Counting permutations by congruence class of major index H el` - - PowerPoint PPT Presentation

Counting permutations by congruence class

• f major index

H´ el` ene Barcelo (Arizona State University), Bruce Sagan (Michigan State University), and Sheila Sundaram (Bard College) www.math.msu.edu/˜sagan March 20, 2006

The major index The inversion number Shuffles The case k = ℓ

Outline

The major index The inversion number Shuffles The case k = ℓ

Let [n] = {1, 2, . . . , n}, and Sn = symmetric group of permutations of [n].

Let [n] = {1, 2, . . . , n}, and Sn = symmetric group of permutations of [n]. Then π = a1a2 . . . an ∈ Sn has major index maj π =

• ai>ai+1

i.

Let [n] = {1, 2, . . . , n}, and Sn = symmetric group of permutations of [n]. Then π = a1a2 . . . an ∈ Sn has major index maj π =

• ai>ai+1

i. Ex. If π = 1 2 3 4 5 6 2 5>3 6>1 4

Let [n] = {1, 2, . . . , n}, and Sn = symmetric group of permutations of [n]. Then π = a1a2 . . . an ∈ Sn has major index maj π =

• ai>ai+1

i. Ex. If π = 1 2 3 4 5 6 2 5>3 6>1 4 then maj π = 2 + 4 = 6.

Let [n] = {1, 2, . . . , n}, and Sn = symmetric group of permutations of [n]. Then π = a1a2 . . . an ∈ Sn has major index maj π =

• ai>ai+1

i. Ex. If π = 1 2 3 4 5 6 2 5>3 6>1 4 then maj π = 2 + 4 = 6.

Theorem

If q is an indeterminate then

• π∈Sn

qmaj π = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1).

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

• History. Implicit in GORDON (1963) and ROSELLE (1974).

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

• History. Implicit in GORDON (1963) and ROSELLE (1974).

Explicit in BARCELO, MAULE, and SUNDARAM (2002).

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

• History. Implicit in GORDON (1963) and ROSELLE (1974).

Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) (1) Prove the special case k = 1: m1,ℓ

n (i, j) = n!/ℓ

∀i, j.

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

• History. Implicit in GORDON (1963) and ROSELLE (1974).

Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) (1) Prove the special case k = 1: m1,ℓ

n (i, j) = n!/ℓ

∀i, j. (2) Use (1) to prove the case k = n: mn,ℓ

n (i, j) = n!/(nℓ)

∀i, j.

Given k, ℓ we let mk,ℓ

n

be the k × ℓ matrix with (i, j) entry mk,ℓ

n (i, j) = #{π ∈ Sn : maj π ≡ i (mod k), maj π−1 ≡ j (mod ℓ)}.

Ex. Suppose n = 5.

Theorem

If k, ℓ ≤ n and gcd(k, ℓ) = 1 then mk,ℓ

n (i, j) = n!

kℓ ∀i, j.

• History. Implicit in GORDON (1963) and ROSELLE (1974).

Explicit in BARCELO, MAULE, and SUNDARAM (2002). Combinatorial Proof (BSS) (1) Prove the special case k = 1: m1,ℓ

n (i, j) = n!/ℓ

∀i, j. (2) Use (1) to prove the case k = n: mn,ℓ

n (i, j) = n!/(nℓ)

∀i, j. (3) Use (2) and induction on n to prove the final case n > k.

Outline

The major index The inversion number Shuffles The case k = ℓ

Let imaj π = maj π−1.

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4 =

• i + 1 left of i

i.

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4 =

• i + 1 left of i

i. The inversion number of π is inv π = #{(ai, aj) : i < j and ai > aj}.

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4 =

• i + 1 left of i

i. inv π = #{21, 53, 51, 54, 31, 61, 64} = 7. The inversion number of π is inv π = #{(ai, aj) : i < j and ai > aj}.

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4 =

• i + 1 left of i

i. inv π = #{21, 53, 51, 54, 31, 61, 64} = 7. The inversion number of π is inv π = #{(ai, aj) : i < j and ai > aj}.

Theorem

• π∈Sn

qinv π = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1).

Let imaj π = maj π−1.

• Ex. If π = 1

2 3 4 5 6 2 5 3 6 1 4 then π−1 = 1 2 3 4 5 6 5>1 3 6>2 4 ∴ imaj π = 1 + 4 =

• i + 1 left of i

i. inv π = #{21, 53, 51, 54, 31, 61, 64} = 7. The inversion number of π is inv π = #{(ai, aj) : i < j and ai > aj}.

