Download From tilings by Pythagorean triangles to Dyck paths: a

From tilings by Pythagorean triangles to Dyck paths
From tilings by Pythagorean triangles to Dyck
paths: a generalization of the Erdős-Hooley
function
José Manuel Rodrı́guez Caballero
(joint work with J. E. Blazek)
Université du Québec à Montréal
rodriguez caballero.jose [email protected]
April 4, 2017
From tilings by Pythagorean triangles to Dyck paths
Division of the presentation
1
Statement of the problem and historical motivations (it will be
presented in the form of a dialogue, so that the audience can
never get bored).
2
Our contribution (it will be presented in definition-theorem
style). Only the main result will be proved, assuming the
auxiliary results (all the proofs will be presented in a future
paper).
3
Examples, conjectures and miscellaneous results.
From tilings by Pythagorean triangles to Dyck paths
Inspirational quotation
“It has been remarked, my Hermippus, that though the ancient
philosophers conveyed most of their instruction in the form of
dialogue, this method of composition has been little practised in
later ages, and seldom succeeded in the hands of those, who have
attempted it”
David Hume
A Treatise on Human Nature
(Pamphilus to Hermippus)
From tilings by Pythagorean triangles to Dyck paths
Statement of the problem
Pythagoras: A Pythagorean triangle is a right triangle with
integer side lengths.
Euripides: Suppose that I have a wire of length 2 n, where n ≥ 1
is an integer. I want to construct, using this wire, a Pythagorean
triangle which also tiles the plane. Can I do it ?
Pythagoras: Any triangle tiles the plane. So, you can do it if and
only if n is the semiperimeter of a Pythagorean triangle. So, your
problem is trivial.
Euripides: I will make it harder...
From tilings by Pythagorean triangles to Dyck paths
Statement of the problem
As usual in Euripides plays...
A dragon-drawn chariot sent by the sun god comes here with a message:
Consider the polynomials generated by the following infinite
product1
Y
m≥1
∞
2
X
(1 − t m )
Pn (q) n
2
=
1
+
(q
−
1)
t .
(1 − q t m ) (1 − q −1 t m )
qn
n=1
Prove that Pn (q) has a coefficient > 1 if and only if n is the
semiperimeter of a Pythagorean triangle.
Yours,
Pythios
1
These polynomials were introduced in the paper: Kassel, C., &
Reutenauer, C. On the zeta function of a punctual Hilbert scheme of the
two-dimensional torus. arXiv preprint arXiv:1505.07229.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: Did I miss something in the history of mathematics ?
Euripides: Don’t worry, I will help you.
As usual in Euripides plays...
Erdős: I came here in a time machine.
Pythagoras: Oh, strange man from the future, what do you
known about the semiperimeters of Pythagorean triangles ?
Erdős: In 1948 I conjectured that almost all positive integers are
the semi-perimeter of a Pythagorean triangle2 .
2
Erdős, P. (1948). On the density of some sequences of integers. Bulletin of
the American Mathematical Society, 54(8), 685-692.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: What do you means by “almost all” ?
Erdős: Let Ω be the set of semiperimeters of Pythagorean
triangles. My conjecture is that
lim
n→+∞
#{n ∈ Z :
n∈Ω &
N
1 ≤ n ≤ N}
= 1.
Euripides: So, my Pythagorean triangle having semiperimeter n
almost surely will tile the plane.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: Why you were interested in this problem ?
Erdős: Because in the early 1930s, Erich Bessel-Hagen3 asked
whether abundant numbers4 have a natural density5 .
Pythagoras: Did you solve that problem ?
3
All the historical references come from: Tenenbaum, G. (2013). Some of
Erdős’ unconventional problems in number theory, thirty-four years later. In
Erds Centennial (pp. 651-681). Springer Berlin Heidelberg.
4
A number n is abundant if the sum of its proper divisors is larger than itself.
