1. Introduction to a new Type of Matching 2. The polynomials U ,k,n - PDF document

1. Introduction to a new Type of Matching 2. The polynomials U Υ ,k,n ( x ) 3. Background and Previous Results 4. Extreme Coefficients 5. 2 Closed Forms from Recursions 6. Q-Analogues 7. Q-statistic 8. Unanswered Questions

Notion of a τ -match Given σ = σ 1 · · · σ n , we define red ( σ ) be the permutation that results by replacing the i -th largest integer that appears in the sequence σ by i . Example: If σ = 2 7 5 4, then red ( σ ) = 1 4 3 2. Given a permutation τ ∈ S j , we define τ - mch ( σ ) = { i | red ( σ i · · · σ i + j − 1 ) = τ } . Example: If τ = 1 3 2 4, and σ = 1 3 2 4 6 5 7, then τ - mch ( σ ) = { 1 , 4 } . When | τ | = 2, then | τ - mch ( σ ) | is familiar. If τ = 2 1, then des ( σ ) = | τ - mch ( σ ) | . If τ = 1 2,then rise ( σ ) = | τ - mch ( σ ) | .

Notion of a Υ-match and n-lap More generally, if Υ is a set of permutations of length j , then we say that a permutation σ = σ 1 · · · σ n ∈ S n has a Υ match at place i provided there is a τ ∈ Υ such that red ( σ i · · · σ i + j − 1 ) = τ . Define Υ- mch ( σ ) to be the number of Υ matches in the permutation σ . Let τ - nlap ( σ ) and Υ − nlap ( σ ) be the maximum number of non-overlapping τ -matches and Υ matches in σ respectively.

A more refined matching condition Suppose we define τ - k - mch ( σ ) = { i | red ( σ i · · · σ i + j − 1 ) = red ( τ ) and for 0 ≤ s ≤ j − 1, σ i + s = τ 1+ s mod k } . Example: If τ = 1 2 and σ = 5 1 7 4 3 6 8 2, then τ - mch ( σ ) = { 2 , 5 , 6 } , but τ -2- mch ( σ ) = { 5 } . Let τ - k - emch ( σ ) = | τ - k - mch ( σ ) | .

Notion of a Υ-k-match More generally, if Υ is a set of sequences of distinct integers of length j , then we say that a permutation σ = σ 1 · · · σ n ∈ S n has a Υ- k - equivalence match at place i provided there is a τ ∈ Υ such that red ( σ i · · · σ i + j − 1 ) = red ( τ ) and for all s ∈ { 0 , . . . , j − 1 } , σ i + s = τ 1+ s mod k . Let Υ- k - emch ( σ ) be the number of Υ- k -equivalence matches in the permutation σ . We shall review the study of the polynomials n x Υ - k - emch ( σ ) = U s Υ ,k,n x s . � � U Υ ,k,n ( x ) = s =0 σ ∈ S n

What exactly will we study? In particular, we shall focus on certain special cases of these polynomials where we consider only patterns of length 2. Fix k ≥ 2 and let A k equal the set of all se- quences we could consider for Ascents. For ex- ample, A 4 = { 1 2 , 1 3 , 1 4 , 1 5 , 2 3 , 2 4 , 2 5 , 2 6 , 3 4 , 3 5 , 3 6 , 3 7 , 4 5 , 4 6 , 4 7 , 4 8 } . Let D k = { b a : a b ∈ A k } and E k = A k ∪ D k .

What has already been studied? Kitaev and Remmel found explicit formulas for the coefficients U s Υ ,k,n in certain special cases. In particular, they studied descents according to the equivalence class mod k of either the first or second element in a descent pair. That is, for any set X ⊆ { 0 , 1 , 2 , . . . } , define • ← − − Des X ( σ ) = { i : σ i > σ i +1 & σ i ∈ X } and ← − des X ( σ ) = |← − − Des X ( σ ) | • − − → Des X ( σ ) = { i : σ i > σ i +1 & σ i +1 ∈ X } and − → des X ( σ ) = |− − → Des X ( σ ) |

Kitaev and Remmel studied ← − des kN ( σ ) = � ⌊ n k ⌋ 1. A ( k ) j =0 A ( k ) j,n x j n ( x ) = � σ ∈ S n x and − → 2. B ( k ) des kN ( σ ) z χ ( σ 1 ∈ kN ) ( x, z ) = � σ ∈ S n x n = � ⌊ n k ⌋ i =0 B ( k ) � 1 i,j,n z i x j . j =0 where kN = { 0 , k, 2 k, . . . } .

