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Alternating Permutations Richard P. Stanley M.I.T. Alternating Permutations p. 1 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 alternating if

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SLIDE 1

Alternating Permutations

Richard P. Stanley M.I.T. Alternating Permutations – p. 1

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SLIDE 2

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

Definitions

A sequence a1, a2, . . . , ak of distinct integers is alternating if a1 > a2 < a3 > a4 < · · · , and reverse alternating if a1 < a2 > a3 < a4 > · · · . Sn: symmetric group of all permutations of 1, 2, . . . , n

Alternating Permutations – p. 2

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SLIDE 4

En

En = #{w ∈ Sn : w is alternating} = #{w ∈ Sn : w is reverse alternating} via a1a2 · · · an → n + 1 − a1, n + 1 − a2, . . . , n + 1 − an

Alternating Permutations – p. 3

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SLIDE 5

En

En = #{w ∈ Sn : w is alternating} = #{w ∈ Sn : w is reverse alternating} via a1a2 · · · an → n + 1 − a1, n + 1 − a2, . . . , 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

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 En is an Euler number, E2n a secant number, E2n−1 a tangent number.

Alternating Permutations – p. 4

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SLIDE 8

Naive proof

52439

← − alternating

1826

  • −

→ alternating Alternating Permutations – p. 5

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SLIDE 9

Naive proof

52439

← − alternating

1826

  • −

→ alternating

Obtain w that is either alternating or reverse alternating.

Alternating Permutations – p. 5

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SLIDE 10

Naive proof

52439

← − alternating

1826

  • −

→ alternating

Obtain w that is either alternating or reverse alternating. ⇒ 2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

Alternating Permutations – p. 5

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SLIDE 11

Completion of proof

2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

Alternating Permutations – p. 6

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SLIDE 12

Completion of proof

2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

2

  • n≥0

En+1 xn n! = 2y′

  • n≥0

n

  • k=0

n k

  • EkEn−k
  • xn

n! = 1 + y2

Alternating Permutations – p. 6

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SLIDE 13

Completion of proof

2En+1 =

n

  • k=0

n k

  • EkEn−k, n ≥ 1

2

  • n≥0

En+1 xn n! = 2y′

  • n≥0

n

  • k=0

n k

  • EkEn−k
  • xn

n! = 1 + y2 ⇒ 2y′ = 1 + y2, etc.

Alternating Permutations – p. 6

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SLIDE 14

A generalization

∃ more sophisticated approaches. E.g., let fk(n) = #{w ∈ Sn : w(r) < w(r + 1) ⇔ k|r} f2(n) = En

Alternating Permutations – p. 7

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SLIDE 15

A generalization

∃ more sophisticated approaches. E.g., let fk(n) = #{w ∈ Sn : w(r) < w(r + 1) ⇔ k|r} f2(n) = En Theorem.

  • k≥0

fk(kn) xkn (kn)! =

  • n≥0

(−1)n xkn (kn)! −1 .

Alternating Permutations – p. 7

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SLIDE 16

Combinatorial trigonometry

Define sec x and tan x by sec x + tan x =

  • n≥0

En xn n! .

Alternating Permutations – p. 8

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SLIDE 17

Combinatorial trigonometry

Define sec x and tan x by sec x + tan x =

  • n≥0

En xn n! .

  • Exercise. sec2 x = 1 + tan2 x

Alternating Permutations – p. 8

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SLIDE 18

Combinatorial trigonometry

Define sec x and tan x by sec x + tan x =

  • n≥0

En xn n! .

  • Exercise. sec2 x = 1 + tan2 x
  • Exercise. tan(x + y) =

tan x + tan y 1 − (tan x)(tan y)

Alternating Permutations – p. 8

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SLIDE 19

Some occurences 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. 9

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SLIDE 20

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. 10

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SLIDE 21

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. 10

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SLIDE 22

Simsun permutations

  • Definition. A permutation w = a1 · · · an ∈ Sn is

simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk

  • f w to 1, 2, . . . , k has no double descents

bi−1 > bi > bi+1.

Alternating Permutations – p. 11

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SLIDE 23

Simsun permutations

  • Definition. A permutation w = a1 · · · an ∈ Sn is

simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk

  • f w to 1, 2, . . . , k has no double descents

bi−1 > bi > bi+1.

