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 = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n!

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn.

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 Avn(π) = {σ ∈ Sn : σ avoids π}.

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 Avn(π) = {σ ∈ Sn : σ avoids π}.

• Ex. If π ∈ Sk then Avk(π)

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 Avn(π) = {σ ∈ Sn : σ avoids π}.

• Ex. If π ∈ Sk then Avk(π) = Sk − {π}.

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 Avn(π) = {σ ∈ Sn : σ avoids π}.

• Ex. If π ∈ Sk then Avk(π) = Sk − {π}.

Say that π and π′ are Wilf equivalent, π ≡ π′, if for all n ≥ 0 # Avn(π) = # Avn(π′).

Two sequences of distinct integers π = a1a2 . . . ak and σ = b1b2 . . . bk are order isomorphic if, for all i and j, ai < aj ⇐ ⇒ bi < bj.

• Ex. The sequences π = 132 and σ = 485 are order isomorphic.

Let Sn be the symmetric group of all permutations of {1, . . . , n}. #Sn = n! Let S = ∪n≥0Sn. 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 Avn(π) = {σ ∈ Sn : σ avoids π}.

• Ex. If π ∈ Sk then Avk(π) = Sk − {π}.

Say that π and π′ are Wilf equivalent, π ≡ π′, if for all n ≥ 0 # Avn(π) = # Avn(π′).

Theorem (Knuth, 1973)

For any π ∈ S3 and n ≥ 0: # Avn(π) = Cn (Catalan number).

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn.

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123}

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3.

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3. Similarly # Av3(π) = 5 = C3, for any π ∈ S3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3. Similarly # Av3(π) = 5 = C3, for any π ∈ S3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem

For n ≥ 1: Cn = C0Cn−1 + C1Cn−2 + C2Cn−3 + · · · + Cn−1C0.

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3. Similarly # Av3(π) = 5 = C3, for any π ∈ S3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem

For n ≥ 1: Cn = C0Cn−1 + C1Cn−2 + C2Cn−3 + · · · + Cn−1C0.

• Ex. C3 = C0C2 + C1C1 + C2C0

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3. Similarly # Av3(π) = 5 = C3, for any π ∈ S3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem

For n ≥ 1: Cn = C0Cn−1 + C1Cn−2 + C2Cn−3 + · · · + Cn−1C0.

• Ex. C3 = C0C2 + C1C1 + C2C0 = 1 · 2 + 1 · 1 + 2 · 1

Theorem (Knuth, 1973)

For any π ∈ S3 and all n ≥ 0 we have # Avn(π) = Cn. The nth Catalan number is Cn = 1 n + 1 2n n

• .

So C0 = 1, C1 = 1, C2 = 2, C3 = 5, C4 = 14, C5 = 42, . . .

• Ex. Av3(123) = {132, 213, 231, 312, 321} = S3 − {123} so

# Av3(123) = 5 = C3. Similarly # Av3(π) = 5 = C3, for any π ∈ S3 so we have shown Knuth’s Theorem holds for n = 3.

Theorem

For n ≥ 1: Cn = C0Cn−1 + C1Cn−2 + C2Cn−3 + · · · + Cn−1C0.

• Ex. C3 = C0C2 + C1C1 + C2C0 = 1 · 2 + 1 · 1 + 2 · 1 = 5.

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2.

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 =

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m.

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = R90(132) = = 231 The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m.

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = R90(132) = = 231 The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m. Note that for any ρ ∈ D4: σ contains π ⇐ ⇒ ρ(σ) contains ρ(π),

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = R90(132) = = 231 The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m. Note that for any ρ ∈ D4: σ contains π ⇐ ⇒ ρ(σ) contains ρ(π), ∴ σ avoids π ⇐ ⇒ ρ(σ) avoids ρ(π).

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = R90(132) = = 231 The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m. Note that for any ρ ∈ D4: σ contains π ⇐ ⇒ ρ(σ) contains ρ(π), ∴ σ avoids π ⇐ ⇒ ρ(σ) avoids ρ(π). ∴ # Avn(π) = # Avn(ρ(π)).

