Many Looks on the Fibonacci Polynomials · Many Looks on the Fibonacci Polynomials T. Amdeberhan Tulane University IMA Workshop November 11, 2014

Many Looks on the Fibonacci Polynomials

T. AmdeberhanTulane University

IMA Workshop

November 11, 2014

Dialog through the times ...

I ”What’s enumeration?” - Socrates (400 BC)

I ”To count, or not to count.” - Shakespeare (1600’s AD)

I ”Oh, rabbits!” - Fibonacci

I ”Huh?” - Anonymous.

Dialog through the times ...

I ”What’s enumeration?” - Socrates (400 BC)

I ”To count, or not to count.” - Shakespeare (1600’s AD)

I ”Oh, rabbits!” - Fibonacci

I ”Huh?” - Anonymous.

Dialog through the times ...

I ”What’s enumeration?” - Socrates (400 BC)

I ”To count, or not to count.” - Shakespeare (1600’s AD)

I ”Oh, rabbits!” - Fibonacci

I ”Huh?” - Anonymous.

Dialog through the times ...

I ”What’s enumeration?” - Socrates (400 BC)

I ”To count, or not to count.” - Shakespeare (1600’s AD)

I ”Oh, rabbits!” - Fibonacci

I ”Huh?” - Anonymous.

Dialog through the times ...

I ”What’s enumeration?” - Socrates (400 BC)

I ”To count, or not to count.” - Shakespeare (1600’s AD)

I ”Oh, rabbits!” - Fibonacci

I ”Huh?” - Anonymous.

”partners in crime”

Mahir Can

Xi Chen

Melanie. Jensen

Victor Moll

Bruce Sagan

Zero, One, One, Two, Three, Five, Eight, ...

F0 = 0,F1 = 1 andFn = Fn−1 + Fn−2.

{0}s,t = 0, {1}s,t = 1 and

{n}s,t = s{n − 1}s,t + t{n − 2}s,t .

Examples:{2} = s, {3} = s2 + t,

{4} = s3 + 2st, {5} = s4 + 3s2t + t2.

Zero, One, One, Two, Three, Five, Eight, ...

F0 = 0,F1 = 1 andFn = Fn−1 + Fn−2.

{0}s,t = 0, {1}s,t = 1 and

{n}s,t = s{n − 1}s,t + t{n − 2}s,t .

Examples:{2} = s, {3} = s2 + t,

{4} = s3 + 2st, {5} = s4 + 3s2t + t2.

Zero, One, One, Two, Three, Five, Eight, ...

F0 = 0,F1 = 1 andFn = Fn−1 + Fn−2.

{0}s,t = 0, {1}s,t = 1 and

{n}s,t = s{n − 1}s,t + t{n − 2}s,t .

Examples:{2} = s, {3} = s2 + t,

{4} = s3 + 2st, {5} = s4 + 3s2t + t2.

analogue of Binet

Let X and Y be roots of

z2 − sz − t = 0.

Then

{n} =X n − Y n

X − Y,

〈n〉 = X n + Y n.

Tilings and explicit forms

Proposition. LetLn = {T : T a linear tiling of a row of n squares}. Then

{n + 1} =∑T∈Ln

wt T .

Proposition.

{n} =∑k≥0

(n − k − 1

k

)sn−2k−1tk .

Fibotorials and Fibonomials

{n}s,t ! =n∏

i=1

{i}s,t ,

{n

k

}s,t

={n}s,t !

{k}s,t !{n − k}s,t !.

Theorem (Sagan-Savage). Combinatorial interpretation ofFibonomials.

Proof. Tilings of a k × (n − k) rectangle containing a partition.�

Very useful

Theorem.

{m + n} = {m}{n + 1}+ t{m − 1}{n}.

Theorem (Hoggart-Long)

gcd({m}, {n}) = {gcd(m, n)}.

a cute identity

Theorem. For s, t ∈ P, we have

∞∑n=0

t(s + t − 1){n}s,t(s + t)n+1

= 1.

Proof. Generating function∑n≥0{n}zn =

z

1− sz − tz2.

enter arithmetic

The d-adic valuation

νd(n) =

{the highest power of d dividing n, n 6= 0

∞ n = 0..

Theorem. Let s, t be odd. Then

ν2({3k}) =

{1 + δE (k) · (ν2({6})− 2), t = 1 mod 4

ν2(k{3}) t = 3 mod 4.

enter arithmetic

The d-adic valuation

νd(n) =

{the highest power of d dividing n, n 6= 0

∞ n = 0..

