 played an iportant role in the work of Brown and Erey that iproved known bounds for the oduli of the roots of chroatic polynoials for dense graphs [BE15]. We will use this result to ake progress on Conjecture 4.3. Because of recursion (4) we consider the Newton bases with respect to the nodes 0, 1, 2, 3,..., which is z k = z(z 1) (z k + 1), k 0. This is known as the falling factorial basis. Expanding d(i; z) in ters of this basis and using the previous theore iediately gives us the following bounds on the roots of d(i; z). Lea 4.6. Suppose d(i; z) = k=0 c kz k. Then the roots of d(i; z) lie in the union of the discs D k = {z C z k ρ(i)}, where k = 0,..., 1 and ρ(i) is the Cauchy bound of the polynoial k=0 c kz k. We now present bounds, linear in, for roots of descent polynoials in the special cases when #I 2, and bounds which appear to be less tight for general I. We begin by revisitng the case when #I = 1. Theore 4.7. Let I = {} and ρ = e e. Then the roots of d(i; z) lie in the union of the discs where k = 0,..., 1. D k = {z C z k ρ }, 13
Proof. By Lea 4.6, it suffices to show that ρ(i) ρ. Since d(i; z) = ( z ) 1 which has the sae roots as z!, it suffices show that ρ is an upper bound for the unique positive real solution to the equation z =!. This solution is!, and using upper Rieann sus to estiate the function ln! fro ln x dx establishes that! +1 /e 1. The result follows. We can use the previous result to derive soewhat different bounds fro those in Theore 4.4 for the special case #I = 1. Corollary 4.8. If I = {} and d(i; z 0 ) = 0 then 1. z 0 ρ + 1, 2. R(z 0 ) ρ, and 3. I(z 0 ) ρ. Furtherore, for all 1, we have e < ρ. Proof. Assertions 1, 2 and 3 follow iediately fro the description of the discs in Theore 4.7. To obtain the bounds on ρ, consider the function f() = e. Taking the derivative gives f () = e ln 2 0, for 1. So f() is decreasing on the interval [1, ) and thus is bounded above by f(1) = e. Applying l Hôpital s Rule shows that li f() = 1 and this liit is a lower bound. The desired inequalities follow fro observing ρ = f()/e. We note that close to the iaginary axis this corollary gives a tighter bound on I(z 0 ) than Theore 4.4 since ρ, reducing the area being considered in the earlier theore by roughly half for large. We now turn to the case #I = 2. Theore 4.9. Let I = {l, } with 1 l <. Then the roots of d(i; z) lie in the union of the discs D k = {z C z k }, for k = 0,..., 1. Proof. We established through coputation that the result is true for 4 so we assue 5. By definition, I = {l}, so by repeatedly applying equation (4) we have ( ) z d(i; z) = d(i ; ) d(i ; z) ( ) (( ) ) ( ) z z = 1 + 1 l l = 1 (( ) ) 1 z 1! l l! z l +1. 14
Multiplying the previous equation by l! and using Lea 4.6, the roots of d(i; z) are contained in the union of the discs D k = {z C z k ρ}, k = 0, 1,..., 1, where ρ is any upper bound on the unique positive real solution to the equation (( ) ) l! 1 z = z l + l!.! l Since ( ) ( l 1 ) l /2, replacing the forer expression by the latter in the previous displayed equation just increases the unique positive real solution. Rewriting the result, it suffices to show that is an upper bound for the positive real solution of ( ) 1 z l 1 2 ( l)! z l 1 = l!. To do so, observe that l > l! and l > 4. So evaluating the left side of the previous ( l)! equality at z = gives ( ) ( ) 1 l 1 1 2 ( l)! l 1 > l! 2 4 1 = l! and so ust exceed the unique positive real solution. Siilar to Corollary 4.8, we can use Theore 4.9 to bound the nor, real and iaginary parts of roots of d(i; z) when #I = 2. Corollary 4.10. If #I = 2 and d(i; z 0 ) = 0 then 1. z 0 2 1, 2. R(z 0 ), and 3. I(z 0 ). Siilar bounds on the roots of d(i; z) can be established when #I = 3 by first repeatedly applying equation (4) to express d(i; z) as a linear cobination of the falling factorials, and then applying a strategy like the one in the proof of Theore 4.9. But applying these techniques as #I grows becoes increasingly coplicated, so it is not clear that this ethod will be able to produce a linear bound in general. We now discuss how to find general bounds on the roots of d(i; z) regardless of the size of I. We begin with the following result. Lea 4.11. We have where for all k I {0}. d(i; z) = c 0 + k I 1 k! c k 1, c k z k, 15