They showed that For all 0 ≤ j ≤ k − 1 and all n ≥ 0, we have A ( k ) s,kn + j (( k − 1) n + j )! s ( − 1) s − r � ( k − 1) n + j + r �� kn + j + 1 � � = s − r r r =0 n − 1 � ( r + 1 + j + ( k − 1) i ) × i =0 n − s ( − 1) n − s − r � ( k − 1) n + j + r �� kn + j + 1 � � = n − s − r r r =0 n � × ( r + ( k − 1) i ) i =1

Special case And using an identity were able to show: � n � 2 ( n !) 2 A (2) s, 2 n = s and 1 � 2 (( n + 1)!) 2 � n A (2) s, 2 n +1 = s + 1 s We will now turn to another special case of U s Υ ,k,n . In particular, we will compute explicit formulas for U s Υ ,k,n where Υ = { (1 k ) } .

Finding formulas: Obtaining a recursion Let ∆ kn + j : x s → sx s − 1 + ( kn + j − s ) x s Γ kn + k : x s → (( k − 1) n + k + s − 1) x s +( n − s +1) x s +1 The polynomials U { (1 k ) } ,k,n ( x ) satisfy the fol- lowing recursions. 1. U { (1 k ) } ,k, 1 ( x ) = 1, 2. For j = 1 , . . . , k − 1, U { (1 k ) } ,k,kn + j ( x ) = ∆ kn + j ( U { (1 k ) } ,k,kn + j − 1 ( x )), and 3. U { (1 k ) } ,k,kn + k ( x ) = Γ kn + k ( U { (1 k ) } ,k,kn + k − 1 ( x )).

The basic recursions This gives rise to the following recursions for the coefficients. For 1 ≤ j ≤ k − 1, U s ( kn + j − s ) U s = { (1 k ) } ,k,kn + j { (1 k ) } ,k,kn + j − 1 +( s + 1) U s +1 { (1 k ) } ,k,kn + j − 1 and { (1 k ) } ,k,kn + k = ( n − s + 2) U s − 1 U s { (1 k ) } ,k,kn + k − 1 +(( k − 1) n + s + k − 1) U s { (1 k ) } ,k,kn + k − 1

Extreme coefficients We have for j = 0 , . . . , k − 1 , U 0 { (1 k ) } ,k,kn + j = (( k − 1) n + j )!(( k − 1) n + j ) n U n { (1 k ) } ,k,kn + j = (( k − 1) n + j )! Also note that U m { (1 k ) } ,k,kn + j = 0 for m > n .

Closed form 1 Starting with the formula for U 0 { (1 k ) } ,k,kn + j we can use the recursions to prove: For all 0 ≤ j ≤ k − 1 and all n such that kn + j > 0, we have U s { (1 k ) } ,k,kn + j = (( k − 1) n + j )! � s r =0 ( − 1) s − r (( k − 1) n + r + j ) n � ( k − 1) n + r + j �� kn + j +1 � s − r r

Closed form 2 Starting with the formula for U n { (1 k ) } ,k,kn + j we can use the recursions to prove: For all 0 ≤ j ≤ k − 1 and all n such that kn + j > 0, we have U s { (1 k ) } ,k,kn + j = (( k − 1) n + j )! � n − s r =0 ( − 1) n − s − r (1 + r ) n � ( k − 1) n + r + j �� kn + j +1 � r n − s − r

Karlsson-Minton hypergeometric series A hypergeometric series is defined by ∞ � � ( a 1 ) k . . . ( a p ) k a 1 , a 2 , . . . , a p z k � b q ; z = p F q b 1 , b 2 , . . . , k !( b 1 ) k . . . ( b q ) k k =0 A Karlsson-Minton hypergeometric series is de- fined by ∞ � � ( c 1 ) k . . . ( c t +1 ) k c 1 , c 2 , . . . , c t +1 z k � ; z = t +1 F t b 1 , . . . , b t k !( b 1 ) k . . . ( b t ) k k =0

