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Maximal D-Avoiding Subsets of Z Maximal D-avoiding subsets of Z Nathan Ramesh Mentor: Christian Gaetz PRIMES Conference May 19, 2018 Example I f1; 4; 5; 8; 11g is f2; 5g-avoiding Example I f1; 3; 5; 7; · · · g is f1; 5g-avoiding Introduction I D is a finite set of positive integers I S ⊂ N called D-avoiding if there do not exist x; y 2 S such that x − y 2 D Example I f1; 3; 5; 7; · · · g is f1; 5g-avoiding Introduction I D is a finite set of positive integers I S ⊂ N called D-avoiding if there do not exist x; y 2 S such that x − y 2 D Example I f1; 4; 5; 8; 11g is f2; 5g-avoiding Introduction I D is a finite set of positive integers I S ⊂ N called D-avoiding if there do not exist x; y 2 S such that x − y 2 D Example I f1; 4; 5; 8; 11g is f2; 5g-avoiding Example I f1; 3; 5; 7; · · · g is f1; 5g-avoiding I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " I If Aq > Bq when q = 1 − "; we write A B Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 I =) A B Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I If Aq > Bq when q = 1 − "; we write A B Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 I =) A B Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 I =) A B Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " I If Aq > Bq when q = 1 − "; we write A B 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 I =) A B Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " I If Aq > Bq when q = 1 − "; we write A B Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g I =) A B Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " I If Aq > Bq when q = 1 − "; we write A B Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 Subset Generating Function I for a subset S; define the generating function X n Sq = q n2S I to compare the size of two subsets A and B; compare the − germs of Aq and Bq as q ! 1 ; i. e. when q = 1 − " I If Aq > Bq when q = 1 − "; we write A B Example I A = f0; 2; 4; 6; · · · g I B = f1; 3; 5; 7; · · · g 2 4 1 I Aq = 1 + q + q + ··· = 1−q2 3 5 q I Bq = q + q + q + ··· = 1−q2 I =) A B Propp's Theorem Theorem (Propp) Every germ-maximal D-avoiding subset S of N is eventually periodic. Example For S = f0; 1; 3; 4; 6; 7; 9; 10; · · · g; 1 + q S = 1 + q + q3 + q4 + q6 + q7 + ··· = q 1 − q3 Periodicity Implies Rationality Lemma Eventual perioicity implies that the associated Sq is a rational function. Periodicity Implies Rationality Lemma Eventual perioicity implies that the associated Sq is a rational function. Example For S = f0; 1; 3; 4; 6; 7; 9; 10; · · · g; 1 + q S = 1 + q + q3 + q4 + q6 + q7 + ··· = q 1 − q3 Theorem (Extension of Propp) Every germ-maximal D-avoiding subset S of Z has rational Sq. I Conjectures I Any germ-maximal subset of Z is completely periodic. I Not true in N: I When D = f1; 4; 7g; f0; 1; 3; 6; 9; 15; 18; · · · g f0; 3; 6; 9; 12; 15; 18; · · · g I Any germ-maximal subset of Z contains 0: Our Extension: Germ Maximality in Z I generating function workaround: X jnj Sq = q n2S I Conjectures I Any germ-maximal subset of Z is completely periodic. I Not true in N: I When D = f1; 4; 7g; f0; 1; 3; 6; 9; 15; 18; · · · g f0; 3; 6; 9; 12; 15; 18; · · · g I Any germ-maximal subset of Z contains 0: Our Extension: Germ Maximality in Z I generating function workaround: X jnj Sq = q n2S