1 47 6 11 Journal of Integer Sequences Vol. 1 010 Article 10.6.8 Counting Peas and Valleys in a Partition of a Set Toufi Mansour Department of Mathematics University of Haifa 1905 Haifa Israel toufi@math.haifa.ac.il Mar Shattuc Department of Mathematics University of Tennessee Knoxville TN 7996 USA shattuc@math.ut.edu Abstract A partition π of the set [n] {1...n} is a collection {B 1 B...B } of nonempty disjoint subsets of [n] called blocs whose union equals [n]. In this paper we find an explicit formula for the generating function for the number of partitions of [n] with exactly blocs according to the number of peas valleys in terms of Chebyshev polynomials of the second ind. Furthermore we calculate explicit formulas for the total number of peas and valleys in all the partitions of [n] with exactly blocs providing both algebraic and combinatorial proofs. 1 Introduction A partition Π of the set [n] {1...n} is a collection {B 1 B...B } of nonempty disjoint subsets of [n] called blocs whose union equals [n]. We assume that blocs are listed in increasing order of their minimal elements that is minb 1 < min B < < min B. The 1
set of all partitions of [n] with blocs is denoted by Pn. The cardinality of Pn is the well-nown Stirling number of the second ind [1] which is usually denoted by S n. Any partition Π can be written in the canonical sequential form π 1 π π n where i B πi for all i see e.g. [7]. Throughout we will identify each partition with its canonical sequential form. For example if Π {1 4} { 5 7} {} {6} is a partition of [7] then its canonical sequential form is π 114 and in such a case we write Π π. Several authors have studied different subword patterns on the set of partitions of [n]. For example Mansour and Munagi [9] studied the number of partitions of [n] according to the number levels rises and descents. Recall that a level respectively rise descent of a word π π 1 π π n is an index i such that π i π i+1 respectively π i < π i+1 π i > π i+1. A general question concerns counting the elements in a subset of Pn which have no occurrences of a particular pattern. On this subject the reader is referred to the wors of Sagan [1] Mansour and Severini [10] Chen et al. [1] and Jelíne and Mansour [4] as well as to the references therein. Let [] n denote the set of all words of length n over the alphabet []. Given π π 1 π π n [] n a pea respectively valley is an entry π j 1 j n of π such that π j < π j+1 > π j+ respectively π j > π j+1 < π j+ ; that is a pea is a rise followed immediately by a descent and a valley is a descent followed immediately by a rise. In this case we say that a pea or valley occurs at j in π. We denote the number of peas respectively valleys in π by peaπ respectively valleyπ. That there are n 1 permutations of [n] without peas valleys was shown [5] and later extended [6] by Kitaev. Comparable results on words are given by Heubach and Mansour [] where they find an explicit formula for the generating function for the number of elements of [] n according to the number of peas valleys see [] for further results. See also the paper by Mansour [8] for analogous results on Dyc paths. The aim of this paper is to find an explicit formula for the generating function for the number of partitions of [n] with exactly blocs expressed canonically according to the number of peas valleys where is fixed. The case of peas respectively valleys is treated in Section respectively Section. Our main approach is to first compute the generating functions for the number of peas or valleys for words over [] of length n and then relate this to the corresponding