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Alternating Permutations Richard P. Stanley M.I.T. Alternating Permutations – p.

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Page 1: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Alternating PermutationsRichard P. Stanley

M.I.T.

Alternating Permutations – p. 1

Page 2: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Basic definitions

A sequence a1, a2, . . . , ak of distinct integers isalternating if

a1 > a2 < a3 > a4 < · · · ,

and reverse alternating if

a1 < a2 > a3 < a4 > · · · .

Alternating Permutations – p. 2

Page 3: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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}

E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

Alternating Permutations – p. 3

Page 4: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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}

E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

Alternating Permutations – p. 3

Page 5: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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}

E.g., E4 = 5 : 2143, 3142, 3241, 4132, 4231

Alternating Permutations – p. 3

Page 6: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

André’s theorem

Theorem (Désiré André, 1879)

y:=∑

n≥0

Enxn

n!= sec x + tan x

E2n is a secant number.

E2n+1 is a tangent number.

⇒ combinatorial trigonometry

Alternating Permutations – p. 4

Page 7: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

André’s theorem

Theorem (Désiré André, 1879)

y:=∑

n≥0

Enxn

n!= sec x + tan x

E2n is a secant number.

E2n+1 is a tangent number.

⇒ combinatorial trigonometry

Alternating Permutations – p. 4

Page 8: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

André’s theorem

Theorem (Désiré André, 1879)

y:=∑

n≥0

Enxn

n!= sec x + tan x

E2n is a secant number.

E2n+1 is a tangent number.

⇒ combinatorial trigonometry

Alternating Permutations – p. 4

Page 9: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Example of combinatorial trig.

sec2 x = 1 + tan2 x

Equate coefficients of x2n/(2n)!:

n∑

k=0

(

2n

2k

)

E2kE2(n−k)

=n−1∑

k=0

(

2n

2k + 1

)

E2k+1E2(n−k)−1.

Prove combinatorially (exercise).

Alternating Permutations – p. 5

Page 10: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Example of combinatorial trig.

sec2 x = 1 + tan2 x

Equate coefficients of x2n/(2n)!:

n∑

k=0

(

2n

2k

)

E2kE2(n−k)

=n−1∑

k=0

(

2n

2k + 1

)

E2k+1E2(n−k)−1.

Prove combinatorially (exercise).

Alternating Permutations – p. 5

Page 11: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Example of combinatorial trig.

sec2 x = 1 + tan2 x

Equate coefficients of x2n/(2n)!:

n∑

k=0

(

2n

2k

)

E2kE2(n−k)

=n−1∑

k=0

(

2n

2k + 1

)

E2k+1E2(n−k)−1.

Prove combinatorially (exercise).

Alternating Permutations – p. 5

Page 12: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Proof of André’s theorem

y :=∑

n≥0

Enxn

n!= sec x + tan x

Naive proof.

2En+1 =n∑

k=0

(

n

k

)

EkEn−k, n ≥ 1

⇒ 2y′ = 1 + y2, etc.

(details omitted)

Alternating Permutations – p. 6

Page 13: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Proof of André’s theorem

y :=∑

n≥0

Enxn

n!= sec x + tan x

Naive proof.

2En+1 =n∑

k=0

(

n

k

)

EkEn−k, n ≥ 1

⇒ 2y′ = 1 + y2, etc.

(details omitted)

Alternating Permutations – p. 6

Page 14: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Some occurences of Euler numbers

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

Alternating Permutations – p. 7

Page 15: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Five vertices

1 1

3 34554

2 2

1 1

2 24554

3 3

1

5

1

34435

11

354

534

2 2 2 2

1111

33224 5 5 4 4 5 5 4

2233

1

5

1

3 4 4 35

1 1

3 54

5 34

2222

Slightly more complicated for E2n

Alternating Permutations – p. 8

Page 16: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Five vertices

1 1

3 34554

2 2

1 1

2 24554

3 3

1

5

1

34435

11

354

534

2 2 2 2

1111

33224 5 5 4 4 5 5 4

2233

1

5

1

3 4 4 35

1 1

3 54

5 34

2222

Slightly more complicated for E2n

Alternating Permutations – p. 8

Page 17: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Simsun permutations

(2) b1b2 · · · bk has a double descent if for some1 < i < n,

bi−1 > bi > bi+1.

w = a1a2 · · · an ∈ Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 ≤ k ≤ n.

Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

Alternating Permutations – p. 9

Page 18: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Simsun permutations

(2) b1b2 · · · bk has a double descent if for some1 < i < n,

bi−1 > bi > bi+1.

w = a1a2 · · · an ∈ Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 ≤ k ≤ n.

Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

Alternating Permutations – p. 9

Page 19: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Simsun permutations

(2) b1b2 · · · bk has a double descent if for some1 < i < n,

bi−1 > bi > bi+1.

w = a1a2 · · · an ∈ Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 ≤ k ≤ n.

Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

Alternating Permutations – p. 9

Page 20: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Simsun permutations

(2) b1b2 · · · bk has a double descent if for some1 < i < n,

bi−1 > bi > bi+1.

w = a1a2 · · · an ∈ Sn is a simsun permutation ifthe subsequence with elements 1, 2, . . . , k has nodouble descents, 1 ≤ k ≤ n.

Example. 3241 is not simsun: the subsequencewith 1, 2, 3 is 321.

Theorem (R. Simion & S. Sundaram) Thenumber of simsun permutations in Sn is En+1.

Alternating Permutations – p. 9

Page 21: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Orbits of mergings

(3) 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. 10

Page 22: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Orbits of mergings

(3) 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. 10

Page 23: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Orbits of mergings

(3) 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. 10

Page 24: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Orbits of mergings

(3) 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. 10

Page 25: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Orbits of mergings

(3) 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. 10

Page 26: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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. 11

Page 27: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Volume of a polytope

(4) 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. 12

Page 28: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Volume of a polytope

(4) 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. 12

Page 29: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The “nicest” proof

Triangulate En so that the maximal simplicesσw are indexed by alternating permutationsw ∈ Sn.

Show Vol(σw) = 1/n!.

Alternating Permutations – p. 13

Page 30: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The “nicest” proof

Triangulate En so that the maximal simplicesσw are indexed by alternating permutationsw ∈ Sn.

Show Vol(σw) = 1/n!.

Alternating Permutations – p. 13

Page 31: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Tridiagonal matrices

An n × n matrix M = (mij) is tridiagonal ifmij = 0 whenever |i − j| ≥ 2.

doubly-stochastic: mij ≥ 0, row and columnsums equal 1

Tn: set of n × n tridiagonal doubly stochasticmatrices

Alternating Permutations – p. 14

Page 32: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Polytope structure of Tn

Easy fact: the map

Tn → Rn−1

M 7→ (m12,m23, . . . ,mn−1,n)

is a (linear) bijection from Tn to En.

Application (Diaconis et al.): random doublystochastic tridiagonal matrices and random walkson Tn

Alternating Permutations – p. 15

Page 33: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Polytope structure of Tn

Easy fact: the map

Tn → Rn−1

M 7→ (m12,m23, . . . ,mn−1,n)

is a (linear) bijection from Tn to En.

Application (Diaconis et al.): random doublystochastic tridiagonal matrices and random walkson Tn

Alternating Permutations – p. 15

Page 34: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Distribution of is(w)

Yesterday: is(w) = length of longest increasingsubsequence of w ∈ Sn

E(n) ∼ 2√

n

For fixed t ∈ R,

limn→∞

Prob

(

isn(w) − 2√

n

n1/6≤ t

)

= F (t),

the Tracy-Widom distribution.

Alternating Permutations – p. 16

Page 35: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Distribution of is(w)

Yesterday: is(w) = length of longest increasingsubsequence of w ∈ Sn

E(n) ∼ 2√

n

For fixed t ∈ R,

limn→∞

Prob

(

isn(w) − 2√

n

n1/6≤ t

)

= F (t),

the Tracy-Widom distribution.

Alternating Permutations – p. 16

Page 36: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Analogues of distribution of is(w)

Length of longest alternating subsequence ofw ∈ Sn

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

The first is much easier!