Theorem

• π∈Sn

qinv π = 1(1 + q)(1 + q + q2) · · · (1 + q + · · · + qn−1). We say maj and inv are equidistributed, i.e., have the same generating function. So are (maj, imaj) and (inv, imaj).

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j).

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j).

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j since then m(1, j) = m(2, j) = . . . = m(n, j).

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j since then m(1, j) = m(2, j) = . . . = m(n, j). Also, m(1, j) + m(2, j) + · · · + m(n, j)

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j since then m(1, j) = m(2, j) = . . . = m(n, j). Also, m(1, j) + m(2, j) + · · · + m(n, j) = m1,ℓ

n (i, j)

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j since then m(1, j) = m(2, j) = . . . = m(n, j). Also, m(1, j) + m(2, j) + · · · + m(n, j) = m1,ℓ

n (i, j) = n!

ℓ .

Proof of (2): If m1,ℓ

n (i, j) = n!

ℓ then mn,ℓ

n (i, j) = n!

nℓ (∀i, j). Let Mk,ℓ

n (i, j) = {π ∈ Sn : inv π ≡ i (mod k), imaj π ≡ j (mod ℓ)}.

∴ mk,ℓ

n (i, j) = #Mk,ℓ n (i, j).

Let M(i, j) = Mn,ℓ

n (i, j)

and m(i, j) = #M(i, j). It suffices to find a bijection M(i, j) ← → M(i + 1, j) ∀i, j since then m(1, j) = m(2, j) = . . . = m(n, j). Also, m(1, j) + m(2, j) + · · · + m(n, j) = m1,ℓ

n (i, j) = n!

ℓ . So m(i, j) = n! nℓ.

Outline

The major index The inversion number Shuffles The case k = ℓ

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ.

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ. If #π = #I = ℓ and #σ = m then the I-shuffle of π and σ is τ = π

I σ

where #τ = ℓ + m, τ|I = π, τ|[ℓ+m]−I = σ.

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ. If #π = #I = ℓ and #σ = m then the I-shuffle of π and σ is τ = π

I σ

where #τ = ℓ + m, τ|I = π, τ|[ℓ+m]−I = σ.

• Ex. If π = 3 1 4 2, I = {1, 3, 4, 6}, and σ = 6 7 5 then

π

I σ = 1

2 3 4 5 6 7 3 6 1 4 7 2 5 .

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ. If #π = #I = ℓ and #σ = m then the I-shuffle of π and σ is τ = π

I σ

where #τ = ℓ + m, τ|I = π, τ|[ℓ+m]−I = σ.

• Ex. If π = 3 1 4 2, I = {1, 3, 4, 6}, and σ = 6 7 5 then

π

I σ = 1

2 3 4 5 6 7 3 6 1 4 7 2 5 . Define f : Sn → Sn as follows. If τ ∈ Sn then write τ = π

I σ

where π ∈ Sℓ.

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ. If #π = #I = ℓ and #σ = m then the I-shuffle of π and σ is τ = π

I σ

where #τ = ℓ + m, τ|I = π, τ|[ℓ+m]−I = σ.

• Ex. If π = 3 1 4 2, I = {1, 3, 4, 6}, and σ = 6 7 5 then

π

I σ = 1

2 3 4 5 6 7 3 6 1 4 7 2 5 . Define f : Sn → Sn as follows. If τ ∈ Sn then write τ = π

I σ

where π ∈ Sℓ. Now let f(τ) = τ ′ where τ ′ = π

I+1 σ

with I + 1 = {i1 + 1, . . . , iℓ + 1} (mod n).

If τ = a1a2 . . . an is a sequence and I = {i1, i2, . . . , iℓ} then τ|I = ai1ai2 . . . aiℓ. If #π = #I = ℓ and #σ = m then the I-shuffle of π and σ is τ = π

I σ

where #τ = ℓ + m, τ|I = π, τ|[ℓ+m]−I = σ.

• Ex. If π = 3 1 4 2, I = {1, 3, 4, 6}, and σ = 6 7 5 then

π

I σ = 1

2 3 4 5 6 7 3 6 1 4 7 2 5 . Define f : Sn → Sn as follows. If τ ∈ Sn then write τ = π

I σ

where π ∈ Sℓ. Now let f(τ) = τ ′ where τ ′ = π

I+1 σ

with I + 1 = {i1 + 1, . . . , iℓ + 1} (mod n).

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 then τ = 3 1 4 2

I 6 7 5. So

τ ′ = 3 1 4 2

I+1 6 7 5 = 6 3 7 1 4 5 2.

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.
• Note. (a) f is bijective: for f −1 use I − 1.