The diagram of π = a1 . . . an is (1, a1), . . . , (n, an) ∈ Z2. Ex. 132 = R90(132) = = 231 The dihedral group D4 of symmetries of the square acts on Sn: D4 = {R0, R90, R180, R270, r0, r1, r−1, r∞} where Rθ is rotation counter-clockwise through θ degrees and rm is reflection in a line of slope m. Note that for any ρ ∈ D4: σ contains π ⇐ ⇒ ρ(σ) contains ρ(π), ∴ σ avoids π ⇐ ⇒ ρ(σ) avoids ρ(π). ∴ # Avn(π) = # Avn(ρ(π)).

Proposition

For any ρ ∈ D4 and any permutation π we have ρ(π) ≡ π.

Outline

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

A permutation statistic is st : S → {0, 1, 2, . . .}.

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3.

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3. So I3(q) = 1 + 2q + 2q2 + q3

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3. So I3(q) = 1 + 2q + 2q2 + q3 = (1)(1 + q)(1 + q + q2).

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3. So I3(q) = 1 + 2q + 2q2 + q3 = (1)(1 + q)(1 + q + q2).

Theorem (Rodrigues, 1839)

In(q) = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1)

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3. So I3(q) = 1 + 2q + 2q2 + q3 = (1)(1 + q)(1 + q + q2).

Theorem (Rodrigues, 1839)

In(q) = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1) def = [n]q!

A permutation statistic is st : S → {0, 1, 2, . . .}. The inversion number of π = a1 . . . an is inv π = #{(i, j) : i < j and ai > aj}.

• Ex. If π = 24135 then inv π = #{(1, 3), (2, 3), (2, 4)} = 3.

Consider the generating function In(q) =

• σ∈Sn

qinv π.

• Ex. When n = 3,

π : 123 132 213 231 312 321 inv π : 1 1 2 2 3 I3(q) = q0 + q1 + q1 + q2 + q2 + q3. So I3(q) = 1 + 2q + 2q2 + q3 = (1)(1 + q)(1 + q + q2).

Theorem (Rodrigues, 1839)

In(q) = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1) def = [n]q! We call [n]q! a q-analogue of n! since [n]1! = n!.

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ.

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0.

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since In(π; 1) = In(π′; 1)

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since # Avn(π) = In(π; 1) = In(π′; 1)

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since # Avn(π) = In(π; 1) = In(π′; 1) = #Avn(π′).

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since # Avn(π) = In(π; 1) = In(π′; 1) = #Avn(π′). Note that (i, j) is an inversion of π iff the line connecting the corresponding points in the diagram of π has negative slope.

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since # Avn(π) = In(π; 1) = In(π′; 1) = #Avn(π′). Note that (i, j) is an inversion of π iff the line connecting the corresponding points in the diagram of π has negative slope.

Proposition (DDJSS)

Let π ∈ S and ρ ∈ D4. Then inv ρ(π) = inv π ⇐ ⇒ ρ ∈ {R0, R180, r1, r−1}.

Given π ∈ S we have a corresponding inversion polynomial In(π; q) =

• σ∈Avn(π)

qinv σ. Call π and π′ inv-Wilf equivalent, π

inv

≡ π′, if In(π; q) = In(π′; q) for all n ≥ 0. This implies π ≡ π′ since # Avn(π) = In(π; 1) = In(π′; 1) = #Avn(π′). Note that (i, j) is an inversion of π iff the line connecting the corresponding points in the diagram of π has negative slope.

Proposition (DDJSS)

Let π ∈ S and ρ ∈ D4. Then inv ρ(π) = inv π ⇐ ⇒ ρ ∈ {R0, R180, r1, r−1}. So for ρ ∈ {R0, R180, r1, r−1} we have ρ(π)

inv

≡ π.

Let [π]inv denote the inv-Wilf equivalence class of π.

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231).

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231). To see that there are no others, note that for π ∈ Sk Ik(π; q)

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231). To see that there are no others, note that for π ∈ Sk Ik(π; q) =

• σ∈Sk−{π}

qinv σ

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231). To see that there are no others, note that for π ∈ Sk Ik(π; q) =

• σ∈Sk−{π}

qinv σ = [k]q! − qinv π.

