SLIDE 1 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
SLIDE 2
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 3
Outline
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 4
Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
SLIDE 5
Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2m rows is repeated at the left and right of the next 2m rows with 0s in between. T T T
SLIDE 6 Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2m rows is repeated at the left and right of the next 2m rows with 0s in between. T T T
Theorem
If m ≥ 0 and 0 ≤ k, n < 2m then n + 2m k
n k
SLIDE 7 Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2m rows is repeated at the left and right of the next 2m rows with 0s in between. ≡
Theorem
If m ≥ 0 and 0 ≤ k, n < 2m then n + 2m k
n k
SLIDE 8 Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2m rows is repeated at the left and right of the next 2m rows with 0s in between. ≡
Theorem
If m ≥ 0 and 0 ≤ k, n < 2m then n + 2m k
n k
SLIDE 9 Here is Pascal’s triangle modulo 2: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 It is fractal: the triangle T which is the first 2m rows is repeated at the left and right of the next 2m rows with 0s in between. ≡
Theorem
If m ≥ 0 and 0 ≤ k, n < 2m then n + 2m k
n k
SLIDE 10
The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.
SLIDE 11 The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are n k
= F1F2 · · · Fn F1F2 · · · FkF1F2 · · · Fn−k .
SLIDE 12 The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are n k
= F1F2 · · · Fn F1F2 · · · FkF1F2 · · · Fn−k . Ex. 6 3
= F1F2 · · · F6 (F1F2F3)2 = 1 · 1 · 2 · 3 · 5 · 8 (1 · 1 · 2)2 = 60.
SLIDE 13 The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are n k
= F1F2 · · · Fn F1F2 · · · FkF1F2 · · · Fn−k . Ex. 6 3
= F1F2 · · · F6 (F1F2F3)2 = 1 · 1 · 2 · 3 · 5 · 8 (1 · 1 · 2)2 = 60. The n
k
SLIDE 14 The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are n k
= F1F2 · · · Fn F1F2 · · · FkF1F2 · · · Fn−k . Ex. 6 3
= F1F2 · · · F6 (F1F2F3)2 = 1 · 1 · 2 · 3 · 5 · 8 (1 · 1 · 2)2 = 60. The n
k
- F are always integers. Modulo 2 we have:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
SLIDE 15 The Fibonacci numbers are defined by F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. For 0 ≤ k ≤ n the Fibonomials are n k
= F1F2 · · · Fn F1F2 · · · FkF1F2 · · · Fn−k . Ex. 6 3
= F1F2 · · · F6 (F1F2F3)2 = 1 · 1 · 2 · 3 · 5 · 8 (1 · 1 · 2)2 = 60. The n
k
- F are always integers. Modulo 2 we have:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Theorem (Chen and S.)
If m ≥ 0 and 0 ≤ k, n < 3 · 2m then n + 3 · 2m k
≡ n k
(mod 2).
SLIDE 16
Outline
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 17
A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos.
SLIDE 18
A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
SLIDE 19 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
SLIDE 20 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
SLIDE 21 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3).
SLIDE 22 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−1 with 0 or 1 fixed points if n ≡ 0 (mod 3) or not, respectively.
SLIDE 23 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−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.
SLIDE 24 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−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 ← →
SLIDE 25 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−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 ← →
SLIDE 26 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−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 ← →
SLIDE 27 A tiling, T, of a row of n squares is a covering of the squares with disjoint dominos and monominos. Let Tn be the set of such.
Proposition
For n ≥ 1 we have Fn = #Tn−1.
Lemma
We have Fn is even if and only if n ≡ 0 (mod 3). Proof Sketch. We construct an involution ι on Tn−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
SLIDE 28 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions.