Alternating Permutations – p. 17

Page 37: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Analogues of distribution of is(w)

Length of longest alternating subsequence ofw ∈ Sn

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

The first is much easier!

Alternating Permutations – p. 17

Page 38: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Analogues of distribution of is(w)

Length of longest alternating subsequence ofw ∈ Sn

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

The first is much easier!

Alternating Permutations – p. 17

Page 39: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Longest alternating subsequences

as(w)= length longest alt. subseq. of w

D(n)=1

n!

w∈Sn

as(w) ∼ ?

w = 56218347 ⇒ as(w) = 5

Alternating Permutations – p. 18

Page 40: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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. 19

Page 41: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The case n = 3

w as(w)

123 1132 2213 3231 2312 3321 2

a1(3) = 1, a2(3) = 3, a3(3) = 2

b1(3) = 1, b2(3) = 4, b3(3) = 6

Alternating Permutations – p. 20

Page 42: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The case n = 3

w as(w)

123 1132 2213 3231 2312 3321 2

a1(3) = 1, a2(3) = 3, a3(3) = 2

b1(3) = 1, b2(3) = 4, b3(3) = 6

Alternating Permutations – p. 20

Page 43: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The main lemma

Lemma. ∀w ∈ Sn ∃ alternating subsequence ofmaximal 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. 21

Page 44: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The main lemma

Lemma. ∀w ∈ Sn ∃ alternating subsequence ofmaximal 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. 21

Page 45: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

The main generating function

B(x, t)=∑

k,n≥0

bk(n)tkxn

n!

Theorem.

B(x, t) =2/ρ

1 − 1−ρt eρx

− 1

ρ,

where ρ=√

1 − t2.

Alternating Permutations – p. 22

Page 46: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Formulas for bk(n)

Corollary.

⇒ b1(n) = 1

b2(n) = n

b3(n) = 14(3

n − 2n + 3)

b4(n) = 18(4

n − (2n − 4)2n)

...

no such formulas for longest increasingsubsequences

Alternating Permutations – p. 23

Page 47: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Formulas for bk(n)

Corollary.

⇒ b1(n) = 1

b2(n) = n

b3(n) = 14(3

n − 2n + 3)

b4(n) = 18(4

n − (2n − 4)2n)

...

no such formulas for longest increasingsubsequences

Alternating Permutations – p. 23

Page 48: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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−ρt eρx

− 1

ρ

)

.

Alternating Permutations – p. 24

Page 49: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

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−ρt eρx

− 1

ρ

)

.

Alternating Permutations – p. 24

Page 50: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Formula for D(n)

n≥0

D(n)xn =∂

∂tA(x, 1)

=6x − 3x2 + x3

6(1 − x)2

= x +∑

n≥2

4n + 1

6xn.

⇒ D(n) =4n + 1

6, n ≥ 2 (why?)

Compare E(n) ∼ 2√

n.

Alternating Permutations – p. 25

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Formula for D(n)

n≥0

D(n)xn =∂

∂tA(x, 1)

=6x − 3x2 + x3

6(1 − x)2

= x +∑

n≥2

4n + 1

6xn.

⇒ D(n) =4n + 1

6, n ≥ 2 (why?)

Compare E(n) ∼ 2√

n.

Alternating Permutations – p. 25

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Formula for D(n)

n≥0

D(n)xn =∂

∂tA(x, 1)

=6x − 3x2 + x3

6(1 − x)2

= x +∑

n≥2

4n + 1

6xn.

⇒ D(n) =4n + 1

6, n ≥ 2 (why?)

Compare E(n) ∼ 2√

n.Alternating Permutations – p. 25

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Variance of as(w)

V (n)=1

n!

w∈Sn

(

as(w) − 4n + 1

6

)2

, n ≥ 2

the variance of as(n) for w ∈ Sn

Corollary.

V (n) =8

45n − 13

180, n ≥ 4

similar results for higher moments

Alternating Permutations – p. 26

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Variance of as(w)

V (n)=1

n!

w∈Sn

(

as(w) − 4n + 1

6

)2

, n ≥ 2

the variance of as(n) for w ∈ Sn

Corollary.

