From tilings by Pythagorean triangles to Dyck paths: a generalization of the Erdős-Hooley function

Transcription

1 From tilings by Pythagorean triangles to Dyck paths: a generalization of the Erdős-Hooley function José Manuel Rodríguez Caballero (joint work with J. E. Blazek) Université du Québec à Montréal rodriguez caballero.jose manuel@uqam.ca April 4, 2017

2 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.

3 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)

4 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...

5 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 product 1 m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) n=1 P n (q) q n t n. Prove that P n (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:

6 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 triangle 2. 2 Erdős, P. (1948). On the density of some sequences of integers. Bulletin of the American Mathematical Society, 54(8),

7 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 #{n Z : n Ω & 1 n N} lim = 1. n + N Euripides: So, my Pythagorean triangle having semiperimeter n almost surely will tile the plane.

8 Historical motivations (Dialogue) Pythagoras: Why you were interested in this problem? Erdős: Because in the early 1930s, Erich Bessel-Hagen 3 asked whether abundant numbers 4 have a natural density 5. 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 ). 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

9 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 multiples 6 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.

10 Historical motivations (Dialogue) Pythagoras: Is your conjecture still open? Erdős: Using so-called Erdős-Hooley function, Maier and Tenenbaum 7 (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),

11 Historical motivations (Dialogue) Erdős: The history of Erdős-Hooley function started when I was thinking about the following classical result 8, 1 lim x x n x σ(n) n where σ(n) is the sum of divisors 9 of n. = π2 6, 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 A divisor of an integer n 1, is an integer d 1 such that n = d k for some integer k 1.

12 Historical motivations (Dialogue) Erdős: In 1974, I proposed the following problem. Problem (Canad. Math. Bull, 17, 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 there is a constant c π2 6 such that 1 lim f (n) = c. x x n x Pythagoras: I see the point, σ(n,n) n provided that f (n) = σ(n,m) m. 10 Simplified and corrected version. = σ(n) n, but in general σ(n,m) m σ(n) n

13 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 ε t 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, Real quote from: Problem 218. Solution by the proposer, Canad. Math. Bull, 17, ISO 690

14 Historical motivations (Dialogue) Erdős: In 1975, Nicolas and me used the function { } σ(n, m) f (n) = max m : 1 m n, in a paper about superabundant numbers 12. Erdős, P., & Nicolas, J. L. (1975). Répartition des nombres superabondants. Bulletin de la Société Mathématique de France, 103, 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.

15 Historical motivations (Dialogue) Erdős: It 1976, Nicolas and I introduced another function F (n) := max{q t (n) : t R & t > 0}, where q t (n) := # { d : d n & 1 2 t < d t} is the number of divisors d of n satisfying 1 2 t < d t. Erdős, P., & Nicolas, J. L. (1976). Méthodes probabilistes et combinatoires en théorie des nombres. Bull. sc. math, 2,

16 Historical motivations (Dialogue) Pythagoras: I remark that both functions, f (n) and F (n) are related by the inequality where 1 2 f (n) F (n) 2, q t (n) = # {d : d n & 12 } t < d t, F (n) = max{q t (n) : t R & t > 0}, σ(n, m) := d, d n d m { σ(n, m) f (n) = max m } : 1 m n. Erdős: Yes, we proved that.

17 Historical motivations (Dialogue) Erdős: Finally, in 1979, Hooley introduced what is now called the Erdős-Hooley function 13, (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. 1, Hooley, C. (1979). On a new technique and its applications to the theory of numbers. Proceedings of the London Mathematical Society, 3(1), 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.

18 Historical motivations (Dialogue) Pythagoras: Do you know the polynomials P n (q)? m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) n=1 P n (q) q n t n 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

19 Historical motivations (Dialogue) Pythagoras: What were their motivations? the Stranger: Kassel and Reutenauer uncovered the family P n (q) of polynomials with remarkable arithmetical properties when counting the ideals of fixed finite codimension 14 n of the algebra of Laurent polynomials in two variables F q [x, y, x 1, y 1 ] over the finite field F q with q elements. Erdős: Interesting. 14 An ideal I has codimension n if the quotient F q[x, y, x 1, y 1 ]/I, seen as a vector space over F q, has dimension n.