5
Given a set of positive integers S, its natural density is the limit of
1
#{n ∈ S : n ≤ x} as x → ∞.
x
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: Davenport (1933), Chowla (1934) and Behrend (1935) all
obtained, working independently, a positive solution of this problem using
(real and complex) moments theory. I solved that problem in 1934
following an elementary approach.
Pythagoras: Any multiple of an abundant number is also an abundant
number. In the same way, any multiple of a semiperimeter of a
Pythagorean triangle is also a the semiperimeter of a Pythagorean
triangle... Does the natural density of a set of multiples6 always exists ?
Erdős: The same question was asked by Chowla, but Besicovitch (1935)
published an example of a set of multiples without natural density. In
response to Besicovitch, I proved in 1948 that the set of semiperimeters
of Pythagorean triangle has natural density, but at that time I didn’t
known its exact value. I conjectured that this density is 1.
6
A set of multiples is a set of positive integers closed for the product.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: Is your conjecture still open ?
Erdős: Using so-called Erdős-Hooley function, Maier and
Tenenbaum7 (1984) were able to prove my conjecture.
Pythagoras: Tell me more about this function. Why does this
function have your name?
7
Maier, H., & Tenenbaum, G. (1984). On the set of divisors of an integer.
Inventiones mathematicae, 76(1), 121-128.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: The history of Erdős-Hooley function started when I was
thinking about the following classical result8 ,
π2
1 X σ(n)
=
,
x→∞ x
n
6
lim
n≤x
where σ(n) is the sum of divisors9 of n.
8
Berger, A. (1883), Sur quelques applications de la fonction Γ à la théorie
des nombres. Vetenskaps-societeten i Upsala. Nova acta Regiae societatis
scientiarum upsaliensis. 1-87
9
A divisor of an integer n ≥ 1, is an integer d ≥ 1 such that n = d k for
some integer k ≥ 1.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: In 1974, I proposed the following problem.
Problem 21810 (Canad. Math. Bull, 17, 621-622. ISO 690).
Consider the function
σ(n, m)
f (n) := max
: 1≤m≤n ,
m
where σ(n, m) is the sum of divisors d of n satisfying d ≤ m. Prove that
2
there is a constant c ≥ π6 such that
1 X
f (n) = c.
x→∞ x
lim
n≤x
Pythagoras: I see the point,
provided that f (n) = σ(n,m)
m .
10
σ(n,n)
n
Simplified and corrected version.
=
σ(n)
n ,
but in general σ(n,m)
≥
m
σ(n)
n
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: Originally when setting the problem I meant that c is
finite, but when I explained my proof to Nicolas he found an error.
I then showed that in fact c = ∞.11 .
Pythagoras: Could you explain to me the main idea of you proof?
Erdős: I use an old theorem of myself [1] which say that the
density εt of the integers having a divisor in the interval [t, 2 t]
tends to zero as t → ∞.
[1] Erdős, P. (1935). On the density of some sequences of
numbers. J. London Math. Soc, 10, 120-125.
11
Real quote from: Problem 218. Solution by the proposer, Canad. Math.
Bull, 17, 621-622. ISO 690
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: In 1975, Nicolas and me used the function
σ(n, m)
f (n) = max
: 1≤m≤n ,
m
in a paper about superabundant numbers12 .
Erdős, P., & Nicolas, J. L. (1975). Répartition des nombres
superabondants. Bulletin de la Société Mathématique de France,
103, 65-90.
An integer n ≥ 1 is a superabundant number if σ(m)
< σ(n)
for all
m
n
1 ≤ m < n. These numbers were introduced by Ramanujan in 1915.
12
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: It 1976, Nicolas and I introduced another function
F (n) := max{qt (n) : t ∈ R & t > 0},
where qt (n) := # d : d|n & 12 t < d ≤ t is the number of
divisors d of n satisfying 12 t < d ≤ t.