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231). To see that there are no others, note that for π ∈ Sk Ik(π; q) =

• σ∈Sk−{π}

qinv σ = [k]q! − qinv π. So if π, π′ ∈ Sk with π

inv

≡ π′ then inv π = inv π′.

Let [π]inv denote the inv-Wilf equivalence class of π.

Theorem (DDJSS)

The inv-Wilf equivalence classes for π ∈ S3 are [123]inv = {123}, [321]inv = {321}, [132]inv = {132, 213}, [231]inv = {231, 312}.

• Proof. The two equivalences follow from the proposition:

213 = R180(132) and 312 = R180(231). To see that there are no others, note that for π ∈ Sk Ik(π; q) =

• σ∈Sk−{π}

qinv σ = [k]q! − qinv π. So if π, π′ ∈ Sk with π

inv

≡ π′ then inv π = inv π′. And inv 123, inv 321, inv 132, inv 231 are all different.

Outline

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

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

Theorem (MacMahon, 1916)

• σ∈Sn

qmaj σ = [n]q!

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

Theorem (MacMahon, 1916)

• σ∈Sn

qmaj σ = [n]q! Given π ∈ S we have a corresponding major index polynomial Mn(π; q) =

• σ∈Avn(π)

qmaj σ.

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

Theorem (MacMahon, 1916)

• σ∈Sn

qmaj σ = [n]q! Given π ∈ S we have a corresponding major index polynomial Mn(π; q) =

• σ∈Avn(π)

qmaj σ. Call π, π′ maj-Wilf equivalent, π

maj

≡ π′, if Mn(π; q) = Mn(π′; q) for all n ≥ 0.

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

Theorem (MacMahon, 1916)

• σ∈Sn

qmaj σ = [n]q! Given π ∈ S we have a corresponding major index polynomial Mn(π; q) =

• σ∈Avn(π)

qmaj σ. Call π, π′ maj-Wilf equivalent, π

maj

≡ π′, if Mn(π; q) = Mn(π′; q) for all n ≥ 0. Let [π]maj denote the maj-Wilf equivalence class of π.

The major index of π = a1 . . . an is maj π =

• ai>ai+1

i.

• Ex. If π = 253614 then maj π = 2 + 4 = 6.

Theorem (MacMahon, 1916)

• σ∈Sn

qmaj σ = [n]q! Given π ∈ S we have a corresponding major index polynomial Mn(π; q) =

• σ∈Avn(π)

qmaj σ. Call π, π′ maj-Wilf equivalent, π

maj

≡ π′, if Mn(π; q) = Mn(π′; q) for all n ≥ 0. Let [π]maj denote the maj-Wilf equivalence class of π. Note: No ρ ∈ D4 preserves the major index.

Theorem (DDJSS)

The maj-Wilf equivalence classes for π ∈ S3 are [123]maj = {123}, [321]maj = {321}, [132]maj = {132, 231}, [213]maj = {213, 312}.

Theorem (DDJSS)

The maj-Wilf equivalence classes for π ∈ S3 are [123]maj = {123}, [321]maj = {321}, [132]maj = {132, 231}, [213]maj = {213, 312}. If π = a1 . . . an and σ1, . . . , σn ∈ S then the inflation of π by the σi is the permutation π[σ1, . . . , σn] whose diagram is obtained from that of π by replacing the ith dot with a copy of σi for all i.

Theorem (DDJSS)

The maj-Wilf equivalence classes for π ∈ S3 are [123]maj = {123}, [321]maj = {321}, [132]maj = {132, 231}, [213]maj = {213, 312}. If π = a1 . . . an and σ1, . . . , σn ∈ S then the inflation of π by the σi is the permutation π[σ1, . . . , σn] whose diagram is obtained from that of π by replacing the ith dot with a copy of σi for all i. Ex. 132 = 132[σ1, σ2, σ3] = σ1 σ2 σ3

Theorem (DDJSS)

The maj-Wilf equivalence classes for π ∈ S3 are [123]maj = {123}, [321]maj = {321}, [132]maj = {132, 231}, [213]maj = {213, 312}. If π = a1 . . . an and σ1, . . . , σn ∈ S then the inflation of π by the σi is the permutation π[σ1, . . . , σn] whose diagram is obtained from that of π by replacing the ith dot with a copy of σi for all i. Ex. 132 = 132[σ1, σ2, σ3] = σ1 σ2 σ3

Conjecture

For all m, n ≥ 0 we have: 132[ιm, 1, δn]

maj

≡ 231[ιm, 1, δn], where ιm = 12 . . . m and δn = n(n − 1) . . . 1.