  • Example. 346251 is not simsun: restriction to

1, 2, 3, 4 is 3421.

Alternating Permutations – p. 11

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SLIDE 24

Simsun permutations

  • Definition. A permutation w = a1 · · · an ∈ Sn is

simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk

  • f w to 1, 2, . . . , k has no double descents

bi−1 > bi > bi+1.

  • Example. 346251 is not simsun: restriction to

1, 2, 3, 4 is 3421. Theorem (R. Simion and S. Sundaram) The number of simsun permutations w ∈ Sn is En.

Alternating Permutations – p. 11

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SLIDE 25

Chains of partitions

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

Alternating Permutations – p. 12

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SLIDE 26

Chains of partitions

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

Alternating Permutations – p. 12

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SLIDE 27

Chains of partitions

(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. 12

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SLIDE 28

Chains of partitions

(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 chains.

Alternating Permutations – p. 12

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SLIDE 29

Chains of partitions

(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 chains.

  • Theorem. The number of Sn-orbits is En

− 1. Alternating Permutations – p. 12

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SLIDE 30

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. 13

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SLIDE 31

A polytope volume

Let Pn be the convex polytope in Rn with equations xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.

Alternating Permutations – p. 14

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SLIDE 32

A polytope volume

Let Pn be the convex polytope in Rn with equations xi ≥ 0, 1 ≤ i ≤ n xi + xi+1 ≤ 1, 1 ≤ i ≤ n − 1.

  • Theorem. vol(Pn) = 1

n!En

Alternating Permutations – p. 14

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SLIDE 33

Ulam’s problem

Let E(n) be the expected length is(w) of the longest increasing subsequence of w.

Alternating Permutations – p. 15

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SLIDE 34

Ulam’s problem

Let E(n) be the expected length is(w) of the longest increasing subsequence of w. w = 5241736 ⇒ is(w) = 3

Alternating Permutations – p. 15

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SLIDE 35

Two highlights

Long story . . .

Alternating Permutations – p. 16

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SLIDE 36

Two highlights

Long story . . . Logan-Shepp, Vershik-Kerov: E(n) ∼ 2√n

Alternating Permutations – p. 16

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SLIDE 37

Two highlights

Long story . . . Logan-Shepp, Vershik-Kerov: E(n) ∼ 2√n Write isn(w) for is(w) when w ∈ Sn. Baik-Deift-Johansson: lim

n→∞ Prob

isn(w) − 2√n n1/6 ≤ t

  • = F(t),

the Tracy-Widom distribution.

Alternating Permutations – p. 16

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SLIDE 38

Alternating subsequences?

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

Alternating Permutations – p. 17

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SLIDE 39

The main lemma

MAIN LEMMA. ∀ w ∈ Sn ∃ alternating subsequence of maximal length that contains n.

Alternating Permutations – p. 18

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SLIDE 40

The main lemma

MAIN LEMMA. ∀ w ∈ Sn ∃ alternating subsequence of maximal length that contains n. ak(n) = #{w ∈ Sn : as(w) = k} bk(n) = a1(n) + a2(n) + · · · + ak(n) = #{w ∈ Sn : as(w) ≤ k}.

Alternating Permutations – p. 18

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SLIDE 41

Recurrence for ak(n)

⇒ 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. 19

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SLIDE 42

The main generating function

Define 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. 20

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SLIDE 43

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. 21

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SLIDE 44

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. 21

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SLIDE 45

Mean (expectation) of as(w)

D(n) = 1 n!

  • w∈Sn

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

Alternating Permutations – p. 22

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SLIDE 46

Mean (expectation) of as(w)

D(n) = 1 n!

  • w∈Sn

as(w), the expectation of as(n) 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. 22

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SLIDE 47

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. 23

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SLIDE 48

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. ⇒ A(n) = 4n + 1 6 , n ≥ 2

Alternating Permutations – p. 23

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SLIDE 49

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. ⇒ A(n) = 4n + 1 6 , n ≥ 2 Recall E(n) ∼ 2√n.

