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Alternating Permutations Richard P. Stanley M.I.T. Alternating - - PowerPoint PPT Presentation

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Alternating Permutations Richard P. Stanley M.I.T. Alternating Permutations p. 1 Basic definitions A sequence a 1 , a 2 , . . . , a k of distinct integers is alternating if a 1 > a 2 < a 3 > a 4 < , and reverse

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SLIDE 1

Alternating Permutations

Richard P. Stanley M.I.T.

Alternating Permutations – p. 1

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SLIDE 2

Basic definitions

A sequence a1, a2, . . . , ak of distinct integers is alternating if a1 > a2 < a3 > a4 < · · · , and reverse alternating if a1 < a2 > a3 < a4 > · · · .

Alternating Permutations – p. 2

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SLIDE 3

Euler numbers

Sn : symmetric group of all permutations of 1, 2, . . . , n

Alternating Permutations – p. 3

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SLIDE 4

Euler numbers

Sn : symmetric group of all permutations of 1, 2, . . . , n Euler number: En = #{w ∈ Sn : w is alternating} = #{w ∈ Sn : w is reverse alternating} (via a1 · · · an → n + 1 − a1, . . . , n + 1 − an)

Alternating Permutations – p. 3

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SLIDE 5

Euler numbers

Sn : symmetric group of all permutations of 1, 2, . . . , n Euler number: En = #{w ∈ Sn : w is alternating} = #{w ∈ Sn : w is reverse alternating} (via a1 · · · an → n + 1 − a1, . . . , n + 1 − an) E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

Alternating Permutations – p. 3

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SLIDE 6

André’s theorem

Theorem (Désiré André, 1879) y :=

  • n≥0

En xn n! = sec x + tan x = 1 + 1x + 1x2 2! + 2x3 3! + 5x4 4! + 16x5 5! + 61x6 6! + 272x7 7! + · · ·

Alternating Permutations – p. 4

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SLIDE 7

André’s theorem

Theorem (Désiré André, 1879) y :=

  • n≥0

En xn n! = sec x + tan x = 1 + 1x + 1x2 2! + 2x3 3! + 5x4 4! + 16x5 5! + 61x6 6! + 272x7 7! + · · · E2n is a secant number. E2n+1 is a tangent number.

Alternating Permutations – p. 4

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SLIDE 8

Proof of André’s theorem

y :=

  • n≥0

En xn n! = sec x + tan x

Alternating Permutations – p. 5

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SLIDE 9

Proof of André’s theorem

y :=

  • n≥0

En xn n! = sec x + tan x Choose S ⊆ {1, 2, . . . , n}, say #S = k.

Alternating Permutations – p. 5

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SLIDE 10

Proof of André’s theorem

y :=

  • n≥0

En xn n! = sec x + tan x Choose S ⊆ {1, 2, . . . , n}, say #S = k. Choose a reverse alternating permutation u = a1a2 · · · ak of S.

Alternating Permutations – p. 5

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SLIDE 11

Proof of André’s theorem

y :=

  • n≥0

En xn n! = sec x + tan x Choose S ⊆ {1, 2, . . . , n}, say #S = k. Choose a reverse alternating permutation u = a1a2 · · · ak of S. Choose a reverse alternating permutation v = b1b2 · · · bn−k of [n] − S.

Alternating Permutations – p. 5

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SLIDE 12

Proof of André’s theorem

y :=

  • n≥0

En xn n! = sec x + tan x Choose S ⊆ {1, 2, . . . , n}, say #S = k. Choose a reverse alternating permutation u = a1a2 · · · ak of S. Choose a reverse alternating permutation v = b1b2 · · · bn−k of [n] − S. Let w = ak · · · a2a1, n + 1, b1b2 · · · bn−k.

Alternating Permutations – p. 5

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SLIDE 13

Proof (continued)

w = ak · · · a2a1, n + 1, b1b2 · · · bn−k Given k, there are: n

k

  • choices for {a1, a2, . . . , ak}

Ek choices for a1a2 · · · ak En−k choices for b1b2 · · · bn−k.

Alternating Permutations – p. 6

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SLIDE 14

Proof (continued)

w = ak · · · a2a1, n + 1, b1b2 · · · bn−k Given k, there are: n

k

  • choices for {a1, a2, . . . , ak}

Ek choices for a1a2 · · · ak En−k choices for b1b2 · · · bn−k. We obtain each alternating and reverse alternating w ∈ Sn+1 once each.