generating functions for partitions of [n] with exactly blocs via the restricted growth function of the partition. All the explicit solutions obtained in Sections and involve Chebyshev polynomials of the second ind see [11]. We also derive explicit formulas for the total number of peas and valleys in all the partitions of [n] with exactly blocs providing both algebraic and combinatorial proofs. Counting peas Let W xq be the generating function for the number of words of length n over the alphabet [] according to the number of peas that is W xq x n. n 0 π [] n q peaπ
Lemma.1. The generating function W xq satisfies the recurrence relation W xq with the initial condition W 0 xq 1. xq 1 + 1 xq 1W 1 xq 1 x1 q1 x xx + q1 xw 1 xq Proof. Let us write an equation for W xq. Since each word in [] n may or may not contain the letter we have W xq W 1 xq + W xq where W xq is the generating function for the number of words π of length n over the alphabet [] according to the number of peas such that π contains the letter. Such words π may be decomposed as either π π π π or π π where π is a nonempty word over the alphabet [ 1] π is a nonempty word over the alphabet [] π is a nonempty word over the alphabet [] which starts with a letter a < and π is a word over the alphabet []. The corresponding generating functions for the number of words of these forms are given by x xw 1 xq 1 xw xq 1 x W 1 xq 1W xq and xqw 1 xq 1W xq xw xq 1 respectively. Summing these five generating functions we obtain which implies which is equivalent to as claimed. W xq xq 1 xq 1W 1 xq + x1 q1 x + x + q1 xw 1 xqw xq W xq xq 1 + 1 xq 1W 1 xq + x1 q1 x W xq + x + q1 xw 1 xqw xq xq 1 + 1 xq 1W 1 xq 1 x1 q1 x xx + q1 xw 1 xq In order to write the explicit formula for the generating function W xq we need the following lemma. Lemma.. Let a n be any sequence defined by the recurrence relation a n A+Ba n 1 C+Da n 1 with the initial condition a 0 1 such that α BC AD 0. Then for all n 0 A+B A α U n 1 B+C U α n B+C α a n α A+B α U n B+C α U n 1 B+C α B where U m is the m-th Chebyshev polynomial of the second ind. A+B α U n 1 B+C U α n B+C α
Proof. We proceed by induction on n. Since U 0 t 1 U 1 t 0 and U t 1 the lemma holds for n 0. Let g n A + B α U n 1 B + C α U n B + C α. Assume the lemma holds for n and let us prove it for n + 1: a n+1 A + Ba n A + B C + Da n C + D Ag n αgn+1 Bg n αgn+1 Ag n Bg n A αg n+1 C αg n+1 BC DAg n Ag n+1 Cg n+1 αg n. Using the fact that the Chebyshev polynomials of the second ind satisfy the recurrence relation U m t tu m 1 t U m t 1 we get g n B+C α g n+1 g n+ which implies a n+1 and completes the induction step. Ag n+1 Cg n+1 B + Cg n+1 + αg n+ Ag n+1 αgn+ Bg n+1 Combining Lemmas.1 and. we obtain the following result. Theorem.. The generating function W xq for the number of words of length n over the alphabet [] according to the number of peas is given by W xq xq 1 U 1 t U t U t U 1 t 1 xq 1 U 1 t U t where t 1 + x 1 q and U m is the m-th Chebyshev polynomial of the second ind. Remar.4. The mapping on [] n given by π 1 π π n + 1 π 1 + 1 π + 1 π n implies that the generating function for the number of words of length n over the alphabet [] according to the number of valleys is also given by W xq. Theorem. for q 0 gives W x 0 which by 1 implies the following corollary. x U 1 t U t U t U 1 t 1 + x U 1 t U t 4