Outline

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

Since # Avn(π) = Cn for any π ∈ S3, the corresponding In(π; q) and Mn(π; q) are q-analogues of the Catalan numbers since setting q = 1 we recover Cn.

Since # Avn(π) = Cn for any π ∈ S3, the corresponding In(π; q) and Mn(π; q) are q-analogues of the Catalan numbers since setting q = 1 we recover Cn. The polynomials Cn(q) = In(132; q) = In(213; q), ˜ Cn(q) = In(231; q) = In(312; q) were introduced by Carlitz and Riordan and studied by numerous authors but the others seem to be new.

Since # Avn(π) = Cn for any π ∈ S3, the corresponding In(π; q) and Mn(π; q) are q-analogues of the Catalan numbers since setting q = 1 we recover Cn. The polynomials Cn(q) = In(132; q) = In(213; q), ˜ Cn(q) = In(231; q) = In(312; q) were introduced by Carlitz and Riordan and studied by numerous authors but the others seem to be new. For n ≥ 1, Cn =

n−1

• k=0

CkCn−k−1.

Since # Avn(π) = Cn for any π ∈ S3, the corresponding In(π; q) and Mn(π; q) are q-analogues of the Catalan numbers since setting q = 1 we recover Cn. The polynomials Cn(q) = In(132; q) = In(213; q), ˜ Cn(q) = In(231; q) = In(312; q) were introduced by Carlitz and Riordan and studied by numerous authors but the others seem to be new. For n ≥ 1, Cn =

n−1

• k=0

CkCn−k−1.

Theorem (DDJSS)

For n ≥ 1 we have In(312; q) =

n−1

• k=0

qkIk(312; q)In−k−1(312; q).

Divisibility properties of Catalan numbers has been a topic of recent interest: Deutsch & Sagan; Eu, Liu, & Yeh; Kauers, Krattenthaler & M¨ uller; Konvalinka; Lin; Liu & Yeh; Postnikov & Sagan; Xin & Xu; Yildiz.

Divisibility properties of Catalan numbers has been a topic of recent interest: Deutsch & Sagan; Eu, Liu, & Yeh; Kauers, Krattenthaler & M¨ uller; Konvalinka; Lin; Liu & Yeh; Postnikov & Sagan; Xin & Xu; Yildiz.

Theorem

We have that Cn is odd if and only if n = 2k − 1 for some k ≥ 0.

Divisibility properties of Catalan numbers has been a topic of recent interest: Deutsch & Sagan; Eu, Liu, & Yeh; Kauers, Krattenthaler & M¨ uller; Konvalinka; Lin; Liu & Yeh; Postnikov & Sagan; Xin & Xu; Yildiz.

Theorem

We have that Cn is odd if and only if n = 2k − 1 for some k ≥ 0. For any polynomial f(q) we let qif(q) = the coefficient of qi in f(q).

Divisibility properties of Catalan numbers has been a topic of recent interest: Deutsch & Sagan; Eu, Liu, & Yeh; Kauers, Krattenthaler & M¨ uller; Konvalinka; Lin; Liu & Yeh; Postnikov & Sagan; Xin & Xu; Yildiz.

Theorem

We have that Cn is odd if and only if n = 2k − 1 for some k ≥ 0. For any polynomial f(q) we let qif(q) = the coefficient of qi in f(q).

Theorem (DDJSS)

For all k ≥ 0 we have qiI2k−1(321; q) = 1 if i = 0, an even number if i ≥ 1.

Outline

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

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312.

