Permutation Patterns and Statistics Bruce Sagan Department of - PowerPoint PPT Presentation

Permutation Patterns and Statistics Bruce Sagan Department of Mathematics Michigan State University East Lansing, MI 48824-1027 sagan@math.msu.edu www.math.msu.edu/ ˜ sagan joint work with T. Dokos (UCLA), T. Dwyer (Dartmouth), B. Johnson (Michigan State), and K. Selsor (U. South Carolina) March 14, 2014

Pattern containment and avoidance Permutation statistics: inversions Permutation statistics: major index q -Catalan numbers Exercises and References

Outline Pattern containment and avoidance Permutation statistics: inversions Permutation statistics: major index q -Catalan numbers Exercises and References

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n !

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485.

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485. We say σ avoids π if σ does not contain π and let Av n ( π ) = { σ ∈ S n : σ avoids π } .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485. We say σ avoids π if σ does not contain π and let Av n ( π ) = { σ ∈ S n : σ avoids π } . Ex. If π ∈ S k then Av k ( π )

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485. We say σ avoids π if σ does not contain π and let Av n ( π ) = { σ ∈ S n : σ avoids π } . Ex. If π ∈ S k then Av k ( π ) = S k − { π } .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485. We say σ avoids π if σ does not contain π and let Av n ( π ) = { σ ∈ S n : σ avoids π } . Ex. If π ∈ S k then Av k ( π ) = S k − { π } . Say that π and π ′ are Wilf equivalent , π ≡ π ′ , if for all n ≥ 0 # Av n ( π ) = # Av n ( π ′ ) .

Two sequences of distinct integers π = a 1 a 2 . . . a k and σ = b 1 b 2 . . . b k are order isomorphic if, for all i and j , a i < a j ⇐ ⇒ b i < b j . Ex. The sequences π = 132 and σ = 485 are order isomorphic. Let S n be the symmetric group of all permutations of { 1 , . . . , n } . # S n = n ! Let S = ∪ n ≥ 0 S n . If π, σ ∈ S then σ contains π as a pattern if there is a subsequence σ ′ of σ order isomorphic to π . Ex. σ = 42183756 contains π = 132 because of σ ′ = 485. We say σ avoids π if σ does not contain π and let Av n ( π ) = { σ ∈ S n : σ avoids π } . Ex. If π ∈ S k then Av k ( π ) = S k − { π } . Say that π and π ′ are Wilf equivalent , π ≡ π ′ , if for all n ≥ 0 # Av n ( π ) = # Av n ( π ′ ) . Theorem (Knuth, 1973) For any π ∈ S 3 and n ≥ 0 : # Av n ( π ) = C n (Catalan number).

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n .

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . .

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 }

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 } so # Av 3 ( 123 ) = 5 = C 3 .

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 } so # Av 3 ( 123 ) = 5 = C 3 . Similarly # Av 3 ( π ) = 5 = C 3 , for any π ∈ S 3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 } so # Av 3 ( 123 ) = 5 = C 3 . Similarly # Av 3 ( π ) = 5 = C 3 , for any π ∈ S 3 so we have shown Knuth’s Theorem holds for n = 3. Theorem For n ≥ 1 : C n = C 0 C n − 1 + C 1 C n − 2 + C 2 C n − 3 + · · · + C n − 1 C 0 .

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 } so # Av 3 ( 123 ) = 5 = C 3 . Similarly # Av 3 ( π ) = 5 = C 3 , for any π ∈ S 3 so we have shown Knuth’s Theorem holds for n = 3. Theorem For n ≥ 1 : C n = C 0 C n − 1 + C 1 C n − 2 + C 2 C n − 3 + · · · + C n − 1 C 0 . Ex. C 3 = C 0 C 2 + C 1 C 1 + C 2 C 0

Theorem (Knuth, 1973) For any π ∈ S 3 and all n ≥ 0 we have # Av n ( π ) = C n . The nth Catalan number is 1 � 2 n � C n = . n + 1 n So C 0 = 1 , C 1 = 1 , C 2 = 2 , C 3 = 5 , C 4 = 14 , C 5 = 42 , . . . Ex. Av 3 ( 123 ) = { 132 , 213 , 231 , 312 , 321 } = S 3 − { 123 } so # Av 3 ( 123 ) = 5 = C 3 . Similarly # Av 3 ( π ) = 5 = C 3 , for any π ∈ S 3 so we have shown Knuth’s Theorem holds for n = 3. Theorem For n ≥ 1 : C n = C 0 C n − 1 + C 1 C n − 2 + C 2 C n − 3 + · · · + C n − 1 C 0 . Ex. C 3 = C 0 C 2 + C 1 C 1 + C 2 C 0 = 1 · 2 + 1 · 1 + 2 · 1