Corollary.5. The generating function W x 0 for the number of words of length n over the alphabet [] without peas is given by W x 0 U 1 t U t 1 xu 1 t U t where t 1 + x and U m is the m-th Chebyshev polynomial of the second ind. Now we consider the number of peas valleys in all the words of length n over the alphabet []. Theorem. gives lim W x U 1 1 U 1 xq lim q 1 q 1 d U dq t U 1 t + U t q1 +x U 1 1 U 1. Since U 1 + 1 and d U dq 1 + x1 q q1 x + we get lim W x xq lim q 1 q 1 x + x 1 1 x which agrees with the generating function for the number of words of length n over the alphabet []. To find the q-partial derivative of W xq at q 1 using Theorem. first let fq xq 1U 1 t U t and gq U t U 1 t 1 xq 1U 1 t U t where t 1+ x1 q. It is readily verified that f1 g1 0 f 1 x g 1 x x f 1 x and g 1 x 4 +1 x where primes denote differentiation with respect to q. Then applying L Hôpital s rule twice to f qgq fqg q yields lim q 1 d dq W xq q1 which implies the following result. g q +1 x 1 x + x 1 x Corollary.6. The number of peas valleys in all the words of length n over the alphabet [] is given by n n +. Now we turn our attention to finding an explicit formula for the generating function PP xq for the number of partitions of [n] with exactly blocs according to the number of peas that is PP xq x n q peaπ. n 0 π Pn 5
Lemma.7. For all 1 PP xq x 1 + x1 qw j xq + qw j xq 1. Proof. If π is any partition of [n] with exactly blocs then π may be written as π π π where π is a partition of [n ] with exactly 1 blocs and π is a word of length n over the alphabet [] such that n + n + 1 n. Since π is either empty starts with the letter or starts with a letter less than we get PP xq xpp 1 xq1 + xw xq + qw xq xw xq 1. The desired result now follows by induction on upon noting the initial condition PP 0 xq 1. Combining Lemma.7 and Theorem. yields the following result. Theorem.8. For all 1 the generating function for the number of partitions of [n] with exactly blocs according to the number of peas is given by PP xq x 1 q 1 + x + x 1 q U jt U j 1 t U j 1 t U j t 1 + x1 q U jt U j 1 t U j 1 t U j t where t 1 + x 1 q and U m is the m-th Chebyshev polynomial of the second ind. The above theorem for q 0 implies that the generating function for the number of partitions of [n] with exactly blocs without peas is given by PP x 0 x x 1 + 1 + x U jt U j 1 t U j 1 t U j t which by 1 is equivalent to PP x 0 x U j 1 t 1 + xu j t 1 xu j 1 t U j t where t 1 + x and U m is the m-th Chebyshev polynomial of the second ind. Table 1 below presents the number of partitions of [n] with exactly blocs without peas. Corollary.9. The generating function for the number of partitions of [n] without peas is given by 1 + PP x 0 x U j 1 t 1 + xu j t 1 xu 1 0 j 1 t U j t where t 1 + x and U m is the m-th Chebyshev polynomial of the second ind. 6
\n 1 4 5 6 7 8 9 10 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 4 8 15 7 48 85 150 64 464 0 0 1 9 7 75 197 50 16 1 7695 4 0 0 0 1 4 16 64 6 818 76 894 86 5 0 0 0 0 1 5 5 15 575 479 1075 41409 Table 1: The number of partitions of [n] with exactly blocs without peas for n 1...1 and 1...5 Now we turn our attention to finding a formula for the total number of peas in all the partitions of [n] with exactly blocs. Theorem.8 gives 1 q 1 + x + x 1 q U jt U j 1 t PP x 1 x U j 1 t U j t lim q 1 1 + x1 q U jt U j 1 t U j 1 t U j t 1 + x U j1 U j 1 1 x U j 1 1 U j 1. x d Uj t U j 1 t dq U j 1 t U j t q1 Using U 1 + 1 and d U dq 1 + x1 q q1 x + we obtain PP x 1 x x x + jx x 1 jx which is in accord with the generating function for the number of partitions of [n] with exactly blocs; see e.g. [1]. Also Theorem.8 gives d dq PP xq q1 PP x 1 lim q 1 d h dq jq h j q where h j q 1 q1+x+x 1 q u j 1+x1 q u j and u j U jt U j 1 t x U j 1 t U j with t 1+ 1 q. Let t u jq d u dq jq and u jq d u dq j q. Since u j 1 1 lim q 1 h j q 1 and lim 1 jx q 1 u jq jx we have lim q 1 d dq h jq xj 1 1 jx x lim 1 u j q 1 qu jq q 1 1 + x1 q u j q xj 1 1 jx 1 1 jx lim q 1 xj 1 1 jx 1 1 jx 1 qu jq 1 + x1 q u j q u j1 x jx. 7