Erdős, P., & Nicolas, J. L. (1976). Méthodes probabilistes et
combinatoires en théorie des nombres. Bull. sc. math, 2, 301-320.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: I remark that both functions, f (n) and F (n) are related by
the inequality
f (n)
1
≤
≤ 2,
2
F (n)
where
qt (n)
=
F (n)
=
σ(n, m)
:=
1
# d : d|n &
t<d ≤t ,
2
max{qt (n) : t ∈ R & t > 0},
X
d,
d|n
d ≤m
f (n)
=
max
Erdős: Yes, we proved that.
σ(n, m)
:
m
1≤m≤n .
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: Finally, in 1979, Hooley introduced what is now called the
Erdős-Hooley function13 ,
X
1,
∆(n) = max
u∈R
d|n
u < ln d ≤ u + 1
which has applications to Waring-type problems, Diophantine
approximation, and Chebyshev’s problem on the greatest prime factor of
polynomial sequences.
Hooley, C. (1979). On a new technique and its applications to the theory
of numbers. Proceedings of the London Mathematical Society, 3(1),
115-151.
13
The natural logarithm appears in this setting because of Hardy-Ramanujan
theorem, which states that the normal order of the number the number of
distinct prime factors of a number n is ln ln n.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: Do you know the polynomials Pn (q) ?
Y
m≥1
∞
X
(1 − t m )2
Pn (q) n
2
= 1 + (q − 1)
t
m
−1
m
(1 − q t ) (1 − q t )
qn
n=1
Erdős: I can’t tell you anything, I am sorry but I don’t know these
polynomials. It seem to me that they belong to XXI century
mathematics.
As in later Platonic Historical motivations (Dialogue)s, a
mysterious character appeared...
the Stranger: I know these polynomials, they were introduced by
Kassel and Reutenauer in 2015...
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: What were their motivations ?
the Stranger: Kassel and Reutenauer uncovered the family Pn (q)
of polynomials with remarkable arithmetical properties when
counting the ideals of fixed finite codimension14 n of the algebra of
Laurent polynomials in two variables Fq [x, y , x −1 , y −1 ] over the
finite field Fq with q elements.
Erdős: Interesting.
An ideal I has codimension n if the quotient Fq [x, y , x −1 , y −1 ]/I , seen as a
vector space over Fq , has dimension n.
14
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
the Stranger: The ideals of codimension n of Fq [x, y , x −1 , y −1 ]
are the Fq -points of the Hilbert scheme
H n := Hilbn A1Fq \{0} × A1Fq \{0}
of n points on the two-dimensional torus (i.e., of the affine plane
minus two distinct straight lines). Kassel and Reutenauer showed
that the local zeta function of H n is
ZH n /Fq =
n
Y
1
1
c
(1 − q n t)cn,0
[(1 − q n+i t) (1 − q n−i t)] n,i
i=1
where the coefficients cn,i are related to Pn (q) via
2
n
(q − 1) Pn (q) = cn,0 q +
n
X
i=1
cn,i q n+i + q n−i .
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Erdős: Stop to talk about science fiction and tell me a concrete
application of the polynomials Pn (q) to elementary number theory.
the Stranger: Do you like the sum of divisors σ(n) ? Well...
Pn (1) = σ(n),
and there is a noncommutative version of Pn (q) such that its evaluation
at q = 1 is the partition function p(n).
Pythagoras: Could you show me an application of Pn (q) to Pythagorean
triangles ?
the Stranger: Yes. It follows from Kassel-Reutenauer’s results that n is
a the hypotenuse of a Pythagorean triangle if and only if
Pn2 (−1) > 1.
From tilings by Pythagorean triangles to Dyck paths
Historical motivations (Dialogue)
Pythagoras: Wonderful. I need one last help from you. Could you
prove that the greatest coefficient of Pn (q) is > 1 if and only if n
is the semiperimeter of a Pythagorean triangle.
the Stranger: Of course, but I need to introduce some definitions
and auxiliary results. I will follows the approach presented in the
following paper:
Caballero, J. M. R., & Blazek, J. E. (2017) Dyck words and the
set of divisors of an integer, (to appears).