Theorem (Extension of Propp) Every germ-maximal D-avoiding subset S of Z has rational Sq. Our Extension: Germ Maximality in Z I generating function workaround: X jnj Sq = q n2S Theorem (Extension of Propp) Every germ-maximal D-avoiding subset S of Z has rational Sq. I Conjectures I Any germ-maximal subset of Z is completely periodic. I Not true in N: I When D = f1; 4; 7g; f0; 1; 3; 6; 9; 15; 18; · · · g f0; 3; 6; 9; 12; 15; 18; · · · g I Any germ-maximal subset of Z contains 0: I Alternatively δ(S) = lim (1 − q) · Sq q!1− Example 1 The density of f0; 2; 4; 6; 8; · · · g is 2 : Density I Density of S defined by jS \ f0; 1; 2; ··· ; ngj δ(S) = lim n!1 n + 1 Example 1 The density of f0; 2; 4; 6; 8; · · · g is 2 : Density I Density of S defined by jS \ f0; 1; 2; ··· ; ngj δ(S) = lim n!1 n + 1 I Alternatively δ(S) = lim (1 − q) · Sq q!1− Density I Density of S defined by jS \ f0; 1; 2; ··· ; ngj δ(S) = lim n!1 n + 1 I Alternatively δ(S) = lim (1 − q) · Sq q!1− Example 1 The density of f0; 2; 4; 6; 8; · · · g is 2 : I Goal: determine µ given D Maximal Density of D-avoiding set I Maximal density of a D-avoiding set is defined by µ(D) = supfδ(S): S is D-avoidingg Maximal Density of D-avoiding set I Maximal density of a D-avoiding set is defined by µ(D) = supfδ(S): S is D-avoidingg I Goal: determine µ given D Proof. Use the following algorithm to greedily build S: 1. Put 0 2 S: 2. Put all x + d 2 S0 for all x 2 S and d 2 D: 3. Put the smallest positive integer not currently in S or S0 into S: Return to step (2). Lower Bound on µ Theorem 1 We have µ(D) ≥ jDj+1 : Lower Bound on µ Theorem 1 We have µ(D) ≥ jDj+1 : Proof. Use the following algorithm to greedily build S: 1. Put 0 2 S: 2. Put all x + d 2 S0 for all x 2 S and d 2 D: 3. Put the smallest positive integer not currently in S or S0 into S: Return to step (2). I Lonely runner number lr(D) = supfjjtDjj : t 2 Rg Conjecture (Lonely Runner) 1 The lonely runner conjecture conjectures that lr(D) ≥ jDj+1 for all D: Lonely Runner Number I jjxjj = min(dxe − x; x − bxc) I jjtDjj = mind2D jjtdjj Conjecture (Lonely Runner) 1 The lonely runner conjecture conjectures that lr(D) ≥ jDj+1 for all D: Lonely Runner Number I jjxjj = min(dxe − x; x − bxc) I jjtDjj = mind2D jjtdjj I Lonely runner number lr(D) = supfjjtDjj : t 2 Rg Lonely Runner Number I jjxjj = min(dxe − x; x − bxc) I jjtDjj = mind2D jjtdjj I Lonely runner number lr(D) = supfjjtDjj : t 2 Rg Conjecture (Lonely Runner) 1 The lonely runner conjecture conjectures that lr(D) ≥ jDj+1 for all D: I Agrees with earlier claim; we have 1 µ(D) ≥ lr(D) ≥ : jDj + 1 Conjecture (Harlambis) For jDj = 3; we have µ(D) = lr(D): Connection to Density I Cantor and Gordon proved that µ(D) ≥ lr(D) for all D: Conjecture (Harlambis) For jDj = 3; we have µ(D) = lr(D): Connection to Density I Cantor and Gordon proved that µ(D) ≥ lr(D) for all D: I Agrees with earlier claim; we have 1 µ(D) ≥ lr(D) ≥ : jDj + 1 Connection to Density I Cantor and Gordon proved that µ(D) ≥ lr(D) for all D: I Agrees with earlier claim; we have 1 µ(D) ≥ lr(D) ≥ : jDj + 1 Conjecture (Harlambis) For jDj = 3; we have µ(D) = lr(D): I Bounding µ from above in terms of lr or some other value; currently we have no way of even quickly determining a maximal upper bound on the value of µ.
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