It is not hard to chec that lim q 1 u jq j+1 x 4 j x 4 which implies j+1 d lim q 1 dq h xj 1 jq 1 jx + x + x j. 1 jx Hence we have d dq PP xq q1 PP x 1 x j+1 + x j 1 jx + xj 1 Since PP x 1 n 1 Snxn and xj 1 x we get the following result upon writing j+1 + j as j j j+1. Corollary.10. The number of peas in all the partitions of [n] with blocs is given by S n 1 + j j j j + 1 f nj where f nj n i ji S n i and S ij is the Stirling number of the second ind. Counting valleys In this section we find an explicit formula for the generating function V P xq for the number of partitions of [n] with exactly blocs according to the number of valleys that is V P xq x n q valleyπ. n 0 π Pn In order to achieve this let W xq respectively W xq be the generating function for the number of words π of length n over the alphabet [] respectively [ 1] according to the number of valleys in π + 1 respectively π. Lemma.1. For all and with W 1xq 1 1 x W xq W xq 1 xw xq W xq W 1xq + xw 1xqW 1xq 1 + xq and W 1 xq 1.. 8
Proof. The initial conditions W 1xq 1 and W 1 x 1 xq 1 hold directly from the definitions. Each word π + 1 such that π is a word over the alphabet [] can be written as either π + 1 where π is a word over the alphabet [ 1] or as π π + 1 where π is a word over the alphabet [ 1] and π is a word over the alphabet []. Note that the number of valleys in the word π + 1 equals the number of valleys in π for any word π over the alphabet [ 1]. Therefore for all which implies W xq W xq + xw xqw xq W xq W xq 1 xw xq. Similarly each word π such that π is a word over the alphabet [ 1] can be written either as π such that π is a word over the alphabet [ ] or as π 1π such that π is a word over the alphabet [ ] and π is a word over the alphabet [ 1]. Note that the number of valleys in the word π 1 equals the number of valleys in 1π 1 for any word π over the alphabet [ ]. Therefore for all W xq W 1xq + xw 1xqW 1xq 1 + xq where xq accounts for the word 1 which yields the result above. Lemma.. For all 1 W xq xq 1A A +1 1 x q 1A where A 1 + x1 xq 1U t U t t 1 + x 1 q and U m is the m-th Chebyshev polynomial of the second ind. Proof. Lemma.1 gives W xq xq 1 + W 1 xq 1 xw 1 xq with initial condition W 1 xq 1. Assume that W xq A B A B A 1 xq 1 + B 1 xa 1 for all 1. Then we have which implies A xq 1B 1 + 1 x q 1A 1 and B B 1 xa 1 with A 1 B 1 1 A 1 + x1 xq 1 and B 1 x. Therefore A ta 1 A A 1 1 A 1 + x1 xq 1. Hence by 1 and induction on we have A A. From the recurrence A xq 1B 1 + 1 x q 1A 1 we get B which completes the proof. 1 A xq 1 +1 1 x q 1A 9
Lemma.1 and imply W xq W +1 xq xq 1. Thus by Lemma. we get which by implies the following result. Lemma.. For all 1 W xq xq 1tA +1 A + A + 1 x q 1A +1 W xq xq 1A A + 1 x q 1A +1 where A 1 + x1 xq 1U t U t t 1 + x 1 q and U m is the m-th Chebyshev polynomial of the second ind. From the fact that each partition π of [n] with exactly blocs may be expressed uniquely as π 1π 1 π π such that π j is a word over the alphabet [j] it follows that the generating function V P xq is given by 1 V P xq x V xq W jxq 4 where V xq is the generating function for the number of words π over the alphabet [] according to the number of valleys in π. Lemma.4. For all 1 V xq 1 1 xw j xq. Proof. Since each word π where π is a word over the alphabet [] can be written either as π where π is a word over the alphabet [ 1] or as π π where π is a word over the alphabet [ 1] and π is a word over the alphabet [] we get V xq V xq + xw xqv