Alternating Permutations – p. 6

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SLIDE 15

Completion of proof

⇒ 2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

Alternating Permutations – p. 7

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SLIDE 16

Completion of proof

⇒ 2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

Multiply by xn+1/(n + 1)! and sum on n ≥ 0: 2y′ = 1 + y2, y(0) = 1.

Alternating Permutations – p. 7

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SLIDE 17

Completion of proof

⇒ 2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

Multiply by xn+1/(n + 1)! and sum on n ≥ 0: 2y′ = 1 + y2, y(0) = 1. ⇒ y = sec x + tan x.

Alternating Permutations – p. 7

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SLIDE 18

A new subject?

  • n≥0

En xn n! = sec x + tan x

Alternating Permutations – p. 8

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SLIDE 19

A new subject?

  • n≥0

En xn n! = sec x + tan x Define tan x =

  • n≥0

E2n+1 x2n+1 (2n + 1)! sec x =

  • n≥0

E2n x2n (2n)!.

Alternating Permutations – p. 8

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SLIDE 20

A new subject?

  • n≥0

En xn n! = sec x + tan x Define tan x =

  • n≥0

E2n+1 x2n+1 (2n + 1)! sec x =

  • n≥0

E2n x2n (2n)!. ⇒ combinatorial trigonometry

Alternating Permutations – p. 8

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SLIDE 21

Exercises on combinatorial trig.

sec2 x = 1 + tan2 x

Alternating Permutations – p. 9

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SLIDE 22

Exercises on combinatorial trig.

sec2 x = 1 + tan2 x tan(x + y) = tan x + tan y 1 − (tan x)(tan y)

Alternating Permutations – p. 9

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SLIDE 23

Exercises on combinatorial trig.

sec2 x = 1 + tan2 x tan(x + y) = tan x + tan y 1 − (tan x)(tan y) EC2, Exercise 5.7

Alternating Permutations – p. 9

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SLIDE 24

Boustrophedon

boustrophedon:

Alternating Permutations – p. 10

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SLIDE 25

Boustrophedon

boustrophedon: an ancient method of writing in which the lines are inscribed alternately from right to left and from left to right.

Alternating Permutations – p. 10

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SLIDE 26

Boustrophedon

boustrophedon: an ancient method of writing in which the lines are inscribed alternately from right to left and from left to right. From Greek boustroph¯ edon (βoυστρoϕηδ´

  • ν),

turning like an ox while plowing: bous, ox + stroph¯ e, a turning (from strephein, to turn)

Alternating Permutations – p. 10

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SLIDE 27

The boustrophedon array

1 → 1 1 ← 1 ← → 1 → 2 → 2 5 ← 5 ← 4 ← 2 ← → 5 → 10 → 14 → 16 → 16 61 ← 61 ← 56 ← 46 ← 32 ← 16 ← 0. · · ·

Alternating Permutations – p. 11

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SLIDE 28

The boustrophedon array

1 → 1 1 ← 1 ← → 1 → 2 → 2 5 ← 5 ← 4 ← 2 ← → 5 → 10 → 14 → 16 → 16 61 ← 61 ← 56 ← 46 ← 32 ← 16 ← 0. · · ·

Alternating Permutations – p. 11

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SLIDE 29

Boustrophedon entries

last term in row n: En−1 sum of terms in row n: En kth term in row n: number of alternating permutations in Sn with first term k, the Entringer number En−1,k−1.

Alternating Permutations – p. 12

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SLIDE 30

Boustrophedon entries

last term in row n: En−1 sum of terms in row n: En kth term in row n: number of alternating permutations in Sn with first term k, the Entringer number En−1,k−1.

  • m≥0
  • n≥0

Em+n,[m,n] xm m! yn n! = cos x + sin x cos(x + y) , [m, n] =

  • m, m + n odd

n, m + n even.

Alternating Permutations – p. 12

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SLIDE 31

Some occurrences of Euler numbers

(1) E2n+1 is the number of complete increasing binary trees on the vertex set [2n + 1] = {1, 2, . . . , 2n + 1}.

