The fractal nature of the Fibonomial triangle Xi Chen School of - PowerPoint PPT Presentation

The fractal nature of the Fibonomial triangle Xi Chen School of Mathematical Sciences, Dalian University of Technology, Dalian City, Liaoning Province, 116024, P. R. China xichen.dut@gmail.com and Bruce E. Sagan Department of Mathematics, Michigan State University, East Lansing, MI 48824-1027, USA sagan@math.msu.edu www.math.msu.edu/˜sagan October 11, 2013

Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems

Outline Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2 m rows is repeated at the left and right of the next 2 m rows with 0s in between. T 0 T T

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2 m rows is repeated at the left and right of the next 2 m rows with 0s in between. T 0 T T Theorem � n + 2 m � � n � If m ≥ 0 and 0 ≤ k , n < 2 m then ≡ ( mod 2) . k k

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2 m rows is repeated at the left and right of the next 2 m rows with 0s in between. ≡ Theorem � n + 2 m � � n � If m ≥ 0 and 0 ≤ k , n < 2 m then ≡ ( mod 2) . k k

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2 m rows is repeated at the left and right of the next 2 m rows with 0s in between. 0 ≡ 0 Theorem � n + 2 m � � n � If m ≥ 0 and 0 ≤ k , n < 2 m then ≡ ( mod 2) . k k

Here is Pascal’s triangle modulo 2: 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2 m rows is repeated at the left and right of the next 2 m rows with 0s in between. ≡ Theorem � n + 2 m � � n � If m ≥ 0 and 0 ≤ k , n < 2 m then ≡ ( mod 2) . k k

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2.

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are � n � F 1 F 2 · · · F n = . F 1 F 2 · · · F k F 1 F 2 · · · F n − k k F

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are � n � F 1 F 2 · · · F n = . F 1 F 2 · · · F k F 1 F 2 · · · F n − k k F � 6 � = F 1 F 2 · · · F 6 ( F 1 F 2 F 3 ) 2 = 1 · 1 · 2 · 3 · 5 · 8 Ex. = 60 . (1 · 1 · 2) 2 3 F

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are � n � F 1 F 2 · · · F n = . F 1 F 2 · · · F k F 1 F 2 · · · F n − k k F � 6 � = F 1 F 2 · · · F 6 ( F 1 F 2 F 3 ) 2 = 1 · 1 · 2 · 3 · 5 · 8 Ex. = 60 . (1 · 1 · 2) 2 3 F � n � The F are always integers. k

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are � n � F 1 F 2 · · · F n = . F 1 F 2 · · · F k F 1 F 2 · · · F n − k k F � 6 � = F 1 F 2 · · · F 6 ( F 1 F 2 F 3 ) 2 = 1 · 1 · 2 · 3 · 5 · 8 Ex. = 60 . (1 · 1 · 2) 2 3 F � n � The F are always integers. Modulo 2 we have: k 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1

The Fibonacci numbers are defined by F 0 = 0, F 1 = 1, and F n = F n − 1 + F n − 2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are � n � F 1 F 2 · · · F n = . F 1 F 2 · · · F k F 1 F 2 · · · F n − k k F � 6 � = F 1 F 2 · · · F 6 ( F 1 F 2 F 3 ) 2 = 1 · 1 · 2 · 3 · 5 · 8 Ex. = 60 . (1 · 1 · 2) 2 3 F � n � The F are always integers. Modulo 2 we have: k 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 Theorem (Chen and S.) If m ≥ 0 and 0 ≤ k , n < 3 · 2 m then � n + 3 · 2 m � � n � ≡ ( mod 2) . k k F F

Outline Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems

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

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

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

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 .

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) .

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively.

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively. If T begins with a domino then ι ( T ) begins with 2 monominos and is otherwise the same.

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively. If T begins with a domino then ι ( T ) begins with 2 monominos and is otherwise the same. Ex. ι ← →

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively. If T begins with a domino then ι ( T ) begins with 2 monominos and is otherwise the same. Unpaired tilings begin with a monomino followed by a domino. Ex. ι ← →

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively. If T begins with a domino then ι ( T ) begins with 2 monominos and is otherwise the same. Unpaired tilings begin with a monomino followed by a domino. Iterate, considering squares 4 and 5, etc. Ex. ι ← →

A tiling , T , of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let T n be the set of such. Ex. T 3 = Proposition For n ≥ 1 we have F n = # T n − 1 . Lemma We have F n is even if and only if n ≡ 0 ( mod 3) . Proof Sketch. We construct an involution ι on T n − 1 with 0 or 1 fixed points if n ≡ 0 ( mod 3) or not, respectively. If T begins with a domino then ι ( T ) begins with 2 monominos and is otherwise the same. Unpaired tilings begin with a monomino followed by a domino. Iterate, considering squares 4 and 5, etc. Ex. ι ← → ι

� n � S and Savage have given a combinatorial interpretation for F by k tiling k × ( n − k ) rectangles containing partitions.

� n � S and Savage have given a combinatorial interpretation for F by k tiling k × ( n − k ) rectangles containing partitions. One can extend the involution ι to such tilings.