Proof. Induct on #I. We have d( ; n) = 1 which satisfies the lea. By induction we can write d(i ; z) = c 0 + k I c k z k, where 1 k! c k 1, for all k I {0}. Now using equation (4) we have that ( ) z d(i; z) = d(i ; ) d(i ; z) = d(i ; ) z! ( c 0 + k I c k z k = c 0 k I c k z k +c z, ) where c = d(i ; )/!. The lea now follows for k < fro the bounds on the c k, and for k = fro the fact that 1 d(i ; )!. The previous lea perits us to find general bounds for the roots of d(i; z). Theore 4.12. Let I satisfy #I 2, and let = ax I. Furtherore let ( ) ρ = in! + 1, (! #I) 1/( ). (14) The roots of d(i; z) all lie in the union of the discs where k = 0, 1,..., 1. In particular, if d(i; z 0 ) = 0 then 1. z 0 ρ + 1, 2. R(z 0 ) ρ, and 3. I(z 0 ) ρ. D k = {z C z k ρ}, (15) Proof. The bounds on z 0, R(z 0 ) and I(z 0 ) all follow fro (15). Define coefficients c k as in Lea 4.11. To prove (15) itself, it suffices to show that ρ is an upper bound for the unique positive real solution of c z = c 0 + c k z k. k I Replacing c by its sallest possible value and the other c k by their largest possible value will only increase the value of the positive solution. So, using the bounds on the c k, it suffices to show that ρ is an upper bound on the unique positive real solution of 1! z = 1 + k I z k. (16) 16
Figure 2: The roots of descent polynoials for I = {1, 3, 4} and I = {1, 2, 4} are plotted as dots and the corresponding bounding discs fro Theore 4.12 are shaded in grey. Applying equation (12) establishes that ρ! + 1. On the other hand, since z k z for all k I and real z 1, ρ is bounded above by the unique positive real solution of the equation z /! = (#I) z, which is (! #I) 1/( ). On the right side of (14) the first arguent achieves the iniu if = 1 since then, using the assued bound on #I yields (! #I) 1/( ) 2! >! + 1. But if 2 then the second arguent is saller since (! #I) 1/( ) (!) 1/2 <! + 1. In fact, if is held constant and then the bound becoes linear. An illustration of these two cases is given in Figure 2, where the graph on the left is for I = {1, 3, 4} and the one on the right is for I = {1, 2, 4}. We now use a technique fro linear algebra to obtain a different sort of restriction on the roots of d(i; n). In fact, we will restrict the position of the zeros of any polynoial whose expansion in the falling factorial basis has nonnegative coefficients. Because of the generality of this result, it will often be less restrictive than Theore 4.12. But along the real axis it will give a linear bound for any I and so it will be an iproveent. Throughout the reainder of this section, we ove freely between a coplex nuber z = x+iy and the vector v = (x, y) R 2. So if z = ρe iθ then we call θ an arguent of v and write arg v = θ. The principle value of v, denoted Arg v, is the arguent of v satisfying π < Arg v π. It will be convenient to let Arg(0, 0) =. Given vectors v 0,..., v we say they are nonnegatively linearly independent if the only linear cobination c 0 v 0 + +c v = (0, 0) with all the c i nonnegative is the trivial cobination where c 0 = = c = 0. Otherwise the vectors are nonnegatively linearly dependent. An open half-plane 17