SLIDE 29 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
SLIDE 30 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
SLIDE 31 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
To finish the proof, use this lemma and the fractal nature of n
k
SLIDE 32 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
To finish the proof, use this lemma and the fractal nature of n
k
For example, when in the third case of the lemma we have n + 3 · 2m k
≡
SLIDE 33 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
To finish the proof, use this lemma and the fractal nature of n
k
For example, when in the third case of the lemma we have n + 3 · 2m k
≡ ⌊2n/3⌋ + 2m+1 ⌊2k/3⌋
SLIDE 34 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
To finish the proof, use this lemma and the fractal nature of n
k
For example, when in the third case of the lemma we have n + 3 · 2m k
≡ ⌊2n/3⌋ + 2m+1 ⌊2k/3⌋
⌊2n/3⌋ ⌊2k/3⌋
SLIDE 35 S and Savage have given a combinatorial interpretation for n
k
tiling k × (n − k) rectangles containing partitions. One can extend the involution ι to such tilings.
Lemma
Let f (n, k) be the number of fixed points of the action of ι so that n
k
- F ≡ f (n, k) (mod 2). Then
f (n, k) = if n ≡ 0 (mod 3) and k ≡ 1 (mod 3), ⌈2n/3⌉ ⌈2k/3⌉
- if n ≡ 1 (mod 3) and k ≡ 0 (mod 3),
⌊2n/3⌋ ⌊2k/3⌋
To finish the proof, use this lemma and the fractal nature of n
k
For example, when in the third case of the lemma we have n + 3 · 2m k
≡ ⌊2n/3⌋ + 2m+1 ⌊2k/3⌋
⌊2n/3⌋ ⌊2k/3⌋
n k
(mod 2).
SLIDE 36
Outline
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 37
A base is an infinite sequence of integers b = (b0 < b1 < b2 < . . . ) with b0 = 1 and bi|bi+1 for i ≥ 0.
SLIDE 38
A base is an infinite sequence of integers b = (b0 < b1 < b2 < . . . ) with b0 = 1 and bi|bi+1 for i ≥ 0. In particular, consider p = (1, p, p2, p3, . . .) and F = (1, 3, 3 · 2, 3 · 22, . . .).
SLIDE 39
A base is an infinite sequence of integers b = (b0 < b1 < b2 < . . . ) with b0 = 1 and bi|bi+1 for i ≥ 0. In particular, consider p = (1, p, p2, p3, . . .) and F = (1, 3, 3 · 2, 3 · 22, . . .). The expansion of n ∈ Z in base b is n = n0b0 + n1b1 + n2b2 + · · · def = (n0, n1, n2, . . .)b where 0 ≤ ni < bi+1/bi for all i.
SLIDE 40 A base is an infinite sequence of integers b = (b0 < b1 < b2 < . . . ) with b0 = 1 and bi|bi+1 for i ≥ 0. In particular, consider p = (1, p, p2, p3, . . .) and F = (1, 3, 3 · 2, 3 · 22, . . .). The expansion of n ∈ Z in base b is n = n0b0 + n1b1 + n2b2 + · · · def = (n0, n1, n2, . . .)b where 0 ≤ ni < bi+1/bi for all i.
Theorem (Lucas)
Let p be prime and n = (n0, n1, . . .)p, k = (k0, k1, . . .)p. Then n k
n0 k0 n1 k1 n2 k2
SLIDE 41 A base is an infinite sequence of integers b = (b0 < b1 < b2 < . . . ) with b0 = 1 and bi|bi+1 for i ≥ 0. In particular, consider p = (1, p, p2, p3, . . .) and F = (1, 3, 3 · 2, 3 · 22, . . .). The expansion of n ∈ Z in base b is n = n0b0 + n1b1 + n2b2 + · · · def = (n0, n1, n2, . . .)b where 0 ≤ ni < bi+1/bi for all i.
Theorem (Lucas)
Let p be prime and n = (n0, n1, . . .)p, k = (k0, k1, . . .)p. Then n k
n0 k0 n1 k1 n2 k2
Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2).
SLIDE 42 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2).
SLIDE 43 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F.
SLIDE 44 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F. Also n + 3 · 2m = (n0, . . . , nm, 1)F.