Alternating Permutations – p. 23

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SLIDE 50

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

Alternating Permutations – p. 24

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SLIDE 51

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

Alternating Permutations – p. 24

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SLIDE 52

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. 24

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SLIDE 53

A new distribution?

P (t) = lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • Alternating Permutations – p. 25
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SLIDE 54

A new distribution?

P (t) = lim

n→∞ Probw∈Sn

as(w) − 2n/3 √n ≤ t

  • Stanley distribution?

Alternating Permutations – p. 25

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SLIDE 55

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. 26

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SLIDE 56

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. 26

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SLIDE 57

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. 27

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SLIDE 58

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. 27

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SLIDE 59

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. 28

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SLIDE 60
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two

Alternating Permutations – p. 29

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SLIDE 61
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two Let f(n) be the number of alternating fixed-point free involutions in S2n.

Alternating Permutations – p. 29

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SLIDE 62
  • Alt. fixed-point free involutions

fixed point free involution w ∈ S2n: all cycles of length two 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. 29

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SLIDE 63

An umbral theorem

Theorem. F (x) =

  • n≥0

f(n)xn

Alternating Permutations – p. 30

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SLIDE 64

An umbral theorem

Theorem. F (x) =

  • n≥0

f(n)xn = 1 + x 1 − x (E2+1)/4

Alternating Permutations – p. 30

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SLIDE 65

An umbral theorem

Theorem. F (x) =

  • n≥0

f(n)xn = 1 + x 1 − x (E2+1)/4

  • Proof. Uses representation theory of the

symmetric group Sn. In particular, there is a “nice” representation of dimension En.

Alternating Permutations – p. 30

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SLIDE 66

Ramanujan’s Lost Notebook

Theorem (Ramanujan, Berndt, implicitly) As x → 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. 31

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SLIDE 67

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. 32

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SLIDE 68

Simple result, hard proof

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

Alternating Permutations – p. 33

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SLIDE 69

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. 33

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SLIDE 70

Special case

  • Corollary. Let p be an odd prime. Then

b(p) = 1 p

  • Ep − (−1)(p−1)/2

.

Alternating Permutations – p. 34

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SLIDE 71

Special case

  • Corollary. Let p be an odd prime. Then

b(p) = 1 p

  • Ep − (−1)(p−1)/2

. Combinatorial proof?

Alternating Permutations – p. 34

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SLIDE 72

Another such result

Recall: the expected number of fixed points of w ∈ Sn is 1.

Alternating Permutations – p. 35

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SLIDE 73

Another such result

Recall: the expected number of fixed points of w ∈ Sn is 1. J(n) = expected number of fixed points of reverse alternating w ∈ Sn

  • Theorem. J(2n) =

1 E2n (E2n − (−1)n). Similar (but more complicated) for n odd or for alternating permutations.

Alternating Permutations – p. 35

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SLIDE 74

Another such result

Recall: the expected number of fixed points of w ∈ Sn is 1. J(n) = expected number of fixed points of reverse alternating w ∈ Sn

  • Theorem. J(2n) =

1 E2n (E2n − (−1)n). Similar (but more complicated) for n odd or for alternating permutations. Combinatorial proof?

Alternating Permutations – p. 35

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SLIDE 75

Descent sets

Let w = a1a2 · · · an ∈ Sn. Descent set of w: D(w) = {i : ai > ai+1} ⊆ {1, . . . , n − 1}

Alternating Permutations – p. 36

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SLIDE 76

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. 36

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SLIDE 77

βn(S)

βn(S) = #{w ∈ Sn : D(w) = S}

Alternating Permutations – p. 37

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SLIDE 78

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. 37

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SLIDE 79

uS

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

  • a, i ∈ S

b, i ∈ S.

Alternating Permutations – p. 38

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SLIDE 80

uS

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

  • a, i ∈ S

b, i ∈ S.

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

uS = abaabba

Alternating Permutations – p. 38

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SLIDE 81

A noncommutative gen. function

Ψn(a, b) =

  • S⊆{1,...,n−1}

βn(S)uS.