consists of all points on one side of a line L through the origin. The corresponding closed half-plane is obtained by also including the points on L. The easy backward direction of the following lea is a well-known tool in the literature. But we present a proof for copleteness as well as showing that the two stateents are actually equivalent. Lea 4.13. Vectors v 0,..., v are nonnegatively linearly independent if and only if they all lie in soe open half-plane. Proof. If the vectors all lie in an open half-plane then clearly so will any nontrivial nonnegative linear cobination. Since the half-plane is open, such a linear cobination can not be zero. Now suppose the vectors do not lie in any open half-plane. There are two cases. If they all lie in a closed half-plane then, since they do not lie in any open half-plane, there ust be two of the vectors, say v 0 and v 1, such that v 0 = cv 1 for soe scalar c > 0. Thus v 0 + cv 1 = (0, 0) and the vectors are nonnegatively linearly dependent. Now suppose that the vectors do not lie in any closed half-plane and consider the vector v 0. We will find two other vectors satisfying a nonnegative linear dependence relation with v 0. Rotating each of v 0,..., v through the angle Arg v 0, we can assue that v 0 lies along the positive x-axis. Since all the vectors do not lie in the half-plane x 0 there ust be soe vector, say v 1, with Arg v 1 > π/2. Consider the line L through v 1. Note that by construction, v 0 and the negative x-axis are on opposite sides of L. And, by the closed half-plane hypothesis again, there ust be soe v 2 on the sae side of L as the negative x-axis but on the opposite side of the x-axis fro v 1. It follows that there is soe nonnegative linear cobination av 1 + bv 2 which lies on the negative x-axis. So av 1 + bv 2 = cv 0 for c > 0 which gives the nonnegative linear dependency cv 0 + av 1 + bv 2 = (0, 0). Since the linear dependencies in the previous proof only involve at ost three vectors, we have actually proved the following result. Lea 4.14. Vectors v 0,..., v are nonnegatively linearly independent if and only any three of the lie in an open half-plane. To ake the connection with roots of polynoials, let P (z) be the vector space of polynoials in a variable z with real coefficients and let B(z) = {b 0 (z),..., b (z)} be a basis for P (z). Consider the subset of P (z) defined by { } P B (z) = f(z) 0 f(z) = c k b k (z) with c k 0 for all k, k=0 where in the above definition 0 represents the zero polynoial. Translating Lea 4.13 into this language we iediately have the following result. Corollary 4.15. The coplex nuber w is not a root of any polynoial in P B (z) if and only if the vectors corresponding to the coplex nubers in B(w) lie in soe open half-plane. We now specialize to the falling factorial basis {z k k 0}. As usual z denotes the coplex conjugate of z, and if S is a set of coplex nubers, then we let S = { z z S}. 18
Theore 4.16. Let F(z) = {z 0,..., z }. The coplex nuber w is not a root of any polynoial in P F (z) if and only if w is in the region R = S S where { } S = z C Arg z 0 and Arg(z i + 1) < π. (17) Proof. Since the coefficients of polynoials f(z) P F (z) are real, we have f(w) = 0 if and only if f( w) = 0. So, letting R be the region of w which are not roots of any such f(z), we have R = S S where S = {z R Arg z 0}. So it suffices to show that S is given as in the stateent of the theore. Equivalently, by the previous corollary, we ust show that for z with Arg z 0 we have z S as defined by equation (17) if and only if the eleents of F(z) lie in an open half-plane. Suppose first that the su inequality in (17) holds for z. Since z 0 = 1, we wish to show that for 1 k the coplex nubers z k lie either on the positive x-axis or in the open half-plane above the x-axis. For then the eleents of F(z) will lie in the open half-plane above the line y = ɛx for a sufficiently sall negative ɛ. Since Arg z 0, we have Arg(z r) 0 for all reals r. Using this and the fact that 1 k, we have 0 k Arg(z i + 1) i=1 i=1 Arg(z i + 1) < π. But z k = k i=1 (z i + 1), so the displayed inequalities iply 0 Arg(z k) < π which is what we wished to show. To coplete the proof we ust show that if i=1 Arg(z i + 1) π then the eleents of F(z) will not all lie in any open half-plane. Fro the arguent in the preceding paragraph we see that s k := k i=1 Arg(z i + 1) is an increasing function of k. And s 0 = 0. Thus there ust be a nonnegative integer l such that s l < π s l+1. If s l+1 = π then