Alternating Permutations – p. 13

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SLIDE 32

Five vertices

1 1 3 3 4 5 5 4 2 2 1 1 2 2 4 5 5 4 3 3 1 5 1 3 4 4 3 5 1 1 3 5 4 5 3 4 2 2 2 2 1 1 1 1 3 3 2 2 4 5 5 4 4 5 5 4 2 2 3 3 1 5 1 3 4 4 3 5 1 1 3 5 4 5 3 4 2 2 2 2

Alternating Permutations – p. 14

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SLIDE 33

Five vertices

1 1 3 3 4 5 5 4 2 2 1 1 2 2 4 5 5 4 3 3 1 5 1 3 4 4 3 5 1 1 3 5 4 5 3 4 2 2 2 2 1 1 1 1 3 3 2 2 4 5 5 4 4 5 5 4 2 2 3 3 1 5 1 3 4 4 3 5 1 1 3 5 4 5 3 4 2 2 2 2

Slightly more complicated for E2n

Alternating Permutations – p. 14

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SLIDE 34

Proof for 2n + 1

b1b2 · · · bm : sequence of distinct integers bi = min{b1, . . . , bm}

Alternating Permutations – p. 15

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SLIDE 35

Proof for 2n + 1

b1b2 · · · bm : sequence of distinct integers bi = min{b1, . . . , bm} Define recursively a binary tree T (b1, . . . , bm) by

bi T b , ..., bm

i+1

( ) T b , ..., b

1

(

i−1 )

Alternating Permutations – p. 15

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SLIDE 36

Completion of proof

  • Example. 439172856

1 3 4 9 7 8 6 5 2

Alternating Permutations – p. 16

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SLIDE 37

Completion of proof

  • Example. 439172856

1 3 4 9 7 8 6 5 2

Let w ∈ S2n+1. Then T(w) is complete if and only if w is alternating, and the map w → T(w) gives the desired bijection.

Alternating Permutations – p. 16

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SLIDE 38

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}.

Alternating Permutations – p. 17

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SLIDE 39

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}.

Alternating Permutations – p. 17

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SLIDE 40

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}. 1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6 125−34−6, 125−346, 123456

Alternating Permutations – p. 17

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SLIDE 41

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}. 1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6 125−34−6, 125−346, 123456 Sn acts on these sequences.

Alternating Permutations – p. 17

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SLIDE 42

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}. 1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6 125−34−6, 125−346, 123456 Sn acts on these sequences.

  • Theorem. The number of Sn-orbits is En

− 1.

Alternating Permutations – p. 17

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SLIDE 43

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}. 1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6 125−34−6, 125−346, 123456 Sn acts on these sequences.

  • Theorem. The number of Sn-orbits is En

− 1.

Proof.

Alternating Permutations – p. 17

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SLIDE 44

Orbits of mergings

(2) Start with n one-element sets {1}, . . . , {n}. Merge together two at a time until reaching {1, 2, . . . , n}. 1−2−3−4−5−6, 12−3−4−5−6, 12−34−5−6 125−34−6, 125−346, 123456 Sn acts on these sequences.

  • Theorem. The number of Sn-orbits is En

− 1.

  • Proof. Exercise.

Alternating Permutations – p. 17

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SLIDE 45

Orbit representatives for n = 5

12−3−4−5 123−4−5 1234−5 12−3−4−5 123−4−5 123−45 12−3−4−5 12−34−5 125−34 12−3−4−5 12−34−5 12−345 12−3−4−5 12−34−5 1234−5

Alternating Permutations – p. 18

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SLIDE 46

Volume of a polytope

(3) Let En be the convex polytope in Rn defined by xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.

Alternating Permutations – p. 19

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SLIDE 47

Volume of a polytope

(3) Let En be the convex polytope in Rn defined by xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.

  • Theorem. The volume of En is En/n!.

Alternating Permutations – p. 19

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SLIDE 48

Naive proof

vol(En) = 1

x1=0

1−x1

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx1 dx2 · · · dxn

Alternating Permutations – p. 20

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SLIDE 49

Naive proof

vol(En) = 1

x1=0

1−x1

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx1 dx2 · · · dxn fn(t) := t

x1=0

1−x1

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx1 dx2 · · · dxn

Alternating Permutations – p. 20

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SLIDE 50

Naive proof

vol(En) = 1

x1=0

1−x1

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx1 dx2 · · · dxn fn(t) := t

x1=0

1−x1

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx1 dx2 · · · dxn f ′

n(t) =

1−t

x2=0

1−x2

x3=0

· · · 1−xn−1

xn=0

dx2 dx3 · · · dxn = fn−1(1 − t).