Gasper proved that � � b 1 + d 1 , b t + d t w, x, . . . = t +2 F t +1 x + c + 1 , b 1 , . . . b t ( b i − x ) di Γ(1+ x + c )Γ(1 − w ) � t i =1 Γ(1+ x − w )Γ( c +1) ( b i ) di � � − c, z − b 1 , z − b t x, . . . × t +2 F t +1 z − w, z − b 1 − d 1 , z − b t − d t . . . where z = 1 + x , w, c, x, b i ∈ C , d i ∈ N , and ℜ ( c − w ) > n − 1.

A Simple Example ( x ) = � ⌊ n 2 ⌋ Let R n ( x ) = A (2) s =0 R s,n x s . Let n ∆ 2 n +1 : x s → sx s − 1 + (2 n − s + 1) x s and Γ 2 n +2 : x s → ( s + 1) x s + (2 n − s + 1) x s +1 . Then Kitaev and Remmel proved the following. The polynomials R n ( x ) n ≥ 1 satisfy the following recursions. 1. R 1 ( x ) = 1, 2. R 2 n +1 ( x ) = ∆ 2 n +1 ( R 2 n ( x )), and 3. R 2 n +2 ( x ) = Γ 2 n +2 ( R 2 n +1 ( x )) .

The basic recursions This fact gave rise to the following recursions for R s,n ( x ). = ( s + 1) R s +1 , 2 n + (2 n − s + 1) R s, 2 n R s, 2 n +1 = ( s + 1) R s, 2 n +1 + (2 n − s + 2) R s − 1 , 2 n +1 R s, 2 n +2 It was through these recursions that Kitaev and Remmel were able to show that � n � 2 ( n !) 2 , and = R k, 2 n k 1 � 2 (( n + 1)!) 2 . � n = R k, 2 n +1 k + 1 k

The q -recursions To prove q -analogues of the results, let 2 n +1 : x s → [ s ] q x s − 1 + q s [2 n − s + 1] q x s ∆ q and 2 n +2 : x s → [ s + 1] q x s + q s +1 [2 n − s + 1] q x s +1 . Γ q Define R q s =0 R q n ( x ) n ≥ 1 = � n s,n x s , via the follow- ing recursions. 1. R q 1 ( x, q ) = 1, 2. R q 2 n +1 ( x, q ) = ∆ q 2 n +1 ( R q 2 n ( x )), and 3. R q 2 n +2 ( x, q ) = Γ q 2 n +2 ( R q 2 n +1 ( x )).

This fact gives rise to the following recursions for the coefficients R q s,n ( x ). s +1 , 2 n + q s [2 n − s + 1] q R q R q [ s + 1] q R q = s, 2 n +1 s, 2 n s, 2 n +1 + q s [2 n − s + 2] q R q R q [ s + 1] q R q = s, 2 n +2 s − 1 , 2 n +1 We can then show that the solution to these recursions are 2 ([ n ] q !) 2 , and 2 ) � n q ( k R q � = k, 2 n k q q ( k +1 2 ) � n 2 ([ n + 1] q !) 2 . R q � = k, 2 n +1 [ k + 1] q k q

A familliar permutation statistic, maj Given σ ∈ S n , maj ( σ ) = � i ∈ Des ( σ ) i . Foata showed that the maj statistic satisfies some simple recursions. That is, for any permuta- tion τ = τ 1 . . . τ n − 1 ∈ S n − 1 , we label the spaces where we can insert n into τ to get a permu- tation in S n as follows. 1. Label the space following τ n − 1 with 0. 2. Next label the spaces that lie between de- scents τ i > τ i +1 from right to left with the integers 1 , . . . , des ( τ ). 3. Finally label the remaining spaces from left to right with the integers des ( τ ) + 1 , . . . , n .

Example: If τ = 3 9 2 8 5 4 1 6 7, then spaces are labeled as follows: 5 3 6 9 4 2 7 8 3 5 2 4 1 1 8 6 9 7 0 . Then Foata proved that if τ ( i ) is the result of inserting n into the space labeled i , then for all i ∈ { 0 , . . . , n } , maj ( τ ( i ) ) = i + maj ( τ ) .