Combinatorial interpretations of binomial coefficient analogues - - PowerPoint PPT Presentation

Combinatorial interpretations of binomial coefficient analogues related to Lucas sequences

Bruce Sagan Department of Mathematics, Michigan State U. East Lansing, MI 48824-1027, sagan@math.msu.edu www.math.msu.edu/˜sagan and Carla Savage Department of Computer Science, North Carolina State U. Raleigh, NC 27695-8206, savage@cayley.csc.ncsu.edu October 2, 2010

Lucanomials Tiling and recursions The combinatorial interpretation

Outline

Lucanomials Tiling and recursions The combinatorial interpretation

Let s, t be variables.

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2.

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t,

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t, {4} = s3 + 2st.

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t, {4} = s3 + 2st. If 0 ≤ k ≤ n, then the corresponding Lucanomial coefficient is n k

• =

{n}! {k}! {n − k}! where {n}! = {1} {2} · · · {n}.

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t, {4} = s3 + 2st. If 0 ≤ k ≤ n, then the corresponding Lucanomial coefficient is n k

• =

{n}! {k}! {n − k}! where {n}! = {1} {2} · · · {n}. Example. 4 2

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t, {4} = s3 + 2st. If 0 ≤ k ≤ n, then the corresponding Lucanomial coefficient is n k

• =

{n}! {k}! {n − k}! where {n}! = {1} {2} · · · {n}. Example. 4 2

• =

{4}! {2}! {2}!

Let s, t be variables. Define polynomials {n} in s, t by {0} = 0, {1} = 1, and {n} = s {n − 1} + t {n − 2} for n ≥ 2. If s, t are integers then the corresponding sequence

• f integers is called a Lucas sequence.
• Example. {2} = s,

{3} = s2 + t, {4} = s3 + 2st. If 0 ≤ k ≤ n, then the corresponding Lucanomial coefficient is n k

• =

{n}! {k}! {n − k}! where {n}! = {1} {2} · · · {n}. Example. 4 2

• =

{4}! {2}! {2}! = s4 + 3s2t + 2t2.

Remarks.

Remarks.

• 1. Special cases:

Remarks.

• 1. Special cases:

1.1 If s = t = 1 then {n} = Fn, the nth Fibonacci number, and n

k

• is a Fibonomial coefficient.

Remarks.

• 1. Special cases:

1.1 If s = t = 1 then {n} = Fn, the nth Fibonacci number, and n

k

• is a Fibonomial coefficient.

1.2 If s = q + 1 and t = −q then {n} = [n]q, the usual q-analogue of n, and n

k

• is a q-binomial coefficient.

Remarks.

• 1. Special cases:

1.1 If s = t = 1 then {n} = Fn, the nth Fibonacci number, and n

k

• is a Fibonomial coefficient.

1.2 If s = q + 1 and t = −q then {n} = [n]q, the usual q-analogue of n, and n

k

• is a q-binomial coefficient.
• 2. It is easy to show that

n

k

• is a polynomial in s, t with

nonnegative coefficients and we will give a simple combinatorial interpretation for them. (There are two in the corresponding paper.)

Remarks.

• 1. Special cases:

1.1 If s = t = 1 then {n} = Fn, the nth Fibonacci number, and n

k

• is a Fibonomial coefficient.

1.2 If s = q + 1 and t = −q then {n} = [n]q, the usual q-analogue of n, and n

k

• is a q-binomial coefficient.
• 2. It is easy to show that

n

k

• is a polynomial in s, t with

nonnegative coefficients and we will give a simple combinatorial interpretation for them. (There are two in the corresponding paper.)

• 3. Other, more involved, combinatorial interpretations for

Fibonomial coefficients have been given by Gessel and Viennot, as well as by Benjamin and Plott.

Outline

Lucanomials Tiling and recursions The combinatorial interpretation

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T = s3 + st + st The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T = s3 + st + st = {4} . The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T = s3 + st + st = {4} . The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

Proposition

{n + 1} =

• T∈Tn

wt T.

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T = s3 + st + st = {4} . The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

Proposition

{n + 1} =

• T∈Tn

wt T. Proof The sum has the same recursion as {n + 1}

A tiling, T, is a covering of a row of n squares with disjoint dominos and monominos. Let Tn be the set of such tilings. Example. T3 :

r r r r r r r r r

• T∈T3

wt T = s3 + st + st = {4} . The weight of a tiling T is wt T = s# of monominos in Tt# of dominos in T.

Proposition

{n + 1} =

• T∈Tn

wt T. Proof The sum has the same recursion as {n + 1} since Tn = {T ∈ Tn : T begins with a monomino} ⊎ {T ∈ Tn : T begins with a domino}.