Alternating Permutations – p. 20

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SLIDE 51

F (y)

f ′

n(t) = fn−1(1 − t), f0(t) = 1, fn(0) = 0 (n > 0)

Alternating Permutations – p. 21

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SLIDE 52

F (y)

f ′

n(t) = fn−1(1 − t), f0(t) = 1, fn(0) = 0 (n > 0)

F (y) =

  • n≥0

fn(t)yn ⇒ ∂2 ∂t2F(y) = −y2F(y), etc.

Alternating Permutations – p. 21

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SLIDE 53

Conclusion of proof

F(y) = (sec y)(cos(t − 1)y + sin ty) ⇒ F(y)|t=1 = sec y + tan y.

Alternating Permutations – p. 22

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SLIDE 54

Tridiagonal matrices

An n × n matrix M = (mij) is tridiagonal if mij = 0 whenever |i − j| ≥ 2. doubly-stochastic: mij ≥ 0, row and column sums equal 1 Tn: set of n × n tridiagonal doubly stochastic matrices

Alternating Permutations – p. 23

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SLIDE 55

Polytope structure of Tn

Easy fact: the map Tn → Rn−1 M → (m12, m23, . . . , mn−1,n) is a (linear) bijection from T to En−1.

Alternating Permutations – p. 24

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SLIDE 56

Polytope structure of Tn

Easy fact: the map Tn → Rn−1 M → (m12, m23, . . . , mn−1,n) is a (linear) bijection from T to En−1. Application (Diaconis et al.): random doubly stochastic tridiagonal matrices and random walks

  • n Tn

Alternating Permutations – p. 24

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SLIDE 57

A modification

Let Fn be the convex polytope in Rn defined by xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 + xi+2 ≤ 1, 1 ≤ i ≤ n − 2. Vn = vol(Fn)

Alternating Permutations – p. 25

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SLIDE 58

A modification

Let Fn be the convex polytope in Rn defined by xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 + xi+2 ≤ 1, 1 ≤ i ≤ n − 2. Vn = vol(Fn) n 1–3 4 5 6 7 8 9 10 n!Vn 1 2 5 14 47 182 786 3774

Alternating Permutations – p. 25

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SLIDE 59

A “naive” recurrence

Vn = fn(1, 1), where f0(a, b) = 1, fn(0, b) = 0 for n > 0 ∂ ∂afn(a, b) = fn−1(b − a, 1 − a).

Alternating Permutations – p. 26

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SLIDE 60

fn(a, b) for n ≤ 3

f1(a, b) = a f2(a, b) = 1 2(2ab − a2) f3(a, b) = 1 6(a3 − 3a2 − 3ab2 + 6ab)

Alternating Permutations – p. 27

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SLIDE 61

fn(a, b) for n ≤ 3

f1(a, b) = a f2(a, b) = 1 2(2ab − a2) f3(a, b) = 1 6(a3 − 3a2 − 3ab2 + 6ab) Is there a “nice” generating function for fn(a, b) or Vn = fn(1, 1)?

Alternating Permutations – p. 27

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SLIDE 62

Distribution of is(w)

is(w) = length of longest increasing subsequence of w ∈ Sn

Alternating Permutations – p. 28

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SLIDE 63

Distribution of is(w)

is(w) = length of longest increasing subsequence of w ∈ Sn is(48361572) = 3

Alternating Permutations – p. 28

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SLIDE 64

Distribution of is(w)

is(w) = length of longest increasing subsequence of w ∈ Sn is(48361572) = 3

Alternating Permutations – p. 28

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SLIDE 65

Distribution of is(w)

is(w) = length of longest increasing subsequence of w ∈ Sn is(48361572) = 3 Vershik-Kerov, Logan-Shepp: E(n) := 1 n!

  • w∈Sn

is(w) ∼ 2√n

Alternating Permutations – p. 28

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SLIDE 66

Limiting distribution of is(w)

Baik-Deift-Johansson: For fixed t ∈ R, lim

n→∞ Prob

isn(w) − 2√n n1/6 ≤ t

  • = F(t),

the Tracy-Widom distribution.