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 2. For 0 ≤ k ≤ n define the q-binomial coefficients to be

n k

• q

= [n]q! [k]q![n − k]q!. (a) Prove that n

k

• 1 =

n

k

• .

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 2. For 0 ≤ k ≤ n define the q-binomial coefficients to be

n k

• q

= [n]q! [k]q![n − k]q!. (a) Prove that n

k

• 1 =

n

k

• .

(b) Prove that n

• q =

n

n

• q = 1.

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 2. For 0 ≤ k ≤ n define the q-binomial coefficients to be

n k

• q

= [n]q! [k]q![n − k]q!. (a) Prove that n

k

• 1 =

n

k

• .

(b) Prove that n

• q =

n

n

• q = 1.

(c) Prove that for 0 < k < n we have n k

• q

= n − 1 k − 1

• q

+ qk n − 1 k

• q

= qn−k n − 1 k − 1

• q

+ n − 1 k

• q

.

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 2. For 0 ≤ k ≤ n define the q-binomial coefficients to be

n k

• q

= [n]q! [k]q![n − k]q!. (a) Prove that n

k

• 1 =

n

k

• .

(b) Prove that n

• q =

n

n

• q = 1.

(c) Prove that for 0 < k < n we have n k

• q

= n − 1 k − 1

• q

+ qk n − 1 k

• q

= qn−k n − 1 k − 1

• q

+ n − 1 k

• q

. (d) Show n

k

• q is a polynomial in q with coefficients in Z≥0.

• 1. (a) Use symmetries of the square to prove 123 ≡ 321 and

132 ≡ 213 ≡ 231 ≡ 312. (b) Use induction to show that # Avn(132) = Cn.

• 2. For 0 ≤ k ≤ n define the q-binomial coefficients to be

n k

• q

= [n]q! [k]q![n − k]q!. (a) Prove that n

k

• 1 =

n

k

• .

(b) Prove that n

• q =

n

n

• q = 1.

(c) Prove that for 0 < k < n we have n k

• q

= n − 1 k − 1

• q

+ qk n − 1 k

• q

= qn−k n − 1 k − 1

• q

+ n − 1 k

• q

. (d) Show n

k

• q is a polynomial in q with coefficients in Z≥0.

(e) Let P(n, k) denote the set of w permutations of k zeros and n − k ones. Defining inv similarly to Sn, prove

• w∈P(n,k)

qinv w = n k

• q

.

References.

• 1. Leonard Carlitz and James Riordan, Two element lattice

permutation numbers and their q-generalization, Duke

• Math. J., 31 (1964) 371–388.
• 2. Szu-En Cheng, Sergi Elizalde, Anisse Kasraoui, and Bruce
• E. Sagan, Inversion polynomials for 321-avoiding

permutations, Duke Math. J., submitted.

• 3. Emeric Deutsch and Bruce E. Sagan, Congruences for

Catalan and Motzkin numbers and related sequences, J. Number Theory, 117 (2006), 191–215.

• 4. Theodore Dokos, B. Tim Dwyer, Brian P

. Johnson, Bruce

• E. Sagan, and Kim Selsor, Permutation Patterns and

Statistics, Discrete Math., 312 (2012), 2760–2775.

• 5. Percy A. MacMahon, Combinatory Analysis, volumes 1

and 2, reprint of the 1916 original, Dover, New York (2004).

• 6. Alexander Postnikov and Bruce E. Sagan, What power of

two divides a weighted Catalan number?, J. Combin. Theory, Ser. A, 114 (2007), 970–977.

• 7. Oscar Rodrigues, Note sur les inversions, ou

d´ erangements produits dans les permutations, J. Math., 4 (1839), 236–240.

• 8. Bruce E. Sagan, and Carla D. Savage, Mahonian Pairs, J.
• Combin. Theory, Ser. A, 119 (2012), 526–545.
• 9. Rodica Simion and Frank W. Schmidt, Restricted

permutations, European J. Combin., 6 (1985), 383–406.

• 10. Richard P

. Stanley, Enumerative combinatorics, volume 1, second edition. Cambridge Studies in Advanced Mathematics, 49, Cambridge University Press, Cambridge (2012).

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