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} .

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

• Theorem

m + n m

• = {n + 1}

m + n − 1 m − 1

• +t {m − 1}

m + n − 1 n − 1

• .

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

• Theorem

m + n m

• = {n + 1}

m + n − 1 m − 1

• +t {m − 1}

m + n − 1 n − 1

• .

Proof Using the definition of the Lucanomials m + n m

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

• Theorem

m + n m

• = {n + 1}

m + n − 1 m − 1

• +t {m − 1}

m + n − 1 n − 1

• .

Proof Using the definition of the Lucanomials m + n m

• ={m + n} {m + n − 1}!

{m}! {n}!

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

• Theorem

m + n m

• = {n + 1}

m + n − 1 m − 1

• +t {m − 1}

m + n − 1 n − 1

• .

Proof Using the definition of the Lucanomials m + n m

• ={m + n} {m + n − 1}!

{m}! {n}! ={n + 1} {m} {m + n − 1}! {m}! {n}! + t {m − 1} {n} {m + n − 1}! {m}! {n}!

Lemma

{m + n} = {n + 1} {m} + t {m − 1} {n} . Proof idea Use {m + n} =

T∈Tm+n−1 wt T.

• Theorem

m + n m

• = {n + 1}

m + n − 1 m − 1

• +t {m − 1}

m + n − 1 n − 1

• .

Proof Using the definition of the Lucanomials m + n m

• ={m + n} {m + n − 1}!

{m}! {n}! ={n + 1} {m} {m + n − 1}! {m}! {n}! + t {m − 1} {n} {m + n − 1}! {m}! {n}! ={n + 1} m + n − 1 m − 1

• + t {m − 1}

m + n − 1 n − 1

• .

Outline

Lucanomials Tiling and recursions The combinatorial interpretation

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n.

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i.

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i. Example. λ = (3, 2, 2) =

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i. A tiling of λ is a tiling of each of its parts; the set of these is denoted Tλ. Example. λ = (3, 2, 2) =

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i. A tiling of λ is a tiling of each of its parts; the set of these is denoted Tλ. Example. λ = (3, 2, 2) = In Tλ:

r r r r r r r

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i. A tiling of λ is a tiling of each of its parts; the set of these is denoted Tλ. The set of such tilings where each part begins with a domino is denoted Dλ. Example. λ = (3, 2, 2) = In Tλ:

r r r r r r r

A partition of n is a weakly decreasing sequence of positive integers λ = (λ1, λ2, . . . , λr) with

i λi = n. The λi are parts.

• Example. Partitions of 4: (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

The Ferrers diagram of λ is an array of r left-justified rows of boxes with λi boxes in row i. A tiling of λ is a tiling of each of its parts; the set of these is denoted Tλ. The set of such tilings where each part begins with a domino is denoted Dλ. Example. λ = (3, 2, 2) = In Tλ:

r r r r r r r

In Dλ:

r r r r r r r

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n.

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

and λ∗ = (3, 2).

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

and λ∗ = (3, 2).

Theorem

m + n m

• =
• λ⊆m×n
• (T,T ∗)∈Tλ×Dλ∗

wt T wt T ∗.

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

and λ∗ = (3, 2).

r r r r r r r r r r r r

∈ Tλ × Dλ∗

Theorem

m + n m

• =
• λ⊆m×n
• (T,T ∗)∈Tλ×Dλ∗

wt T wt T ∗.

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

and λ∗ = (3, 2).

r r r r r r r r r r r r

∈ Tλ × Dλ∗

Theorem

m + n m

• =
• λ⊆m×n
• (T,T ∗)∈Tλ×Dλ∗

wt T wt T ∗. Proof idea The sum has the same recursion as m+n

m

• since

λ = (λ1, . . . , λr) fits in an m × n rectangle, written λ ⊆ m × n, if r ≤ m and λ1 ≤ n. The complement is λ∗ = (λ∗

1, . . . , λ∗ s) where

the λ∗

j are the column lengths of the skew partition (m × n)/λ.

• Example. λ = (3, 2, 2) ⊆ 3 × 4

and λ∗ = (3, 2).

r r r r r r r r r r r r

∈ Tλ × Dλ∗

Theorem

m + n m

• =
• λ⊆m×n
• (T,T ∗)∈Tλ×Dλ∗

wt T wt T ∗. Proof idea The sum has the same recursion as m+n

m

• since
• λ⊆m×n

Tλ×Dλ∗ =  

λ1=n

Tλ × Dλ∗  

• 



λ∗

1=m

Tλ × Dλ∗   .

THANKS FOR LISTENING!