z 0 and z l+1 are nonnegatively linearly dependent and we are done by Lea 4.13. If s l+1 > π then we ust have 0 < Arg z < π. It follows that 0 < Arg(z l) < π. Since z l+1 = (z l)z l, the previous inequalities force a point on the negative x-axis to be a nonnegative linear cobination of z l and z l+1. So, together with z 0 = 1 we have a nonnegative linear dependency in this case as well. This concludes the proof of the theore. Finally, we return to descent polynoials. If S is any set of coplex nubers and w C then let S + w = {z + w z S}. Corollary 4.17. Let I be a finite set of positive integers. Then any eleent of R + where R is defined as in Theore 4.16 is not a root of d(i; z). Proof. By Theore 3.3, we can write ( ) z d(i; z) = a k (I) k k=0 i=1 = a k (I) (z ) k, k! where a k (I)/k! 0 for all k. So f(z) := d(i; z + ) P F (z). Applying the previous theore and using the fact that z R + if and only if z R finishes the proof. 19
Figure 3: Roots of descent polynoials d(i; n) with I [4] plotted inside the two bounding regions and close-up view of the region R + 4 near the real axis. Figure 3 plots all of the roots of descent polynoials corresponding to subsets I [4] as sall dots, the worst-case bounds described in Theore 4.12 for such roots are shaded in light grey and the dark grey arc is the region R + 4 where R is as described in Theore 4.16. The iage on the right gives a close-up view of the region R + 4 near the real-axis. While in the first iage the region R + 4 looks to be bounded by a curve passing through the real-axis near z = 6.65, it actually passes through the real-axis at z = 7 and then curves back to include coplex nubers whose real parts are less than 7. We can use the previous corollary to get our best bound for the size of roots along the positive x-axis which holds for general I. Proposition 4.18. If z 0 is a real root of d(i; z) then z 0 2 1. Proof. For a real nuber z 0 we have Arg z 0 = 0 if z 0 > 0 and Arg z 0 = π if z 0 < 0. So to be in the region S of equation (17) we ust have z 0 > 1. Applying Corollary 4.17 we see that if z 0 > 2 1 then it can not be a zero of d(i; z) and the result follows. 5 Other Coxeter groups Recall that for any finite Coxeter syste (W, S), the (right) descent set of w W is Des w = {s S l(ws) < l(w)}, (18) where l is the length function. In this section we will consider the Coxeter groups B n and D n. We will use sybols near the beginning of the Greek alphabet for eleents of B n and D n to distinguish the fro the perutations in A n 1 = S n. We view B n as the group of signed perutations β = β 1... β n where β i {±1,..., ±n} for all i Z and the sequence β 1... β n is a perutation in A n 1, and we view D n as the subgroup of B n consisting of all β = β 1... β n where there are an even nuber of β i in { 1, 2,..., n}. Since D n is a subgroup of B n, the notation defined below in ters of B n also applies to D n. We 20
will use the coon convention that b will be written as b. For exaple two eleents of B 6 are β = 34 1 562 and γ = 3 4 1 562, and the second eleent is also an eleent of the subgroup D 6, whereas the first is not. The siple reflections in B n are S B = S A {s 0 } where s 0 = (1, 1) and S A denotes the set of adjacent transpositions generating the Coxeter group of type A n 1. Identifying reflections and subscripts as we have done in the syetric group, we see that for β B n we have Des β [n 1] {0}. Because of this, it will be convenient to extend perutations in B n by writing β = β 0 β 1... β n where β 0 = 0. In this notation, our previous exaples would be written β = 0 34 1 562 and γ = 0 3 4 1 562. Translating definition (18) using our conventions, we see that if β = β 0 β 1... β n B n then Des β = {i 0 β i > β i+1 }, (19) where we are using the usual order on the integers for the inequalities. To continue our exaples in B 6, we have Des β = {0, 2, 3, 5} and Des γ = {0, 1, 3, 5}. Now given a finite set of nonnegative integers I and n > where continues to be defined