Alternating Permutations – p. 29

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SLIDE 67

Alternating analogues

Length of longest alternating subsequence of w ∈ Sn

Alternating Permutations – p. 30

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SLIDE 68

Alternating analogues

Length of longest alternating subsequence of w ∈ Sn Length of longest increasing subsequence of an alternating permutation w ∈ Sn.

Alternating Permutations – p. 30

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SLIDE 69

Alternating analogues

Length of longest alternating subsequence of w ∈ Sn Length of longest increasing subsequence of an alternating permutation w ∈ Sn. The first is much easier!

Alternating Permutations – p. 30

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SLIDE 70

Longest alternating subsequences

as(w)= length of longest alt. subseq. of w w = 56218347 ⇒ as(w) = 5

Alternating Permutations – p. 31

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SLIDE 71

Longest alternating subsequences

as(w)= length of longest alt. subseq. of w w = 56218347 ⇒ as(w) = 5 D(n)= 1 n!

  • w∈Sn

as(w) ∼ ?

Alternating Permutations – p. 31

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SLIDE 72

Definitions of ak(n) and bk(n)

ak(n) = #{w ∈ Sn : as(w) = k} bk(n) = a1(n) + a2(n) + · · · + ak(n) = #{w ∈ Sn : as(w) ≤ k}

Alternating Permutations – p. 32

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SLIDE 73

The case n = 3

w as(w) 123 1 132 2 213 3 231 2 312 3 321 2

Alternating Permutations – p. 33

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SLIDE 74

The case n = 3

w as(w) 123 1 132 2 213 3 231 2 312 3 321 2 a1(3) = 1, a2(3) = 3, a3(3) = 2 b1(3) = 1, b2(3) = 4, b3(3) = 6

Alternating Permutations – p. 33

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SLIDE 75

The main lemma

  • Lemma. ∀ w ∈ Sn ∃ alternating subsequence of

maximal length that contains n.

Alternating Permutations – p. 34

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SLIDE 76

The main lemma

  • Lemma. ∀ w ∈ Sn ∃ alternating subsequence of

maximal length that contains n. Corollary. ⇒ ak(n) =

n

  • j=1

n − 1 j − 1

  • 2r+s=k−1

(a2r(j − 1) + a2r+1(j − 1)) as(n − j)

Alternating Permutations – p. 34

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SLIDE 77

The main generating function

B(x, t)=

  • k,n≥0

bk(n)tkxn n! Theorem. B(x, t) = 2/ρ 1 − 1−ρ

t eρx − 1

ρ, where ρ= √ 1 − t2.

Alternating Permutations – p. 35

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SLIDE 78

Formulas for bk(n)

Corollary. ⇒ b1(n) = 1 b2(n) = n b3(n) =

1 4(3n − 2n + 3)

b4(n) =

1 8(4n − (2n − 4)2n)

. . .

Alternating Permutations – p. 36

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SLIDE 79

Formulas for bk(n)

Corollary. ⇒ b1(n) = 1 b2(n) = n b3(n) =

1 4(3n − 2n + 3)

b4(n) =

1 8(4n − (2n − 4)2n)

. . . no such formulas for longest increasing subsequences

Alternating Permutations – p. 36

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SLIDE 80

Mean (expectation) of as(w)

D(n) = 1 n!

  • w∈Sn

as(w) = 1 n!

n

  • k=1

k · ak(n), the expectation of as(w) for w ∈ Sn

Alternating Permutations – p. 37

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SLIDE 81

Mean (expectation) of as(w)

D(n) = 1 n!

  • w∈Sn

as(w) = 1 n!

n

  • k=1

k · ak(n), the expectation of as(w) for w ∈ Sn Let A(x, t) =

  • k,n≥0

ak(n)tkxn n! = (1 − t)B(x, t) = (1 − t)

  • 2/ρ

1 − 1−ρ

t eρx − 1

ρ

  • .

Alternating Permutations – p. 37

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SLIDE 82

Formula for D(n)

  • n≥0

D(n)xn = ∂ ∂tA(x, 1) = 6x − 3x2 + x3 6(1 − x)2 = x +

  • n≥2

4n + 1 6 xn.