xq 5 where V xq is the generating function for the number of words π over the alphabet [ 1] according to the number of valleys in π. Similarly each word π where π is a word over the alphabet [ 1] can be written either as π where π is a word over the alphabet [ ] or as π 1π where π is a word over the alphabet [ ] and π is a word over the alphabet [ 1]. Also the number of valleys in π π 1 equals the number of valleys in 1π 1π 1 for any word π over the alphabet [ ]. Therefore V xq V 1xq + xw 1xqV 1 xq. 6 Hence by 5 and 6 we get V xq V 1 xq xw xqv xq for all 1. Applying this recurrence relation times we have V xq 1 xq as claimed. 10 1 xw j
Now we can state the main result of this section. Theorem.5. The generating function V P xq for the number of partitions of [n] with exactly blocs is given by V P xq x 1 1 xu 1 t U t U j 1 t 1 xq 1U j t 1 xu j 1 t U j t where t 1 + x 1 q and U m is the m-th Chebyshev polynomial of the second ind. Proof. Lemma.4 and 4 yield V P xq x Thus by Lemmas. and. we have 1 1 1 xw j xq W jxq. V P xq x A +1 1 x q 1A A +1 A 1 xq 1Aj A j+1 1 x q 1A j A j+1 A j A j+ 1 x q 1A j+1 x A 1 x q 1A 1 A +1 A 1 xq 1A j A j+1 A j where A 1 + x1 xq 1U t U t. Using the initial conditions A 1 1 A 1 + x1 xq 1 and the fact that we get V P xq x+1 q 1 A +1 A A j+1 A j xq 11 xu j 1 t U j t 1 xq 1A j A j+1 A j x 1 1 xu 1 t U t 1 + x1 xq 1U j t U j t 1 xu j 1 t U j t which by the recurrence U m t tu m 1 t U m t is equivalent to as claimed. V P xq x 1 1 xu 1 t U t 11 U j 1 t 1 xq 1U j t 1 xu j 1 t U j t
\n 1 4 5 6 7 8 9 10 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 6 11 0 6 64 11 199 50 615 0 0 1 5 16 45 11 15 799 199 491 109 4 0 0 0 1 7 1 119 40 1484 495 15984 5088 5 0 0 0 0 1 9 51 49 116 4914 04 81510 Table : The number of partitions of [n] with exactly blocs without valleys for n 1...1 and 1...5 The generating function for the number of partitions of [n] without valleys is given by 1 + 1 V P x 0 0 x 1 xu 1 t U t 1 U j 1 t 1 + xu j t 1 xu j 1 t U j t where t 1 + x and U m is the m-th Chebyshev polynomial of the second ind. Table presents the number of partitions of [n] with exactly blocs without valleys. Theorem.5 together with the fact U j 1 j + 1 yields V P x 1 x 1 U j 1 1 U j 1 1 xu 1 1 U 1 1 xu j 1 1 U j 1 x 1 jx which is the generating function for the number of partitions of [n] with exactly blocs; see e.g. [1]. Also Theorem.5 gives d dq V P xq q1 V P x 1 1 1 d U lim j 1 t 1 xq 1U j t q 1 dq 1 xu j 1 t U j t 1 1 jx 1 xu 1 1 U 1 1 x U j 11 + xu j 1 U j 11 jx 1 xu j 11 U j 1 1 jx 1 xu 1 1 U 1 1 x 1
which by the fact that U j1 : d U dq jt q1 x j+ implies that the generating function for the number of valleys in all the partitions of [n] with exactly blocs is given by d dq V P xq q1 x 1 1 jx x 1 x 1 jx This yields the following explicit formula. j xj 1 x + x + x j x j x j+1 1 jx x j+1 1 jx Corollary.6. The number of valleys in all the partitions of [n] with exactly blocs is given by 1 j j + 1 S n 1 S n + f nj f nj where f nj n i ji S n i f nj n i ji S n i and S ij is the Stirling number of the second ind. 4 Combinatorial proofs In this section we provide direct combinatorial proofs of Corollaries.6.6 and.10. We start with Corollary.6. Proof of Corollary.6. Proof. To show that there are n n + total peas valleys it is enough to show for each i 1 i n that there are a total of n + peas at i within all the words w 1 w w n in [] n. There are n such peas at i for which w i w i+ within all the members of [] n. To see this note that for any three distinct numbers a < b < c in [] there are n words of length n in the alphabet [] whose i + 1 st letter