Pythagoras: No problem, take the whole screen if you want.
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Definition
The Dyck language D is defined by the grammar consisting of two
terminal symbols ( and ), one nonterminal symbol S and the
production rules
S → S S,
S → ( S ),
S → ( ).
A word belonging to the Dyck language is called a Dyck worda .
a
Informally, a Dyck word is just a nonempty well-formed parentheses
(1)
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Definition
The sign of the letters ( and ) are defined as follows,
sign (() := +1,
(2)
sign ()) := −1.
(3)
The height of a Dyck word w is
height(w ) := max
( j
X
)
sign (wi ) :
0≤j ≤k −1 ,
i=0
where w = w0 w1 w2 ... wk−1 and each wi is a letter, with
0 ≤ i ≤ k − 1, is a letter.
(4)
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Example
The expression w = ( ( ) ( ) ( ( ) ) ) is an example of Dyck word.
Its corresponding Dyck path is shown in the following picture. The
height of this Dyck word is 3.
Figure: Dyck path of length 10.
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Definition
A strict Ferrers diagram is a vector of positive integers
α = (α0 , α1 , α2 , ..., αk−1 )
(5)
satisfying α0 > α1 > α2 > ... > αk−1 . The integer k is called the
number of parts of α.
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Example
The strict Ferrers diagram α = (15, 14, 12, 10, 9, 8, 7, 5, 4, 1) can
be represented as follows (using French notation),
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Definition
Consider the Dyck work w = w0 w1 w2 ... wk−1 , where each wj is a
letter (a parenthesis), and a strict Ferrers diagram α having exactly
k parts as shown in (5). A Ferrers-Dyck polynomial is defined as
k−1
X
α 1
:=
sign (wj ) q αj ,
w q
(q
−
1)
q
j=0
where sign(wj ) is the sign of the letter wj .
(6)
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Example
(15, 14, 12, 10, 9, 8, 7, 5, 4, 1)
(()()(()))
:= q 13 + 2q 12 + 2q 11 + q 10
q
+q 9 + 2q 8 + q 7 + 2q 6 + 3q 5 + 3q 4 + 2q 3 + q 2 + q + 1.
From tilings by Pythagorean triangles to Dyck paths
Ferrers-Dyck polynomials
Lemma (Lemma 8)
Let k ≥ 2 be an even integer. Given a Dyck word w of length k
and a strict Ferrers diagram α having exactly k parts,
(i) the expression (6) defines a polynomial whose coefficients are
non-negative integers,
(ii) the greatest coefficient of the polynomial (6) is the height of
the Dyck word w .
From tilings by Pythagorean triangles to Dyck paths
Inequalities for divisors
Definition
Given a finite set of positive rational numbers S and a real number
λ > 1, the λ-class of S is the following word
hhSiiλ := w0 w1 w2 ..., wk−1 ,
(7)
wherea
S
i
λS
:= {λ s :
λS
{µ0 < µ1 < ... < µk−1 } ,
( if µi ∈ S,
:=
) if µi ∈ λ S,
wi
s ∈ S} ,
=
for all 0 ≤ i ≤ k − 1.
a
Given two sets A and B, we define A
a
B := (A ∪ B) \ (A ∩ B).
(8)
(9)
(10)
From tilings by Pythagorean triangles to Dyck paths
Inequalities for divisors
Example
For S = {1, 2, 3, 6, 7} and λ = e (Euler’s number), the word hhSiiλ
can be computed using the inequality
1( < 2( < e) < 3( < 2 e) < 6( < 7( < 3 e) < 6 e) < 7 e) .
From tilings by Pythagorean triangles to Dyck paths
Inequalities for divisors
Lemma (Proposition 11)
Given a finite set of positive rational numbers S and a real number
λ > 1:
(i) hhSiiλ is a Dyck word,
(ii) the height of the Dyck word hhSiiλ is the greatest value of h for
which there are s1 , s2 , ..., sh ∈ S satisfying the inequality
s1 < s2 < ... < sh < λ s1 .