V (n) =8

45n − 13

180, n ≥ 4

similar results for higher moments

Alternating Permutations – p. 26

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Variance of as(w)

V (n)=1

n!

w∈Sn

(

as(w) − 4n + 1

6

)2

, n ≥ 2

the variance of as(n) for w ∈ Sn

Corollary.

V (n) =8

45n − 13

180, n ≥ 4

similar results for higher moments

Alternating Permutations – p. 26

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A new distribution?

P (t) = limn→∞

Probw∈Sn

(

as(w) − 2n/3√n

≤ t

)

Stanley distribution?

Alternating Permutations – p. 27

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A new distribution?

P (t) = limn→∞

Probw∈Sn

(

as(w) − 2n/3√n

≤ t

)

Stanley distribution?

Alternating Permutations – p. 27

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Limiting distribution

Theorem (Pemantle, Widom, (Wilf)).

limn→∞

Probw∈Sn

(

as(w) − 2n/3√n

≤ t

)

=1√π

∫ t√

45/4

−∞e−s2

ds

(Gaussian distribution)

Alternating Permutations – p. 28

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Limiting distribution

Theorem (Pemantle, Widom, (Wilf)).

limn→∞

Probw∈Sn

(

as(w) − 2n/3√n

≤ t

)

=1√π

∫ t√

45/4

−∞e−s2

ds

(Gaussian distribution)

Alternating Permutations – p. 28

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Umbral enumeration

Umbral formula: involves Ek, where E is anindeterminate (the umbra). Replace Ek with theEuler 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. 29

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Umbral enumeration

Umbral formula: involves Ek, where E is anindeterminate (the umbra). Replace Ek with theEuler 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. 29

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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))t

2 + · · ·

= 1 + t +1

2(1 − 1)t2 + · · ·

= 1 + t + O(t3).

Alternating Permutations – p. 30

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Alt. fixed-point free involutions

fixed-point free involution w ∈ S2n: all cyclesof length two

Let f(n) be the number of alternatingfixed-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. 31

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Alt. fixed-point free involutions

fixed-point free involution w ∈ S2n: all cyclesof length two

Let f(n) be the number of alternatingfixed-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. 31

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Alt. fixed-point free involutions

fixed-point free involution w ∈ S2n: all cyclesof length two

Let f(n) be the number of alternatingfixed-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. 31

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An umbral theorem

Theorem.F (x) :=

n≥0

f(n)xn

=

(

1 + x

1 − x

)(E2+1)/4

Alternating Permutations – p. 32

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An umbral theorem

Theorem.F (x) :=

n≥0

f(n)xn

=

(

1 + x

1 − x

)(E2+1)/4

Alternating Permutations – p. 32

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Proof idea

Proof. Uses representation theory of thesymmetric 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 combinatorialproperties of characters of Sn.

Alternating Permutations – p. 33

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Proof idea

Proof. Uses representation theory of thesymmetric 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 combinatorialproperties of characters of Sn.

Alternating Permutations – p. 33

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Proof idea

Proof. Uses representation theory of thesymmetric 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 combinatorialproperties of characters of Sn.

Alternating Permutations – p. 33

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Ramanujan’s Lost Notebook

Theorem (Ramanujan, Berndt, implicitly) Asx → 0+,

2∑

n≥0

(

1 − x

1 + x

)n(n+1)

∼∑

k≥0

f(k)xk = F (x),

an analytic (non-formal) identity.

Alternating Permutations – p. 34

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A formal identity

Corollary (via Ramanujan, Andrews).

F (x) = 2∑

n≥0

qn

∏nj=1(1 − q2j−1)∏2n+1

j=1 (1 + qj),

where q =(

1−x1+x

)2/3, a formal identity.

Alternating Permutations – p. 35

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Simple result, hard proof

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

Theorem. Let b(n) be the number ofalternating n-cycles in Sn. Then if n is odd,

b(n) =1

n

d|nµ(d)(−1)(d−1)/2En/d

∼ En/n.