is c and whose i th and i + nd letters comprise the set {ab}. Upon letting a b in the preceding one sees that there are a total of n peas at i for which w i w i+. Let π π 1 π π n be any partition represented by its canonical sequence. Recall that a rise descent is said to occur at i if π i < π i+1 π i > π i+1. We will call a rise or a descent at i non-trivial if neither i nor i + 1 are the smallest elements of their respective blocs. The following lemma provides an explicit formula for the total number of non-trivial rises in all of the members of Pn. Lemma 4.1. The number of non-trivial rises descents in all of the partitions of [n] with exactly blocs is given by j f nj j 1 j.
where f nj n i ji S n i. Proof. Given i and j where i n and j consider all the members of Pn which may be decomposed uniquely as π π jαβ 7 where π is a partition with j 1 blocs α is a word of length i in the alphabet [j] whose last two letters form a rise and β is possibly empty. For example if i 5 j and π 1114 P1 4 then π 1 α 11 and β 4. The total number of non-trivial rises can then be obtained by finding the number of partitions which may be expressed as in 7 for each i and j and then summing over all possible values of i and j. And there are j j i S n i members of Pn which may be expressed as in 7 since there are j i choices for the first i letters of α j choices for the final two letters in α as the last letter must exceed its predecessor and S n i choices for the remaining letters π jβ which necessarily constitute a partition of an n i-set into blocs. If π is a partition of [n] and i [n] then we will say that i is minimal if i is the smallest element of a bloc within π i.e. if the i th slot of the canonical representation of π corresponds to the left-most occurrence of a letter. We now provide direct combinatorial proofs of Corollaries.6 and.10. Proof of Corollary.6. Proof. By Lemma 4.1 the first sum j j f nj counts the total number of non-trivial rises within all of the members of Pn. From these we subtract all non-trivial rises at r r where r 1 either goes in the same bloc as r j j fnj total such rises or goes in a bloc to the left of r j j fnj total such rises. Thus the second sum j+1 j fnj counts all non-trivial rises at r in which r 1 fails to appear in a bloc to the right of r upon noting j + j j+1. Therefore the difference between the two sums counts all valleys at m in which m + is not minimal. We ll call valleys at m where m+ is minimal trivial. So to complete the proof we must show that the total number of trivial valleys at m m 1 is given by 1 Sn 1 which we rewrite as Sn + 1 Sn Sn 1 using the fact that S n 1 S n +S n 1. First there are Sn trivial valleys at m where m is not minimal. To see this pic three numbers a < b < c in [] and let t denote the smallest element of the c th bloc of λ Pn. Increase all members of [tn ] {tt+1...n } by two within λ leaving all numbers within their current blocs. Then add t to bloc b and t+1 to bloc a to obtain a member of Pn in which a trivial valley occurs at t where t is not minimal. For example if n 10 5a 1b and c 4 with λ {1 4} { } {5} {6 8} {7} belonging to P8 5 then t 6 and the resulting member λ {1 4 7} { } {5 6} {8 10} {9} of P10 5 has a trivial valley at 6. There are also Sn trivial valleys at m where m is minimal and not occurring as a singleton bloc. To see this pic numbers a < b < c in [] and let t denote the smallest element of the b th bloc of λ Pn. Increase all members of [tn ] by two within λ and add t to bloc c and t+1 to bloc a. The resulting member of Pn will have a trivial 14