(11)
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Definition
The sequence (Pn (q))n≥1 of Kassel-Reutenauer polynomials is
given by the following generating functiona ,
Y
m≥1
∞
X
(1 − t m )2
Pn (q) n
2
=
1
+
(q
−
1)
t . (12)
m
−1
m
(1 − q t ) (1 − q t )
qn
n=1
a
See: Kassel, C., & Reutenauer, C. (2015). On the zeta function of a
punctual Hilbert scheme of the two-dimensional torus. arXiv preprint
arXiv:1505.07229.
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Definition (Definition 13)
The Erdős-Nicolas function F (n) is the greatest value of h such
that
d1 < d2 < d3 < ... < dh < 2 d1 ,
(13)
for some divisors d1 , d2 , d3 , ..., dh of n.
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Definition
(i) Define the function
g (x) :=
jn
−
x
(ii) For each integer n ≥ 1, define
xk
+ n + 1.
λ
JnKλ := (g (µ0 ) , g (µ1 ) , ..., g (µk−1 )) ,
hhniiλ := hhSiiλ ,
where S is theaset of divisors of n and µ0 , µ1 , ..., µk−1 are the
elements of S λ S given in increasing order as shown in (9).
(14)
(15)
(16)
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Lemma (Proposition 15)
For each integer n ≥ 1,
JnK2
Pn (q) = hhnii2
.
q
(17)
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Example
J6K2 = (12, 7, 6, 1) ,
hh6ii2 = ( ( ) ),
(12, 7, 6, 1)
P6 (q) = (())
4
3
= q 10 + q 9 + q 8 + q 7 + q 6 + 2q 5
q
2
+q + q + q + q + 1.
From tilings by Pythagorean triangles to Dyck paths
Kassel-Reutenauer polynomials
Theorem
For each integer n ≥ 1, the greatest coefficient of Pn (q) is F (n).
Proof.
By Proposition 15, Pn (q) can be expressed as a Ferrers-Dyck
polynomial given by (17). By Lemma 8, the greater coefficient of
the Ferrers-Dyck polynomial in (17) is the height of the Dyck work
hhnii2 (this is a Dyck word because of Proposition 11(i)). By
Proposition 11. (ii), the height of hhnii2 = hhSii2 , with S being the
set of divisors of n, is the greatest value of h such that there are
s1 , s2 , ..., sh ∈ S satisfying the inequality (11) for λ = 2. The
greatest value of such h is F (n) by Definition 13. Therefore, the
greatest coefficient of Pn (q) is F (n).
From tilings by Pythagorean triangles to Dyck paths
Solution of our problem
the Stranger: It follows, as a corollary, that Pn (q) has a
coefficient > 1 if and only if F (n) > 1, i.e. if n is the
semiperimeter of a Pythagorean triangle.
Pythios: Ok, I need some time to digest the information. Could
you show me more examples ?
the Stranger: Yes, surely.
From tilings by Pythagorean triangles to Dyck paths
Erdős-Nicolas function
Proposition (Blazek-Caballero, 2017). For each integer n ≥ 1,
the height of the Dyck path hhnii2 is F (n).
Figure: Dyck path of hh126ii2 = ( ( ) ( ( ) ( ) ) ( ) ).
1
qt (n) := # d : d|n &
t<d ≤t ,
2
F (n) := max{qt (n) : t ∈ R & t > 0},
F (126)
=
3.
From tilings by Pythagorean triangles to Dyck paths
Erdős-Hooley function
Proposition (Blazek-Caballero, 2017). For each integer n ≥ 1,
the height of the Dyck path hhniie is ∆(n).
Figure: Dyck path of hh126iie = ( ( ) ( ) ( ( ) ( ( )( )( ) ) ( ) ) ( ) ( ) ).
∆(n) = max
u∈R
X
d|n
u < ln d ≤ u + 1
1,
∆(126) = 4.