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.
• Note. (a) f is bijective: for f −1 use I − 1.

(b) inv τ ′ − inv τ ≡ ℓ(mod n).

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.
• Note. (a) f is bijective: for f −1 use I − 1.

(b) inv τ ′ − inv τ ≡ ℓ(mod n). (c) imaj τ ′ − imaj τ = 0, ±ℓ ≡ 0 (mod ℓ).

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.
• Note. (a) f is bijective: for f −1 use I − 1.

(b) inv τ ′ − inv τ ≡ ℓ(mod n). (c) imaj τ ′ − imaj τ = 0, ±ℓ ≡ 0 (mod ℓ). Now (b) and (c) imply that f restricts to f : M(i, j) → M(i + ℓ, j).

Define f : Sn → Sn by f(τ) = τ ′ where τ = π

I σ

imples τ ′ = π

I+1 σ

for π ∈ Sℓ.

• Ex. If ℓ = 4 and τ = 3 6 1 4 7 2 5 implies τ ′ = 6 3 7 1 4 5 2.
• Note. (a) f is bijective: for f −1 use I − 1.

(b) inv τ ′ − inv τ ≡ ℓ(mod n). (c) imaj τ ′ − imaj τ = 0, ±ℓ ≡ 0 (mod ℓ). Now (b) and (c) imply that f restricts to f : M(i, j) → M(i + ℓ, j). But gcd(n, ℓ) = 1, so iterating f gives a bijection M(i, j) ← → M(i + 1, j).

Outline

The major index The inversion number Shuffles The case k = ℓ

Theorem

If gcd(k, ℓ) = 1 and d ≥ 1 with kd, ℓd ≤ n then mkd,ℓd

n

is composed of d × d blocks all equal to 1 kℓmd,d

n

.

Theorem

If gcd(k, ℓ) = 1 and d ≥ 1 with kd, ℓd ≤ n then mkd,ℓd

n

is composed of d × d blocks all equal to 1 kℓmd,d

n

. Let µ and φ be the number-theoretic M¨

• bius and Euler

functions, respectively.

Theorem

If 1 ≤ i, j ≤ n then mn,n

n (i, j) = 1

n2

• d|n

dn/d n d

• ! φ(d)2 µ
• d

gcd(i,d)

• µ
• d

gcd(j,d)

• φ
• d

gcd(i,d)

• φ
• d

gcd(j,d)

.

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n),

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n), ω = primitive nth root of unity,

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n), ω = primitive nth root of unity, χi = the G-character for ωi induced up to Sn. The right side of (∗) is the inner product χi, χj by a formula of FOULKES (1972).

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n), ω = primitive nth root of unity, χi = the G-character for ωi induced up to Sn. The right side of (∗) is the inner product χi, χj by a formula of FOULKES (1972). Let f λ

i

= # of standard Young tableaux T of shape λ with maj T ≡ i (mod n).

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n), ω = primitive nth root of unity, χi = the G-character for ωi induced up to Sn. The right side of (∗) is the inner product χi, χj by a formula of FOULKES (1972). Let f λ

i

= # of standard Young tableaux T of shape λ with maj T ≡ i (mod n). By decomposing into irreducibles, STANLEY (1999) shows χi, χj =

• λ⊢n

f λ

i f λ j

Proof of mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

Let G = group generated by the cycle (1, 2, . . . , n), ω = primitive nth root of unity, χi = the G-character for ωi induced up to Sn. The right side of (∗) is the inner product χi, χj by a formula of FOULKES (1972). Let f λ

i

= # of standard Young tableaux T of shape λ with maj T ≡ i (mod n). By decomposing into irreducibles, STANLEY (1999) shows χi, χj =

• λ⊢n

f λ

i f λ j RSK

= mn,n

n (i, j).

mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

If k = ℓ = pr for p prime, then (∗) simplifies. Let Jk,ℓ = the k × ℓ matrix of all ones.

mn,n

n (i, j) (∗)

=

1 n2

• d|n dn/d n

d

• ! φ(d)2 µ

“

d gcd(i,d)

” µ “

d gcd(j,d)

” φ “

d gcd(i,d)

” φ “

d gcd(j,d)

”.

If k = ℓ = pr for p prime, then (∗) simplifies. Let Jk,ℓ = the k × ℓ matrix of all ones. By induction on n we can prove the following result.

Theorem

For each prime p, there are sequences (qn)n≥1, (rn)n≥1, and (sn)n≥1 such that mp,p

np =

qn J1,1 rn J1,p−1 rn Jp−1,1 sn Jp−1,p−1

• .