SLIDE 45 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F. Also n + 3 · 2m = (n0, . . . , nm, 1)F. n + 3 · 2m k
≡
SLIDE 46 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F. Also n + 3 · 2m = (n0, . . . , nm, 1)F. n + 3 · 2m k
≡ n0 k0
· · · nm km
1
(mod 2)
SLIDE 47 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F. Also n + 3 · 2m = (n0, . . . , nm, 1)F. n + 3 · 2m k
≡ n0 k0
· · · nm km
1
(mod 2) = n0 k0
· · · nm km
SLIDE 48 Theorem (Chen and S)
Let n = (n0, n1, . . .)F and k = (k0, k1, . . .)F. Then n k
≡ n0 k0
n1 k1
n2 k2
· · · (mod 2). To finish the proof of the main theorem, if 0 ≤ n, k < 3 · 2m then n = (n0, . . . , nm)F and k = (k0, . . . , km)F. Also n + 3 · 2m = (n0, . . . , nm, 1)F. n + 3 · 2m k
≡ n0 k0
· · · nm km
1
(mod 2) = n0 k0
· · · nm km
≡ n k
(mod 2).
SLIDE 49
Outline
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 50 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
.
SLIDE 51 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
. To finish the proof of the main theorem when 0 < n, k < 3 · 2m, recall that the Fn have period three modulo two.
SLIDE 52 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
. To finish the proof of the main theorem when 0 < n, k < 3 · 2m, recall that the Fn have period three modulo two. So n + 3 · 2m k
SLIDE 53 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
. To finish the proof of the main theorem when 0 < n, k < 3 · 2m, recall that the Fn have period three modulo two. So n + 3 · 2m k
= Fn−k+3·2m+1 n + 3 · 2m − 1 k − 1
+ Fk−1 n + 3 · 2m − 1 k
SLIDE 54 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
. To finish the proof of the main theorem when 0 < n, k < 3 · 2m, recall that the Fn have period three modulo two. So n + 3 · 2m k
= Fn−k+3·2m+1 n + 3 · 2m − 1 k − 1
+ Fk−1 n + 3 · 2m − 1 k
≡ Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
(mod 2)
SLIDE 55 Proposition
We have
n k
= Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
. To finish the proof of the main theorem when 0 < n, k < 3 · 2m, recall that the Fn have period three modulo two. So n + 3 · 2m k
= Fn−k+3·2m+1 n + 3 · 2m − 1 k − 1
+ Fk−1 n + 3 · 2m − 1 k
≡ Fn−k+1 n − 1 k − 1
+ Fk−1 n − 1 k
(mod 2) = n k
.
SLIDE 56
Outline
Fractals and Fibonomials A combinatorial proof A Lucas’ congruence proof An inductive proof Open problems
SLIDE 58
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p.
SLIDE 59
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p. If T is the triangle consisting of the first pm rows then the first pm+1 rows are
1
1
1
2
2
1
2
2
. . .
SLIDE 60
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p. If T is the triangle consisting of the first pm rows then the first pm+1 rows are
1
1
1
2
2
1
2
2
. . . For p = 3, the Fibonomial triangle can be described in terms of triangles of side 4 · 3m.
SLIDE 61
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p. If T is the triangle consisting of the first pm rows then the first pm+1 rows are
1
1
1
2
2
1
2
2
. . . For p = 3, the Fibonomial triangle can be described in terms of triangles of side 4 · 3m. But the result is much more complicated and we only have an inductive proof.
SLIDE 62
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p. If T is the triangle consisting of the first pm rows then the first pm+1 rows are
1
1
1
2
2
1
2
2
. . . For p = 3, the Fibonomial triangle can be described in terms of triangles of side 4 · 3m. But the result is much more complicated and we only have an inductive proof. What can be said for other p?
SLIDE 63
- 1. Other moduli. Pascal’s triangle is fractal modulo p for any
prime p. If T is the triangle consisting of the first pm rows then the first pm+1 rows are
1
1
1
2
2
1
2
2
. . . For p = 3, the Fibonomial triangle can be described in terms of triangles of side 4 · 3m. But the result is much more complicated and we only have an inductive proof. What can be said for other p? Note that it is an open problem to determine the period of the seqence Fn, n ≥ 0, modulo p.
SLIDE 65
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3.
