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 of 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 of 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , { 4 } = s 3 + 2 st . 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , { 4 } = s 3 + 2 st . Example. { 2 } = s , If 0 ≤ k ≤ n , then the corresponding Lucanomial coefficient is � n { n } ! � = { k } ! { n − k } ! 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , { 4 } = s 3 + 2 st . Example. { 2 } = s , If 0 ≤ k ≤ n , then the corresponding Lucanomial coefficient is � n { n } ! � = { k } ! { n − k } ! 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , { 4 } = s 3 + 2 st . Example. { 2 } = s , If 0 ≤ k ≤ n , then the corresponding Lucanomial coefficient is � n { n } ! � = { k } ! { n − k } ! k where { n } ! = { 1 } { 2 } · · · { n } . Example. � 4 � { 4 } ! = 2 { 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 of integers is called a Lucas sequence . { 3 } = s 2 + t , { 4 } = s 3 + 2 st . Example. { 2 } = s , If 0 ≤ k ≤ n , then the corresponding Lucanomial coefficient is � n { n } ! � = { k } ! { n − k } ! k where { n } ! = { 1 } { 2 } · · · { n } . Example. � 4 � { 4 } ! { 2 } ! { 2 } ! = s 4 + 3 s 2 t + 2 t 2 . = 2

Remarks.

Remarks. 1. Special cases:

Remarks. 1. Special cases: 1.1 If s = t = 1 then { n } = F n , the n th Fibonacci number, and � n � is a Fibonomial coefficient. k

Remarks. 1. Special cases: 1.1 If s = t = 1 then { n } = F n , the n th Fibonacci number, and � n � is a Fibonomial coefficient. k 1.2 If s = q + 1 and t = − q then { n } = [ n ] q , the usual � n � q -analogue of n , and is a q -binomial coefficient. k

Remarks. 1. Special cases: 1.1 If s = t = 1 then { n } = F n , the n th Fibonacci number, and � n � is a Fibonomial coefficient. k 1.2 If s = q + 1 and t = − q then { n } = [ n ] q , the usual � n � q -analogue of n , and is a q -binomial coefficient. k � n � 2. It is easy to show that is a polynomial in s , t with k 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 } = F n , the n th Fibonacci number, and � n � is a Fibonomial coefficient. k 1.2 If s = q + 1 and t = − q then { n } = [ n ] q , the usual � n � q -analogue of n , and is a q -binomial coefficient. k � n � 2. It is easy to show that is a polynomial in s , t with k 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 T n be the set of such tilings.

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 :

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : 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 T n be the set of such tilings. Example. T 3 : r r r r r r r r r The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T .

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r r r r � wt T T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T .

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r r r r � s 3 wt T = + st + st T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T .

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r r r r � s 3 wt T = + st + st = { 4 } . T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T .

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r r r r � s 3 wt T = + st + st = { 4 } . T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T . Proposition � { n + 1 } = wt T . T ∈T n

A tiling , T , is a covering of a row of n squares with disjoint dominos and monominos. Let T n be the set of such tilings. Example. T 3 : r r r r r r r r r � s 3 wt T = + st + st = { 4 } . T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T . Proposition � { n + 1 } = wt T . T ∈T n 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 T n be the set of such tilings. Example. T 3 : r r r r r r r r r � s 3 wt T = + st + st = { 4 } . T ∈T 3 The weight of a tiling T is wt T = s # of monominos in T t # of dominos in T . Proposition � { n + 1 } = wt T . T ∈T n Proof The sum has the same recursion as { n + 1 } since T n = { T ∈ T n : T begins with a monomino } ⊎ { T ∈ T n : 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 } . Use { m + n } = � Proof idea T ∈T m + n − 1 wt T . �

Lemma { m + n } = { n + 1 } { m } + t { m − 1 } { n } . Use { m + n } = � Proof idea T ∈T m + n − 1 wt T . � Theorem � m + n � � m + n − 1 � � m + n − 1 � = { n + 1 } + t { m − 1 } . m − 1 n − 1 m

Lemma { m + n } = { n + 1 } { m } + t { m − 1 } { n } . Use { m + n } = � Proof idea T ∈T m + n − 1 wt T . � Theorem � m + n � � m + n − 1 � � m + n − 1 � = { n + 1 } + t { m − 1 } . m − 1 n − 1 m Proof Using the definition of the Lucanomials � m + n � m