From tilings by Pythagorean triangles to Dyck paths
Comparison
Dyck paths of hh126ii2 , hh126iie , hh126ii3 , hh126ii4 from above to below.
From tilings by Pythagorean triangles to Dyck paths
Dyck paths of hh7!ii2 , hh7!iie and hh7!ii3
From tilings by Pythagorean triangles to Dyck paths
n = 2m
hhnii2 = ( )
=⇒
Proposition. If n is a power of 2 then hhnii2 = ( ).
Proof. Suppose that n = 2m . Then
S
2S
S
i
2S
{1, 2, 4, 8, 16, ..., 2m } ,
:= 2, 4, 8, 16, ..., 2m , 2m+1 ,
=
1 < 2m+1 .
=
Hence, hhnii2 = ( ).
Corollary. If n is a power of 2 then Pn (q) =
q 2 n−1 −1
q−1 .
Corollary. If n is a power of 2 then σ(n) = 2 n − 1.
From tilings by Pythagorean triangles to Dyck paths
hhnii2 = ( )
=⇒
n = 2m
Proposition. If hhnii2 = ( ) then n is a power
of 2.
a
Proof. Suppose that hhnii2 = ( ). So, S 2 S = {1 < 2 n}, where
S = {d1 < d2 < ... < dm } is the set of divisors of n. On the other
hand, 2 S = {2 d1 < 2 d2 < ... < 2 dm }. So,
d2 = 2 d1 , d3 = 2 d2 , ..., dm = 2 dm−1 . Therefore,
n = dm = 2m .
Corollary. If Pn (q) =
q 2 n−1 −1
q−1
then n is a power of 2.
Almost perfect numbers’ conjecture.15 . If σ(n) = 2 n − 1 then
n is a power of 2.
15
Guy, R. K. “Almost Perfect, Quasi-Perfect, Pseudoperfect, Harmonic,
Weird, Multiperfect and Hyperperfect Numbers.” §B2 in Unsolved Problems in
Number Theory, 2nd ed. New York: Springer-Verlag, pp. 16 and 45-53, 1994
From tilings by Pythagorean triangles to Dyck paths
hhnii2 = ( )
⇐⇒
n = 2m
Proposition (Blazek-Caballero, 2017). For all n ≥ 1, we have that n is
2 n−1
a power of 2 if and only if Pn (q) = q q−1−1 .
Almost perfect numbers’ conjecture. For all n ≥ 1, we have that n is
a power of 2 if and only if σ(n) = 2 n − 1.
Using our results, the almost perfect numbers’ conjecture can be derived
from our following conjecture.
Conjecture (Blazek-Caballero, 2017). If Pn (1) = 2 n − 1 then
2 n−1
Pn (q) = q q−1−1 .
Y
m≥1
∞
2
X
(1 − t m )
Pn (q) n
2
=
1
+
(q
−
1)
t
m
−1
m
(1 − q t ) (1 − q t )
qn
n=1
From tilings by Pythagorean triangles to Dyck paths
Miscellaneous results
Proposition (Blazek-Caballero, 2017). Let n ≥ 2 be an even
integer. We have that n is a perfect number16 if and only if
n −1
Pn (q) = q n−1 + 1 qq−1
.
Proposition (Blazek-Caballero, 2017). Let n ≥ 3 be an odd
integer. We have that n is a prime number if and only if
3
1 (n+1)
2
Pn (q) = q 2 (n−1) + 1 q q−1 −1 .
Y
m≥1
16
∞
X
(1 − t m )2
Pn (q) n
2
= 1 + (q − 1)
t
m
−1
m
(1 − q t ) (1 − q t )
qn
n is perfect if σ(n) = 2 n.
n=1
From tilings by Pythagorean triangles to Dyck paths
Y
m≥1
∞
X
Pn (q) n
(1 − t m )2
2
=
1
+
(q
−
1)
t
(1 − q t m ) (1 − q −1 t m )
qn
n=1
Questions & Answers
Thank you
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