Alternating Permutations – p. 39

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SLIDE 82

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 Thus Ψ3(a, b) = aa + 2ab + 2ba + bb

Alternating Permutations – p. 39

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SLIDE 83

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 Thus Ψ3(a, b) = aa + 2ab + 2ba + bb = (a + b)2 + (ab + ba)

Alternating Permutations – p. 39

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SLIDE 84

The cd-index

  • Theorem. There exists a noncommutative

polynomial Φn(c, d), called the cd-index of Sn, with nonnegative integer coefficients, such that Ψn(a, b) = Φn(a + b, ab + ba).

Alternating Permutations – p. 40

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SLIDE 85

The cd-index

  • Theorem. There exists a noncommutative

polynomial 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). Therefore Φ3(c, d) = c2 + d.

Alternating Permutations – p. 40

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SLIDE 86

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. 41

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SLIDE 87

The set Sµ

Recall: w = a1 · · · an ∈ Sn is simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk of w to 1, 2, . . . , k has no double descents bi−1 > bi > bi+1.

Alternating Permutations – p. 42

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SLIDE 88

The set Sµ

Recall: w = a1 · · · an ∈ Sn is simsun if for all 1 ≤ k ≤ n, the restriction b1 · · · bk of w to 1, 2, . . . , k has no double descents bi−1 > bi > bi+1. Given a cd-monomial µ, let c → 0, d → 10, and remove final 0. Example. µ = cd2c2d → 01010001, the characteristic vector of Sµ = {2, 4, 8}.

Alternating Permutations – p. 42

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SLIDE 89

The coefficients of Φn(c, d)

Theorem (Simion-Sundaram) Let µ be a cd-monomial of degree n − 1. The coefficient of µ in Φn(c, d) is the number of simsun permutations w ∈ Sn−1 with descent set Sµ.

Alternating Permutations – p. 43

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SLIDE 90

The case n = 4

w D(w) µ 123 ∅ c3 → 00 213 1 dc → 10 312 1 dc → 10 132 2 cd → 01 231 2 cd → 01 ⇒ Φ4(c, d) = c3 + 2dc + 2cd

Alternating Permutations – p. 44

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SLIDE 91

Expansion of cd-monomials

Let m = m(c, d) be a cd-monomial, e.g., c2d2cd. Expand m(a + b, ab + ba): (a + b)2(ab + ba)2(a + b)(ab + ba) = a9 + ba8 + · · · .

Alternating Permutations – p. 45

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SLIDE 92

Expansion of cd-monomials

Let m = m(c, d) be a cd-monomial, e.g., c2d2cd. Expand m(a + b, ab + ba): (a + b)2(ab + ba)2(a + b)(ab + ba) = a9 + ba8 + · · · . Key observation: m(a + b, ab + ba) is a sum of distinct monomials, always including ababab · · · and bababa · · · .

Alternating Permutations – p. 45

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SLIDE 93

Alternating permutations

The coefficient of ababab · · · and of bababa · · · in Ψn(a, b) is βn(2, 4, 6, . . . ) = βn(1, 3, 5, . . . ) = En.

Alternating Permutations – p. 46

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SLIDE 94

Alternating permutations

The coefficient of ababab · · · and of bababa · · · in Ψn(a, b) is βn(2, 4, 6, . . . ) = βn(1, 3, 5, . . . ) = En. Hence:

  • Theorem. (a) Φn(1, 1) = En (can interpret

coefficients combinatorially). (b) For all S ⊆ {1, . . . , n − 1}, βn(S) ≤ βn(1, 3, 5, . . . ) = En.

Alternating Permutations – p. 46

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SLIDE 95

An example

Φ5 = 1c4 + 3c2d + 5cdc + 3dc2 + 4d2 1 + 3 + 5 + 3 + 4 = 16 = E5

Alternating Permutations – p. 47

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SLIDE 96

Comments

  • NOTE. (b) is due originally to Niven and later de

Bruijn (different proofs).

Alternating Permutations – p. 48

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SLIDE 97

Comments

  • NOTE. (b) is due originally to Niven and later de

Bruijn (different proofs).

  • NOTE. Not hard to show that

βn(S) < En unless S = {1, 3, 5, . . . } or S = {2, 4, 6, . . . }.

Alternating Permutations – p. 48

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SLIDE 98

Alternating Permutations – p. 49