Alternating Permutations – p. 38

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SLIDE 83

Formula for D(n)

  • n≥0

D(n)xn = ∂ ∂tA(x, 1) = 6x − 3x2 + x3 6(1 − x)2 = x +

  • n≥2

4n + 1 6 xn. ⇒ D(n) = 4n + 1 6 , n ≥ 2

Alternating Permutations – p. 38

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SLIDE 84

Formula for D(n)

  • n≥0

D(n)xn = ∂ ∂tA(x, 1) = 6x − 3x2 + x3 6(1 − x)2 = x +

  • n≥2

4n + 1 6 xn. ⇒ D(n) = 4n + 1 6 , n ≥ 2 Compare E(n) ∼ 2√n.

Alternating Permutations – p. 38

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SLIDE 85

Variance of as(w)

V (n)= 1 n!

  • w∈Sn
  • as(w) − 4n + 1

6 2 , n ≥ 2 the variance of as(w) for w ∈ Sn

Alternating Permutations – p. 39

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SLIDE 86

Variance of as(w)

V (n)= 1 n!

  • w∈Sn
  • as(w) − 4n + 1

6 2 , n ≥ 2 the variance of as(w) for w ∈ Sn Corollary. V (n) = 8 45n − 13 180, n ≥ 4

Alternating Permutations – p. 39

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SLIDE 87

Variance of as(w)

V (n)= 1 n!

  • w∈Sn
  • as(w) − 4n + 1

6 2 , n ≥ 2 the variance of as(w) for w ∈ Sn Corollary. V (n) = 8 45n − 13 180, n ≥ 4 similar results for higher moments

Alternating Permutations – p. 39

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SLIDE 88

A new distribution?

P (t) = lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • Alternating Permutations – p. 40
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SLIDE 89

A new distribution?

P (t) = lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • Stanley distribution?

Alternating Permutations – p. 40

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SLIDE 90

Limiting distribution

Theorem (Pemantle, Widom, (Wilf)). lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • =

1 √π t

√ 45/4 −∞

e−s2 ds (Gaussian distribution)

Alternating Permutations – p. 41

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SLIDE 91

Limiting distribution

Theorem (Pemantle, Widom, (Wilf)). lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • =

1 √π t

√ 45/4 −∞

e−s2 ds (Gaussian distribution)

Alternating Permutations – p. 41

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SLIDE 92

Umbral enumeration

Umbral formula: involves Ek, where E is an indeterminate (the umbra). Replace Ek with the Euler number Ek. (Technique from 19th century, modernized by Rota et al.)

Alternating Permutations – p. 42

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SLIDE 93

Umbral enumeration

Umbral formula: involves Ek, where E is an indeterminate (the umbra). Replace Ek with the Euler number Ek. (Technique from 19th century, modernized by Rota et al.) Example. (1 + E2)3 = 1 + 3E2 + 3E4 + E6 = 1 + 3E2 + 3E4 + E6 = 1 + 3 · 1 + 3 · 5 + 61 = 80

Alternating Permutations – p. 42

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SLIDE 94

Another example

(1 + t)E = 1 + Et + E 2

  • t2 +

E 3

  • t3 + · · ·

= 1 + Et + 1 2E(E − 1)t2 + · · · = 1 + E1t + 1 2(E2 − E1))t2 + · · · = 1 + t + 1 2(1 − 1)t2 + · · · = 1 + t + O(t3).

Alternating Permutations – p. 43

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SLIDE 95

An umbral quiz

Let B be the Bell number umbra. Then

Alternating Permutations – p. 44

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SLIDE 96

An umbral quiz

Let B be the Bell number umbra. Then (1 + t)B = ??

Alternating Permutations – p. 44

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SLIDE 97

An umbral quiz

Let B be the Bell number umbra. Then (1 + t)B = et

Alternating Permutations – p. 44

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SLIDE 98
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two (number = 1 · 3 · 5 · · · (2n − 1))

Alternating Permutations – p. 45

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SLIDE 99
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two (number = 1 · 3 · 5 · · · (2n − 1)) Let f(n) be the number of alternating fixed-point free involutions in S2n.

Alternating Permutations – p. 45

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SLIDE 100
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two (number = 1 · 3 · 5 · · · (2n − 1)) Let f(n) be the number of alternating fixed-point free involutions in S2n. n = 3 : 214365 = (1, 2)(3, 4)(5, 6) 645231 = (1, 6)(2, 4)(3, 5) f(3) = 2

Alternating Permutations – p. 45

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SLIDE 101

An umbral theorem

Theorem. F (x) =

  • n≥0

f(n)xn

Alternating Permutations – p. 46

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SLIDE 102

An umbral theorem

Theorem. F (x) =

  • n≥0

f(n)xn = 1 + x 1 − x (E2+1)/4

Alternating Permutations – p. 46

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SLIDE 103

Proof idea

  • Proof. Uses representation theory of the

symmetric group Sn.