20 Historical motivations (Dialogue) the Stranger: The ideals of codimension n of F q [x, y, x 1, y 1 ] are the F q -points of the Hilbert scheme (( ) ( )) H n := Hilb n A 1 F q \{0} A 1 F q \{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 1 Z H n /F q = (1 q n t) c n,0 n i=1 1 [(1 q n+i t) (1 q n i t)] c n,i where the coefficients c n,i are related to P n (q) via (q 1) 2 P n (q) = c n,0 q n + n ( c n,i q n+i + q n i). i=1

21 Historical motivations (Dialogue) Erdős: Stop to talk about science fiction and tell me a concrete application of the polynomials P n (q) to elementary number theory. the Stranger: Do you like the sum of divisors σ(n)? Well... P n (1) = σ(n), and there is a noncommutative version of P n (q) such that its evaluation at q = 1 is the partition function p(n). Pythagoras: Could you show me an application of P n (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 P n 2( 1) > 1.

22 Historical motivations (Dialogue) Pythagoras: Wonderful. I need one last help from you. Could you prove that the greatest coefficient of P n (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.

23 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 ( ). (1) A word belonging to the Dyck language is called a Dyck word a. a Informally, a Dyck word is just a nonempty well-formed parentheses

24 Ferrers-Dyck polynomials Definition The sign of the letters ( and ) are defined as follows, The height of a Dyck word w is sign (() := +1, (2) sign ()) := 1. (3) { j } height(w) := max sign (w i ) : 0 j k 1, (4) i=0 where w = w 0 w 1 w 2... w k 1 and each w i is a letter, with 0 i k 1, is a letter.

25 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.

26 Ferrers-Dyck polynomials Definition A strict Ferrers diagram is a vector of positive integers α = (α 0, α 1, α 2,..., α k 1 ) (5) satisfying α 0 > α 1 > α 2 >... > α k 1. The integer k is called the number of parts of α.

27 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),

28 Ferrers-Dyck polynomials Definition Consider the Dyck work w = w 0 w 1 w 2... w k 1, where each w j 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 α w := q 1 q (q 1) k 1 sign (w j ) q α j, (6) j=0 where sign(w j ) is the sign of the letter w j.

29 Ferrers-Dyck polynomials Example (15, 14, 12, 10, 9, 8, 7, 5, 4, 1) ( ( ) ( ) ( ( ) ) ) := q q q 11 + q 10 q +q 9 + 2q 8 + q 7 + 2q 6 + 3q 5 + 3q 4 + 2q 3 + q 2 + q + 1.

30 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.

31 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 where a S λ := w 0 w 1 w 2..., w k 1, (7) for all 0 i k 1. λ S := {λ s : s S}, (8) S λ S = {µ 0 < µ 1 <... < µ k 1 }, (9) { ( if µi S, w i := (10) ) if µ i λ S, a Given two sets A and B, we define A B := (A B) \ (A B).

32 Inequalities for divisors Example For S = {1, 2, 3, 6, 7} and λ = e (Euler s number), the word S λ can be computed using the inequality 1 ( < 2 ( < e ) < 3 ( < 2 e ) < 6 ( < 7 ( < 3 e ) < 6 e ) < 7 e ).

33 Inequalities for divisors Lemma (Proposition 11) Given a finite set of positive rational numbers S and a real number λ > 1: (i) S λ is a Dyck word, (ii) the height of the Dyck word S λ is the greatest value of h for which there are s 1, s 2,..., s h S satisfying the inequality s 1 < s 2 <... < s h < λ s 1. (11)

34 Kassel-Reutenauer polynomials Definition The sequence (P n (q)) n 1 of Kassel-Reutenauer polynomials is given by the following generating function a, m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) n=1 P n (q) q n t n. (12) a See: Kassel, C., & Reutenauer, C. (2015). On the zeta function of a punctual Hilbert scheme of the two-dimensional torus. arxiv preprint arxiv:

35 Kassel-Reutenauer polynomials Definition (Definition 13) The Erdős-Nicolas function F (n) is the greatest value of h such that d 1 < d 2 < d 3 <... < d h < 2 d 1, (13) for some divisors d 1, d 2, d 3,..., d h of n.