Alternating Permutations – p. 36

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Simple result, hard proof

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

Theorem. Let b(n) be the number ofalternating n-cycles in Sn. Then if n is odd,

b(n) =1

n

d|nµ(d)(−1)(d−1)/2En/d

∼ En/n.

Alternating Permutations – p. 36

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Special case

Corollary. Let p be an odd prime. Then

b(p) =1

p

(

Ep − (−1)(p−1)/2)

.

Combinatorial proof?

Alternating Permutations – p. 37

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Special case

Corollary. Let p be an odd prime. Then

b(p) =1

p

(

Ep − (−1)(p−1)/2)

.

Combinatorial proof?

Alternating Permutations – p. 37

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Inc. subsequences of alt. perms.

Recall: is(w) = length of longest increasingsubsequence of w ∈ Sn. Define

C(n) =1

En

w

is(w),

where w ranges over all En alternatingpermutations in Sn.

Alternating Permutations – p. 38

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β

Little is known, e.g., what is

β = limn→∞

log C(n)

log n?

I.e., C(n) = nβ+o(1).

Compare limn→∞log E(n)

log n = 1/2.

Easy: β ≥ 12.

Alternating Permutations – p. 39

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β

Little is known, e.g., what is

β = limn→∞

log C(n)

log n?

I.e., C(n) = nβ+o(1).

Compare limn→∞log E(n)

log n = 1/2.

Easy: β ≥ 12.

Alternating Permutations – p. 39

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Limiting distribution?

What is the (suitably scaled) limiting distributionof is(w), where w ranges over all alternatingpermutations in Sn?

Is it the Tracy-Widom distribution?

Possible tool: ∃ “umbral analogue” of Gessel’sdeterminantal formula.

Alternating Permutations – p. 40

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Limiting distribution?

What is the (suitably scaled) limiting distributionof is(w), where w ranges over all alternatingpermutations in Sn?

Is it the Tracy-Widom distribution?

Possible tool: ∃ “umbral analogue” of Gessel’sdeterminantal formula.

Alternating Permutations – p. 40

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Limiting distribution?

What is the (suitably scaled) limiting distributionof is(w), where w ranges over all alternatingpermutations in Sn?

Is it the Tracy-Widom distribution?

Possible tool: ∃ “umbral analogue” of Gessel’sdeterminantal formula.

Alternating Permutations – p. 40

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Descent sets

Let w = a1a2 · · · an ∈ Sn. Descent set of w:

D(w) = {i : ai > ai+1} ⊆ {1, . . . , n − 1}

D(4157623) = {1, 4, 5}D(4152736) = {1, 3, 5} (alternating)

D(4736152) = {2, 4, 6} (reverse alternating)

Alternating Permutations – p. 41

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Descent sets

Let w = a1a2 · · · an ∈ Sn. Descent set of w:

D(w) = {i : ai > ai+1} ⊆ {1, . . . , n − 1}

D(4157623) = {1, 4, 5}D(4152736) = {1, 3, 5} (alternating)

D(4736152) = {2, 4, 6} (reverse alternating)

Alternating Permutations – p. 41

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βn(S)

βn(S) = #{w ∈ Sn : D(w) = S}

w D(w)

123 ∅213 {1}312 {1}132 {2}231 {2}321 {1, 2}

β3(∅) = 1, β3(1) = 2

β3(2) = 2, β3(1, 2) = 1

Alternating Permutations – p. 42

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βn(S)

βn(S) = #{w ∈ Sn : D(w) = S}

w D(w)

123 ∅213 {1}312 {1}132 {2}231 {2}321 {1, 2}

β3(∅) = 1, β3(1) = 2

β3(2) = 2, β3(1, 2) = 1

Alternating Permutations – p. 42

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uS

Fix n. Let S ⊆ {1, · · · , n − 1}. Let uS = t1 · · · tn−1,where

ti =

{

a, i 6∈ S

b, i ∈ S.