Theorem. Let s, t be odd. Then

ν2({3k}) =

{1 + δE (k) · (ν2({6})− 2), t = 1 mod 4

ν2(k{3}) t = 3 mod 4.

... more generally

Theorem. Let d ≥ 2 be an integer. Then,

νd({n}d ,−1) = δE (n) · νd(dn/2).

Conjecture. Let s ≥ 2 and d ≥ 3 odd. Then s∗, d∗ such that

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

Recently proved by S. Park.

Note: {n}`,−1 linked to Lecture Hall Partitions.

... more generally

Theorem. Let d ≥ 2 be an integer. Then,

νd({n}d ,−1) = δE (n) · νd(dn/2).

Conjecture. Let s ≥ 2 and d ≥ 3 odd. Then s∗, d∗ such that

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

Recently proved by S. Park.

Note: {n}`,−1 linked to Lecture Hall Partitions.

... more generally

Theorem. Let d ≥ 2 be an integer. Then,

νd({n}d ,−1) = δE (n) · νd(dn/2).

Conjecture. Let s ≥ 2 and d ≥ 3 odd. Then s∗, d∗ such that

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

Recently proved by S. Park.

Note: {n}`,−1 linked to Lecture Hall Partitions.

... more generally

Theorem. Let d ≥ 2 be an integer. Then,

νd({n}d ,−1) = δE (n) · νd(dn/2).

Conjecture. Let s ≥ 2 and d ≥ 3 odd. Then s∗, d∗ such that

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

Recently proved by S. Park.

Note: {n}`,−1 linked to Lecture Hall Partitions.

Generalized Lecture Hall Theorem ...

With {n} = {n}`,−1,the proof by Bousquet-Melou and Eriksson utilizes

n∏j=1

{n}+ {n − 1}+ · · ·+ {j}{j}

∈ N.

Conjecture. Parity-splits

n∏j=1

{2n}2k−1 + {2n − 2}2k−1 + · · ·+ {2j}2k−1

{2j}∈ N.

Generalized Lecture Hall Theorem ...

With {n} = {n}`,−1,the proof by Bousquet-Melou and Eriksson utilizes

n∏j=1

{n}+ {n − 1}+ · · ·+ {j}{j}

∈ N.

Conjecture. Parity-splits

n∏j=1

{2n}2k−1 + {2n − 2}2k−1 + · · ·+ {2j}2k−1

{2j}∈ N.

Euler-Cassini

log concave sequences

a2n − an−1an+1 ≥ 0 ⇔ anan+m−1 ≥ an−1an+m.

See works by Brenti, Stanley and Wilf.

Example. For ordinary Fibonacci numbers,

F 2rn − Fr(n+1)Fr(n−1) = (−1)rF 2

r .

Cigler’s q-extension: {0}(q) = 0, {1}(q) = 1 and

{n}(q) = s{n − 1}(q) + tqn−2{n − 2}(q).

Euler-Cassini

log concave sequences

a2n − an−1an+1 ≥ 0 ⇔ anan+m−1 ≥ an−1an+m.

See works by Brenti, Stanley and Wilf.

Example. For ordinary Fibonacci numbers,

F 2rn − Fr(n+1)Fr(n−1) = (−1)rF 2

r .

Cigler’s q-extension: {0}(q) = 0, {1}(q) = 1 and

{n}(q) = s{n − 1}(q) + tqn−2{n − 2}(q).

Euler-Cassini

log concave sequences

a2n − an−1an+1 ≥ 0 ⇔ anan+m−1 ≥ an−1an+m.

See works by Brenti, Stanley and Wilf.

Example. For ordinary Fibonacci numbers,

F 2rn − Fr(n+1)Fr(n−1) = (−1)rF 2

r .

Cigler’s q-extension: {0}(q) = 0, {1}(q) = 1 and

{n}(q) = s{n − 1}(q) + tqn−2{n − 2}(q).

Cassini a la Dodgson

Theorem.

{n}s,t(q) · {n}s,qt(q)−{n−1}s,qt(q) · {n+ 1}s,t(q) = (−t)n−1q(n2).

Proof. Write {n}(q) as a determinant and use

detM =NW · SE − SW · NE

CTRdetM =

NW · SE − SW · NECTR

detM =NW · SE − SW · NE

CTR. �

det

A11 A12 A13

A21 A22 A23

A31 A32 A33

=

det

(A11 A12

A21 A22

)det

(A22 A23

A32 A33

)− det

(A21 A22

A31 A32

)det

(A12 A13

A22 A23

)det(A22

)

Cassini a la Dodgson

Theorem.

{n}s,t(q) · {n}s,qt(q)−{n−1}s,qt(q) · {n+ 1}s,t(q) = (−t)n−1q(n2).

