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