Example. n = 8, S = {2, 5, 6} ⊆ {1, . . . , 7}uS = abaabba

Alternating Permutations – p. 43

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uS

Fix n. Let S ⊆ {1, · · · , n − 1}. Let uS = t1 · · · tn−1,where

ti =

{

a, i 6∈ S

b, i ∈ S.

Example. n = 8, S = {2, 5, 6} ⊆ {1, . . . , 7}uS = abaabba

Alternating Permutations – p. 43

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A noncommutative gen. function

Ψn(a, b) =∑

S⊆{1,...,n−1}βn(S)uS .

Example. Recall

β3(∅) = 1, β3(1) = 2, β3(2) = 2, β3(1, 2) = 1

ThusΨ3(a, b) = aa + 2ab + 2ba + bb

= (a + b)2 + (ab + ba)

Alternating Permutations – p. 44

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A noncommutative gen. function

Ψn(a, b) =∑

S⊆{1,...,n−1}βn(S)uS .

Example. Recall

β3(∅) = 1, β3(1) = 2, β3(2) = 2, β3(1, 2) = 1

ThusΨ3(a, b) = aa + 2ab + 2ba + bb

= (a + b)2 + (ab + ba)

Alternating Permutations – p. 44

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A noncommutative gen. function

Ψn(a, b) =∑

S⊆{1,...,n−1}βn(S)uS .

Example. Recall

β3(∅) = 1, β3(1) = 2, β3(2) = 2, β3(1, 2) = 1

ThusΨ3(a, b) = aa + 2ab + 2ba + bb

= (a + b)2 + (ab + ba)

Alternating Permutations – p. 44

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The cd-index

Theorem. There exists a noncommutativepolynomial Φn(c, d), called the cd-index of Sn,with nonnegative integer coefficients, such that

Ψn(a, b) = Φn(a + b, ab + ba).

Example. Recall

Ψ3(a, b) = aa+2ab+2ba+ b2 = (a+ b)2 +(ab+ ba).

ThereforeΦ3(c, d) = c2 + d.

Alternating Permutations – p. 45

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The cd-index

Theorem. There exists a noncommutativepolynomial Φn(c, d), called the cd-index of Sn,with nonnegative integer coefficients, such that

Ψn(a, b) = Φn(a + b, ab + ba).

Example. Recall

Ψ3(a, b) = aa+2ab+2ba+ b2 = (a+ b)2 +(ab+ ba).

ThereforeΦ3(c, d) = c2 + d.

Alternating Permutations – p. 45

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Small values of Φn(c, d)

Φ1 = 1

Φ2 = c

Φ3 = c2 + d

Φ4 = c3 + 2cd + 2dc

Φ5 = c4 + 3c2d + 5cdc + 3dc2 + 4d2

Φ6 = c5 + 4c3d + 9c2dc + 9cdc2 + 4dc3

+12cd2 + 10dcd + 12d2c.

Alternating Permutations – p. 46

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Let deg c = 1, deg d = 2.

µ: cd-monomial of degree n − 1

Replace each c in µ with 0, each d with 10, andremove final 0. Get the characteristic vector of aset Sµ ⊆ [n − 2].

Example. n = 10:

µ = cd2c2d → 0 · 10 · 10 · 0 · 0 · 10× = 01010001,

the characteristic vector of Sµ = {2, 4, 8} ⊆ [8]

Alternating Permutations – p. 47

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Let deg c = 1, deg d = 2.

µ: cd-monomial of degree n − 1

Replace each c in µ with 0, each d with 10, andremove final 0. Get the characteristic vector of aset Sµ ⊆ [n − 2].

Example. n = 10:

µ = cd2c2d → 0 · 10 · 10 · 0 · 0 · 10× = 01010001,

the characteristic vector of Sµ = {2, 4, 8} ⊆ [8]

Alternating Permutations – p. 47

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Let deg c = 1, deg d = 2.

µ: cd-monomial of degree n − 1

Replace each c in µ with 0, each d with 10, andremove final 0. Get the characteristic vector of aset Sµ ⊆ [n − 2].