36 Kassel-Reutenauer polynomials Definition (i) Define the function g(x) := (ii) For each integer n 1, define n x x λ + n + 1. (14) n λ := (g (µ 0 ), g (µ 1 ),..., g (µ k 1 )), (15) n λ := S λ, (16) where S is the set of divisors of n and µ 0, µ 1,..., µ k 1 are the elements of S λ S given in increasing order as shown in (9).

37 Kassel-Reutenauer polynomials Lemma (Proposition 15) For each integer n 1, P n (q) = n 2 n 2. (17) q

38 Kassel-Reutenauer polynomials Example 6 2 = (12, 7, 6, 1), 6 2 = ( ( ) ), P 6 (q) = (12, 7, 6, 1) ( ( ) ) = q 10 + q 9 + q 8 + q 7 + q 6 + 2q 5 q +q 4 + q 3 + q 2 + q + 1.

39 Kassel-Reutenauer polynomials Theorem For each integer n 1, the greatest coefficient of P n (q) is F (n). Proof. By Proposition 15, P n (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 n 2 (this is a Dyck word because of Proposition 11(i)). By Proposition 11. (ii), the height of n 2 = S 2, with S being the set of divisors of n, is the greatest value of h such that there are s 1, s 2,..., s h 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 P n (q) is F (n).

40 Solution of our problem the Stranger: It follows, as a corollary, that P n (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.

41 Erdős-Nicolas function Proposition (Blazek-Caballero, 2017). For each integer n 1, the height of the Dyck path n 2 is F (n). Figure: Dyck path of = ( ( ) ( ( ) ( ) ) ( ) ). q t (n) := # {d : d n & 12 } t < d t, F (n) := max{q t (n) : t R & t > 0}, F (126) = 3.

42 Erdős-Hooley function Proposition (Blazek-Caballero, 2017). For each integer n 1, the height of the Dyck path n e is (n). Figure: Dyck path of 126 e = ( ( ) ( ) ( ( ) ( ( )( )( ) ) ( ) ) ( ) ( ) ). (n) = max u R d n u < ln d u + 1 1, (126) = 4.

43 Comparison Dyck paths of 126 2, 126 e, 126 3, from above to below.

44 Dyck paths of 7! 2, 7! e and 7! 3

45 n = 2 m = n 2 = ( ) Proposition. If n is a power of 2 then n 2 = ( ). Proof. Suppose that n = 2 m. Then Hence, n 2 = ( ). S = {1, 2, 4, 8, 16,..., 2 m }, 2 S := { 2, 4, 8, 16,..., 2 m, 2 m+1}, S 2 S = { 1 < 2 m+1}. Corollary. If n is a power of 2 then P n (q) = q2 n 1 1 q 1. Corollary. If n is a power of 2 then σ(n) = 2 n 1.

46 n 2 = ( ) = n = 2 m Proposition. If n 2 = ( ) then n is a power of 2. Proof. Suppose that n 2 = ( ). So, S 2 S = {1 < 2 n}, where S = {d 1 < d 2 <... < d m } is the set of divisors of n. On the other hand, 2 S = {2 d 1 < 2 d 2 <... < 2 d m }. So, d 2 = 2 d 1, d 3 = 2 d 2,..., d m = 2 d m 1. Therefore, n = d m = 2 m. Corollary. If P n (q) = q2 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 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

47 n 2 = ( ) n = 2 m Proposition (Blazek-Caballero, 2017). For all n 1, we have that n is a power of 2 if and only if P n (q) = q2 n 1 1 q 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 P n (1) = 2 n 1 then P n (q) = q2 n 1 1 q 1. m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) n=1 P n (q) q n t n

48 Miscellaneous results Proposition (Blazek-Caballero, 2017). Let n 2 be an even integer. We have that n is a perfect number 16 if and only if P n (q) = ( q n ) q n 1 q 1. Proposition (Blazek-Caballero, 2017). Let n 3 be an odd integer. We have that n is a prime number if and only if ( ) P n (q) = q 3 2 (n 1) q 2 (n+1) 1 q 1. m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) n=1 P n (q) q n t n 16 n is perfect if σ(n) = 2 n.

49 m 1 (1 t m ) 2 (1 q t m ) (1 q 1 t m = 1 + (q 1)2 ) Thank you n=1 P n (q) q n t n Questions & Answers ( (. ) (. ) ( (0.0) ( (. )(@.@)(. ) ) (0.0) ) (. ) (. ) )