Proof. Write {n}(q) as a determinant and use

detM =NW · SE − SW · NE

CTRdetM =

NW · SE − SW · NECTR

detM =NW · SE − SW · NE

CTR. �

det

A11 A12 A13

A21 A22 A23

A31 A32 A33

=

det

(A11 A12

A21 A22

)det

(A22 A23

A32 A33

)− det

(A21 A22

A31 A32

)det

(A12 A13

A22 A23

)det(A22

)

tails of Riemann a la Cassini

Fibonacci analogue of the Riemann zeta

ζF (z) =∞∑k=1

1

F zk

.

Theorem. Let s ≥ t ≥ 1 and n, r ∈ P. Then,( ∞∑k=n

1

{rk}s,t

)−1 = {rn}s,t − {r(n − 1)}s,t − δE (r(n − 1)),

( ∞∑k=n

1

{rk}2s,1

)−1 = {rn}2s,1 − {r(n − 1)}2s,1 − δE (r(n − 1)).

Holiday-Komatsu: t = 1, r = 1.

tails of Riemann a la Cassini

Fibonacci analogue of the Riemann zeta

ζF (z) =∞∑k=1

1

F zk

.

Theorem. Let s ≥ t ≥ 1 and n, r ∈ P. Then,( ∞∑k=n

1

{rk}s,t

)−1 = {rn}s,t − {r(n − 1)}s,t − δE (r(n − 1)),

( ∞∑k=n

1

{rk}2s,1

)−1 = {rn}2s,1 − {r(n − 1)}2s,1 − δE (r(n − 1)).

Holiday-Komatsu: t = 1, r = 1.

tails of Riemann a la Cassini

Fibonacci analogue of the Riemann zeta

ζF (z) =∞∑k=1

1

F zk

.

Theorem. Let s ≥ t ≥ 1 and n, r ∈ P. Then,( ∞∑k=n

1

{rk}s,t

)−1 = {rn}s,t − {r(n − 1)}s,t − δE (r(n − 1)),

( ∞∑k=n

1

{rk}2s,1

)−1 = {rn}2s,1 − {r(n − 1)}2s,1 − δE (r(n − 1)).

Holiday-Komatsu: t = 1, r = 1.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

Notations. Ask Stanley what Cn is.

dbbb(n) = # of non-zero digits of n is base bbb.

Well-known. ν2(Cn) = d222(n + 1)− 1. Here 222 = (1, 2, 22, . . . ).

By analogy, define

C{n} =1

{n + 1}

{2n

n

}.

Ekhad noted

C{n} =

{2n − 1

n − 1

}+ t

{2n − 1

n − 2

}.

”Is there a combinatorial interpretation?” - Shapiro.

...what power of 2 divides ...

Theorem. Let s, t be of opposite parity. Then,

ν2(C{n})

{d222(n + 1)− 1, t odd

dFFF (n + 1)− 1, t even.

Theorem. Let FFF = (1, 3, 3 · 2, 3 · 22, 3 · 33, . . . ) and s, t odd.

Then,

ν2(C{n})

{dFFF (n + 1)− ν2({6})− 3

dFFF (n + 1)− 1.

... square-freeness

Theorem. Let 1 6= m divide n. Viewed as polynomials:

{m}2 does not divide {n}.

For example, {n}2 does not divide {n2}.

Introduce flat and sharp analogues

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

.

Theorem. The following are polynomials in N[s, t]:{n

k

}[

,

{n

k

}]

, C [{n}, C ]

{n}.

... square-freeness

Theorem. Let 1 6= m divide n. Viewed as polynomials:

{m}2 does not divide {n}.

For example, {n}2 does not divide {n2}.

Introduce flat and sharp analogues

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

.

Theorem. The following are polynomials in N[s, t]:{n

k

}[

,

{n

k

}]

, C [{n}, C ]

{n}.

... square-freeness

Theorem. Let 1 6= m divide n. Viewed as polynomials:

{m}2 does not divide {n}.

For example, {n}2 does not divide {n2}.

Introduce flat and sharp analogues

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

.

Theorem. The following are polynomials in N[s, t]:{n

k

}[

,

{n

k

}]

, C [{n}, C ]

{n}.

... square-freeness

Theorem. Let 1 6= m divide n. Viewed as polynomials:

{m}2 does not divide {n}.

For example, {n}2 does not divide {n2}.

Introduce flat and sharp analogues

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

.

Theorem. The following are polynomials in N[s, t]:{n

k

}[

,

{n

k

}]

, C [{n}, C ]

{n}.

SQUEEZE ... POUR ...

... and ENJOY!