Example. n = 10:

µ = cd2c2d → 0 · 10 · 10 · 0 · 0 · 10× = 01010001,

the characteristic vector of Sµ = {2, 4, 8} ⊆ [8]

Alternating Permutations – p. 47

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Let deg c = 1, deg d = 2.

µ: cd-monomial of degree n − 1

Replace each c in µ with 0, each d with 10, andremove final 0. Get the characteristic vector of aset Sµ ⊆ [n − 2].

Example. n = 10:

µ = cd2c2d → 0 · 10 · 10 · 0 · 0 · 10× = 01010001,

the characteristic vector of Sµ = {2, 4, 8} ⊆ [8]

Alternating Permutations – p. 47

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cd-index coefficients

Recall: w = a1a2 · · · an ∈ Sn is a simsunpermutation if the subsequence with elements1, 2, . . . , k has no double descents, 1 ≤ k ≤ n.

Theorem (Simion-Sundaram, variant ofFoata-Schützenberger) The coefficient of µ inΦ(c, d) is equal to the number of simsunpermutations in Sn−1 with descent set Sµ.

Alternating Permutations – p. 48

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cd-index coefficients

Recall: w = a1a2 · · · an ∈ Sn is a simsunpermutation if the subsequence with elements1, 2, . . . , k has no double descents, 1 ≤ k ≤ n.

Theorem (Simion-Sundaram, variant ofFoata-Schützenberger) The coefficient of µ inΦ(c, d) is equal to the number of simsunpermutations in Sn−1 with descent set Sµ.

Alternating Permutations – p. 48

Page 101: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

An example

Example. Φ6 =c5+4c3d+9c2dc+9cdc2+4dc3+12cd2+10dcd+12d2c,

dcd → 10 · 0 · 10× ⇒ Sdcd = {1, 4}

The ten simsun permutations w ∈ S5 withD(w) = {1, 4}:

21354, 21453, 31254, 31452, 41253

41352, 42351, 51243, 51342, 52341,

but not 32451.

Alternating Permutations – p. 49

Page 102: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

An example

Example. Φ6 =c5+4c3d+9c2dc+9cdc2+4dc3+12cd2+10dcd+12d2c,

dcd → 10 · 0 · 10× ⇒ Sdcd = {1, 4}

The ten simsun permutations w ∈ S5 withD(w) = {1, 4}:

21354, 21453, 31254, 31452, 41253

41352, 42351, 51243, 51342, 52341,

but not 32451.

Alternating Permutations – p. 49

Page 103: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

An example

Example. Φ6 =c5+4c3d+9c2dc+9cdc2+4dc3+12cd2+10dcd+12d2c,

dcd → 10 · 0 · 10× ⇒ Sdcd = {1, 4}

The ten simsun permutations w ∈ S5 withD(w) = {1, 4}:

21354, 21453, 31254, 31452, 41253

41352, 42351, 51243, 51342, 52341,

but not 32451.

Alternating Permutations – p. 49

Page 104: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Two consequences

Theorem. (a) Φn(1, 1) = En (the number ofsimsum permutations w ∈ Sn).

(b) (Niven, de Bruijn) For all S ⊆ {1, . . . , n − 1},

βn(S) ≤ En,

with equality if and only if S = {1, 3, 5, . . . } orS = {2, 4, 6 . . . }

Alternating Permutations – p. 50

Page 105: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Two consequences

Theorem. (a) Φn(1, 1) = En (the number ofsimsum permutations w ∈ Sn).

(b) (Niven, de Bruijn) For all S ⊆ {1, . . . , n − 1},

βn(S) ≤ En,

with equality if and only if S = {1, 3, 5, . . . } orS = {2, 4, 6 . . . }

Alternating Permutations – p. 50

Page 106: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

An example

Φ5 = 1c4 + 3c2d + 5cdc + 3dc2 + 4d2

1 + 3 + 5 + 3 + 4 = 16 = E5

Alternating Permutations – p. 51

Page 107: Alternating Permutations - MIT Mathematicsrstan/transparencies/altperm-ams.pdf · Simsun permutations (2) b1b2 bk has a double descent if for some 1 < i < n, bi 1 > bi >

Alternating Permutations – p. 52