by equation (1), we let D B (I; n) = {β B n Des(β) = I} and d B (I; n) = #D B (I; n). (20) We will first derive a recursive forula for d B (I; n) analogous to the one for d(i; n) in Proposition 2.1. Theore 5.1. Let I be a nonepty, finite set of nonnegative integers. Then we have d B (I; n) = ( ) n 2 n d B (I ; ) d B (I ; n). (21) Proof. Consider the set P of signed perutations β B n which can be written as a concatenation β = 0β β satisfying 1. #β = and #β = n, and 2. Des β = I and β is increasing. We can write P as the disjoint union of those β where β > β 1 and those where the reverse inequality holds. So #P = d B (I; n) + d B (I ; n). On the other hand, the eleents of P can be constructed as follows. Pick a subset S of eleents of [n] which can be done in ( n ) ways. For a signed perutation fro the eleents of S whose descent set is I which can be done in d B (I ; ) ways. Next choose the sign of the n eleents in [n] S which can be done in 2 n ways. Then arrange the in increasing order to for β which can only be done in only one way. It follows that #P = ( n ) 2 n d B (I ; ). Coparing this with the expression for #P at the end of the previous paragraph copletes the proof. Next we prove the type B analogue of Theore 2.3. To state it, we let I + = I {0}. 21
Also, if J is a set of positive integers then we will let δ 1 (J) denote the first coponent of the coposition δ(j). Note that { in J if J, δ 1 (J) = n if J =. Theore 5.2. If I is a set of nonnegative integers with #I + = k, then d B (I; n) = ( ) { n 2 ( 1) k i n δ 1 (J) if 0 I, δ(j) (2 n 2 n δ1(j) ) if 0 I. i 0 J ( I+ i ) (22) Proof. We first consider the case where 0 / I so that I = I +, and proceed by induction on #I. If I =, then d B (I; n) = 1. In this case, the right-hand side of equation (22) also gives ( n δ( )) = 1. We assue that the result holds for all sets I not containing 0 with #I k. Consider #I = k + 1 and = ax(i). Using recursion (21), and the induction hypothesis we have ( ) n d B (I; n) = 2 n ( ) ( 1) k i 2 δ 1(J) ( ) n ( 1) k i 2 n δ 1(J) δ(j) δ(j) i 0 J ( I i 0 i ) J ( I i ) = ( 1) k+1 i ( ) n 2 n δ1(j) + ( ) n 2 n δ 1(J) δ(j) δ(j) i 0 J ( I i), J J ( I i), / J = ( ) n ( 1) k+1 i 2 n δ1(j). δ(j) i 0 J ( I i) Since I = I + when 0 / I, this copletes the proof for this case. Next we consider when 0 I. If I = {0} then Theore 5.1 shows d B (I; n) = 2 n 1, and the right hand of equation (22) above gives ( n δ( )) (2 n 2 n n ). So equation (22) holds in this case. The induction arguent is exactly the sae as that of the case when 0 / I, but one replaces 2 δ 1(J) with 2 2 δ 1(J) and 2 n δ 1(J) with 2 n 2 n δ 1(J). Using Theores 2.3 and 5.2, we can also give a siple nuerical relationship between the descent forulas in types A and B. Corollary 5.3. Let I be a finite set of positive integers and I 0 = I {0}. Then d B (I; n) + d B (I 0 ; n) = 2 n d(i; n). Since the right-hand side of equation (22) is well defined for all real nubers n, we use it to extend the definition d B (I; n) to R and talk about its roots. The proof of the following theore is siilar to that of Theore 4.1 and so is oitted. Theore 5.4. If I is a set of nonnegative integers and i I then d B (I; i) = 0. 22
The reaining results of this section pertain to the Coxeter group D n. We continue to use all the conventions for B n with this subgroup. In particular, we will use the sae definition of Des β as in equation (19), and the notation D D (I; n) and d D (I; n) is defined exactly as in equation (20) except that β runs over D n rather than B n. Our results in type D n are very siilar to those in type B n except with soe changes iposed by using a different power of two and the interingling of d D and d B in the sae forula. Theore 5.5. Let I be a nonepty, finite set of nonnegative integers. Then d D (I; n) = ( ) n 2 n 1 d B (I ; ) d D (I ; n). (23) Proof. Consider the set P of signed perutations β D n satisfying the sae two conditions as in the proof of Theore 5.1. As before, #P = d D (I; n) + d D (I ; n). An alternative construction of the eleents of P is as follows. Pick eleents fro [n] which can be done in ( n ) ways. Use those eleents to create a type B signed perutation β with descent set I which can be done in d B (I ; ) ways. Since a type D n perutation ust have an even nuber of negative signs, of the reaining n eleents choose the sign of the first n 1 of the; the sign of the last eleent in the set of nubers appearing in β is then deterined by the nuber of negative signs assigned previously. Thus choosing the signs of the eleents appearing in β can be done in 2 n 1 ways. Now for the unique increasing arrangeent of these signed integers to for β. It follows that #P = ( n ) 2 n 1 d B (I ; ) and we are done as in the proof of Theore 5.1. Next we can use Theore 5.5 to prove a Type D n analogue of Theores 2.3 and 5.2. As the proof are siilar to those we have seen before, we oit the. Theore 5.6. If I is a set of nonnegative integers with #I + = k, then ( 1) k + ( ) n ( 1) k i 2 n δ 1(J) 1 if 0 I, δ(j) i>0 J ( I+ i ) d D (I; n) = ( 1) k (2 n 1 1) + ( ) n ( 1) k i (2 n 1 2 n δ1(j) 1 ) if 0 I. δ(j) i>0 forall n > J ( I+ i ) Finally we present the analogues of Corollary 5.3, and Theore 5.4 for type D n. Corollary 5.7. Let I be a nonepty set of positive integers and I 0 = I {0}. Then 1. d D (I; n) + d D (I 0 ; n) = 2 n 1 d(i; n), and 2. d D (I; i) = d D (I 0 ; i) = 0 whenever i I. 23
6 Coents and open questions We end with soe coents about our results. These include avenues for future research and ore conjectures. (1) Consecutive pattern avoidance. One way to unify Theores 1.1 and 1.2 is through the theory of consecutive pattern avoidance. Call two sequences of integers a 1 a 2... a k and b 1 b 2... b k order isoorphic provided a i < a j if and only if b i < b j for all pairs of indices 1 i, j k. Given σ S k called the pattern, we say that π S n contains a consecutive copy of σ at index i if the factor π i π i+1... π i+k 1 is order isoorphic to σ. If π contains no consectutive copies of σ then we say that π consecutively avoids σ. Note that a consecutive copy of 21 is just a descent while a peak is a consective copy of 132 or 231. Given any finite set of patterns Π and a finite set of positive integers I define Π(I; n) = {π S n π has a consecutive copy of soe σ Π precisely at the indices in I}. Also define the function av Π (n) = #Π( ; n), the nuber of perutations in S n consecutively avoiding all perutations in Π. Given Π S k say that Π is nonoverlapping if for any (not necessarily distinct) σ, τ Π and any l with 1 < l < k the prefix of σ of length l is not order isoorphic to the suffix of τ of length l. We will now prove our analogue of Theores 1.1 and 1.2 in this setting. Theore 6.1. Let Π S k be a nonoverlapping set of patterns and let I be a finite set of positive integers. Then for all n + k 1 we have #Π(I; n) V Π where V Π is the vector space of all Q-linear cobinations of functions in the set {n k av Π (n + l) k Z 0, l Z}. Proof. We induct on. We have #Π( ; n) = av Π (n) and so the result clearly holds when = 0. For 1, consider the set P of perutations π S n which can be written as a concatenation π = π π such that π Π(I ; ) and π Π( ; n ). Since Π is nonoverlapping, copies of consecutive patterns fro Π in π occur at the positions in I and possibly also at exactly one of the indices, 1,..., k + 2. It follows that k 2 #P = #Π(I ; n) + #Π(I; n) + #Π(I { i}; n). We can also construct the eleents of P as follows. Pick the eleents of [n] to be in π which can be done in ( n ) ways. Arrange those eleents to have consecutive copies of eleents of Π at the indices of I which can be done in #Π(I ; ) ways. Finally, put the reaining eleents in π so that it avoids consecutive copies of eleents of Π which can be done in av Π (n ) ways. Equating the two counts for P and rearranging ters we get ( ) n k 2 #Π(I; n) = av Π (n )#Π(I ; ) #Π(I ; n) #Π(I { i}; n), fro which the theore follows by induction. 24 i=1 i=1