Alternating Permutations – p. 47

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SLIDE 104

Proof idea

  • Proof. Uses representation theory of the

symmetric group Sn. There is a character χ of Sn (due to H. O. Foulkes) such that for all w ∈ Sn, χ(w) = 0 or ± Ek.

Alternating Permutations – p. 47

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SLIDE 105

Proof idea

  • Proof. Uses representation theory of the

symmetric group Sn. There is a character χ of Sn (due to H. O. Foulkes) such that for all w ∈ Sn, χ(w) = 0 or ± Ek. Now use known results on combinatorial properties of characters of Sn.

Alternating Permutations – p. 47

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SLIDE 106

Ramanujan’s Second Notebook

Theorem (Ramanujan, Berndt, implicitly) As x → 0+, 2

  • n≥0

(−1)n 1 − x 1 + x n(n+1) ∼

  • k≥0

f(k)xk = F(x), an analytic (non-formal) identity.

Alternating Permutations – p. 48

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SLIDE 107

A formal identity

Corollary (via Ramanujan, Andrews). F(x) = 2

  • n≥0

qn n

j=1(1 − q2j−1)

2n+1

j=1 (1 + qj)

, where q = 1−x

1+x

2/3, a formal identity.

Alternating Permutations – p. 49

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SLIDE 108

Simple result, hard proof

Recall: number of n-cycles in Sn is (n − 1)!.

Alternating Permutations – p. 50

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SLIDE 109

Simple result, hard proof

Recall: number of n-cycles in Sn is (n − 1)!.

  • Theorem. Let b(n) be the number of

alternating n-cycles in Sn. Then if n is odd, b(n) = 1 n

  • d|n

µ(d)(−1)(d−1)/2En/d.

Alternating Permutations – p. 50

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SLIDE 110

Special case

  • Corollary. Let p be an odd prime. Then

b(p) = 1 p

  • Ep − (−1)(p−1)/2

.

Alternating Permutations – p. 51

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SLIDE 111

Special case

  • Corollary. Let p be an odd prime. Then

b(p) = 1 p

  • Ep − (−1)(p−1)/2

. Combinatorial proof?

Alternating Permutations – p. 51

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SLIDE 112
  • Inc. subsequences of alt. perms.

Recall: is(w) = length of longest increasing subsequence of w ∈ Sn. Define C(n) = 1 En

  • w

is(w), where w ranges over all En alternating permutations in Sn.

Alternating Permutations – p. 52

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SLIDE 113

β

Crude estimate: what is β = lim

n→∞

log C(n) log n ? I.e., C(n) = nβ+o(1).

Alternating Permutations – p. 53

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SLIDE 114

β

Crude estimate: what is β = lim

n→∞

log C(n) log n ? I.e., C(n) = nβ+o(1). Easy: 1

2 ≤ β ≤ 1

Alternating Permutations – p. 53

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SLIDE 115

Limiting distribution?

What is the (suitably scaled) limiting distribution

  • f is(w), where w ranges over all alternating

permutations in Sn?

Alternating Permutations – p. 54

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SLIDE 116

Limiting distribution?

What is the (suitably scaled) limiting distribution

  • f is(w), where w ranges over all alternating

permutations in Sn? Is it the Tracy-Widom distribution?

Alternating Permutations – p. 54

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SLIDE 117

Limiting distribution?

What is the (suitably scaled) limiting distribution

  • f is(w), where w ranges over all alternating

permutations in Sn? Is it the Tracy-Widom distribution? Possible tool: ∃ “umbral analogue” of Gessel’s determinantal formula.

Alternating Permutations – p. 54

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SLIDE 118

Limiting distribution?

What is the (suitably scaled) limiting distribution

  • f is(w), where w ranges over all alternating

permutations in Sn? Is it the Tracy-Widom distribution? Possible tool: ∃ “umbral analogue” of Gessel’s determinantal formula. . . . . . .

Alternating Permutations – p. 54

slide-119
SLIDE 119

Alternating Permutations – p. 55