SLIDE 66
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3. If one considers the base T = (1, 4, 4 · 3, 4 · 32, . . .)
SLIDE 67
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3. If one considers the base T = (1, 4, 4 · 3, 4 · 32, . . .) and lets n = (n0, n1, . . .)T and k = (k0, k1, . . .)T then it appears that n k
≡ 0 (mod 3) ⇐ ⇒ n0 k0
n1 k1
n2 k2
· · · ≡ 0 (mod 3).
SLIDE 68
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3. If one considers the base T = (1, 4, 4 · 3, 4 · 32, . . .) and lets n = (n0, n1, . . .)T and k = (k0, k1, . . .)T then it appears that n k
≡ 0 (mod 3) ⇐ ⇒ n0 k0
n1 k1
n2 k2
· · · ≡ 0 (mod 3). But we do not always have equality of the Fibonomial above and the product.
SLIDE 69
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3. If one considers the base T = (1, 4, 4 · 3, 4 · 32, . . .) and lets n = (n0, n1, . . .)T and k = (k0, k1, . . .)T then it appears that n k
≡ 0 (mod 3) ⇐ ⇒ n0 k0
n1 k1
n2 k2
· · · ≡ 0 (mod 3). But we do not always have equality of the Fibonomial above and the product. Knuth and Wilf proved an analogue for Fibonomials
- f Kummer’s Theorem describing the highest power of p dividing a
binomial coefficient.
SLIDE 70
- 2. Lucas’ congruence. We were unable to find a Fibonomial
analogue of Lucas’ congruence modulo 3. If one considers the base T = (1, 4, 4 · 3, 4 · 32, . . .) and lets n = (n0, n1, . . .)T and k = (k0, k1, . . .)T then it appears that n k
≡ 0 (mod 3) ⇐ ⇒ n0 k0
n1 k1
n2 k2
· · · ≡ 0 (mod 3). But we do not always have equality of the Fibonomial above and the product. Knuth and Wilf proved an analogue for Fibonomials
- f Kummer’s Theorem describing the highest power of p dividing a
binomial coefficient. This can be used to prove the Fibonomial analogue of Lucas’ congruence modulo 2, but does not give enough information to do the same modulo 3.
SLIDE 72
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
.
SLIDE 73
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
. One can show that Cn,F ∈ Z for all n.
SLIDE 74
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
. One can show that Cn,F ∈ Z for all n. For a base b and n ∈ Z, let ζb(n) = number of nonzero digits of (n)b.
SLIDE 75
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
. One can show that Cn,F ∈ Z for all n. For a base b and n ∈ Z, let ζb(n) = number of nonzero digits of (n)b.
Theorem
The 2-adic valuation of Cn is ν2(Cn) = ζ2(n + 1) − 1.
SLIDE 76
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
. One can show that Cn,F ∈ Z for all n. For a base b and n ∈ Z, let ζb(n) = number of nonzero digits of (n)b.
Theorem
The 2-adic valuation of Cn is ν2(Cn) = ζ2(n + 1) − 1.
Theorem (Amdeberhan, Chen, Moll, S)
The 2-adic valuation of Cn,F is ν2(Cn,F) = ζF(n + 1) if n ≡ 3 or 4 (mod 6), ζF(n + 1) − 1 else.
SLIDE 77
- 3. Catalan numbers. Louis Shapiro suggested defining a
Fibonacci analogue of the Catalan numbers by Cn,F = 1 Fn+1 2n n
. One can show that Cn,F ∈ Z for all n. For a base b and n ∈ Z, let ζb(n) = number of nonzero digits of (n)b.
Theorem
The 2-adic valuation of Cn is ν2(Cn) = ζ2(n + 1) − 1.
Theorem (Amdeberhan, Chen, Moll, S)
The 2-adic valuation of Cn,F is ν2(Cn,F) = ζF(n + 1) if n ≡ 3 or 4 (mod 6), ζF(n + 1) − 1 else. Shapiro also asked for a combinatorial interpretation of Cn,F but none is known.
SLIDE 78
THANKS FOR LISTENING!