valley at t with the singleton bloc {t} not occurring. Finally there are 1 Sn 1 trivial valleys at m in which the bloc {m} occurs. To see this pic a < b in [ 1] and let t denote the smallest element of the b th bloc of λ Pn 1. Increase all members of [tn ] by two within λ. Then add t + 1 to bloc a as well as the singleton bloc {t}. The resulting member of Pn will have a trivial valley at t with the singleton bloc {t} occurring. For example if n 9 4a 1 and b with λ {1 4} { 6} {5 7} P7 then t 5 and the resulting member λ {1 4 6} { 8} {5} {7 9} of P9 4 has a trivial valley at 5. Proof of Corollary.10. Proof. By reasoning similar to that used in the proof of Lemma 4.1 there are a total of j j j fnj non-trivial descents at r + 1 in which r is not minimal in all the members of Pn. Reasoning as in the proof of Corollary.6 above we see that there are a total of j+1 j fnj non-trivial descents at r + 1 in which either r either goes in the same bloc as r + 1 or goes in a bloc to the right of r + 1 where r itself is not minimal. Thus the difference between the two sums counts all peas at r where r + 1 is not minimal. So to complete the proof we ll show that the total number of peas at r where r + 1 is minimal is given by Sn 1. To see this we first observe that peas at r where r + 1 is minimal are synonymous with descents at r + 1 where r + 1 is minimal. To show that Sn 1 counts all descents at i for some i where i is the smallest member of its bloc in some member of Pn first choose two numbers a < b in []. Given λ Pn 1 let t denote the smallest member of bloc b. Increase all members of [t + 1n 1] in λ by one leaving them within their blocs and then add t + 1 to bloc a. This produces a descent between the first element of bloc b and an element of bloc a within some member of Pn. For example if n 7 4a 1 and b then λ {1 5} {} { 4} {6} P6 4 would give rise to the descent between and 4 in {1 4 6} {} { 5} {7} P7 4. References [1] W. C. Chen E. P. Deng R. X. Du R. P. Stanley and C. H. Yan Crossings and nestings of matchings and partitions Trans. Amer. Math. Soc. 594 007 1555 1575. [] S. Heubach and T. Mansour Enumeration of -letter patterns in compositions Integers Conference Celebration of the 70th Birthday of Ron Graham University of West Georgia Carrollton 005. [] S. Heubach and T. Mansour Combinatorics of Compositions and Words CRC Press Boca Raton 009. [4] V. Jelíne and T. Mansour On pattern-avoiding partitions Electron. J. Combin. 15 008 #R9. [5] S. Kitaev Multi-avoidance of generalized patterns Discrete Math. 60 00 89 100. 15
[6] S. Kitaev Partially ordered generalized patterns Discrete Math. 98 005 1 9. [7] M. Klazar On abab-free and abba-free set partitions European J. Combin. 17 1996 5 68. [8] T. Mansour Counting peas at height in a Dyc path J. Integer Seq. 5 00 Article.1.1. [9] T. Mansour and A. Munagi Enumeration of partitions by long rises levels and descents J. Integer Seq. 1 009 Article 9.1.8. [10] T. Mansour and S. Severini Enumeration of -noncrossing partitions Discrete Math. 08 008 4570 4577. [11] T. Rivlin Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory John Wiley New Yor 1990. [1] B. E. Sagan Pattern avoidance in set partitions to appear in Ars Combin. [1] R. P. Stanley Enumerative Combinatorics Vol. 1 Cambridge University Press Cambridge UK 1996. 010 Mathematics Subject Classification: 05A18 05A15 4C05 Keywords: set partition generating function recurrence relation pea valley Concerned with sequences A000110 A00877. Received April 19 010; revised version received June 18 010. Published in Journal of Integer Sequences June 010. Return to Journal of Integer Sequences home page. 16