many looks on the fibonacci polynomials · many looks on the fibonacci polynomials t. amdeberhan...

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.

”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.

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.

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.

{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

)sn−2k−1tk .

Fibotorials and Fibonomials

{n}s,t ! =n∏

{i}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

Proof. Generating function∑n≥0{n}zn =

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

νd({n}d ,−1) = δE (n) · νd(dn/2).

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

... more generally

νd({n}d ,−1) = δE (n) · νd(dn/2).

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

... more generally

νd({n}d ,−1) = δE (n) · νd(dn/2).

νd({n}s,−1) = δd∗Z(n) · νd(s∗n/d∗).

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

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

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

Euler-Cassini

F 2rn − Fr(n+1)Fr(n−1) = (−1)rF 2

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

Euler-Cassini

F 2rn − Fr(n+1)Fr(n−1) = (−1)rF 2

{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

CTR. �

A11 A12 A13

A21 A22 A23

A31 A32 A33

(A11 A12

A21 A22

(A22 A23

A32 A33

)− det

(A21 A22

A31 A32

(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

CTRdetM =

NW · SE − SW · NECTR

CTR. �

A11 A12 A13

A21 A22 A23

A31 A32 A33

(A11 A12

A21 A22

(A22 A23

A32 A33

)− det

(A21 A22

A31 A32

(A12 A13

A22 A23

)det(A22

tails of Riemann a la Cassini

Fibonacci analogue of the Riemann zeta

ζF (z) =∞∑k=1

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

{rk}s,t

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

( ∞∑k=n

{rk}2s,1

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

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

ζF (z) =∞∑k=1

{rk}s,t

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

( ∞∑k=n

{rk}2s,1

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

ζF (z) =∞∑k=1

{rk}s,t

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

( ∞∑k=n

{rk}2s,1

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

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

”Is there a combinatorial interpretation?” - Shapiro.

enter Catalanomials

By analogy, define

C{n} =1

{n + 1}

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

enter Catalanomials

By analogy, define

C{n} =1

{n + 1}

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

enter Catalanomials

By analogy, define

C{n} =1

{n + 1}

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

enter Catalanomials

By analogy, define

C{n} =1

{n + 1}

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

enter Catalanomials

By analogy, define

C{n} =1

{n + 1}

Ekhad noted

C{n} =

{2n − 1

n − 1

{2n − 1

n − 2

...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.

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

, C [{n}, C ]

... square-freeness

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

, C [{n}, C ]

... square-freeness

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

, C [{n}, C ]

... square-freeness

{n}[ = {p1} · · · {pr}, {n}] ={n}{n}[

, C [{n}, C ]

SQUEEZE ... POUR ...

... and ENJOY!

many looks on the fibonacci polynomials · many looks on the fibonacci polynomials t. amdeberhan...

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