On The Discriminator of Lucas Sequences

Transcription

1 On The Discriminator of Lucas Sequences Bernadette Faye Ph.D Student FraZA, Bordeaux November 8, 2017

2 The Discriminator Let a = {a n } n 1 be a sequence of distinct integers. The Discriminator is defined as D a (n) = min{m : a 0,..., a n 1 are pairwise distinct modulo m}.

3 The Discriminator Example For a = {1, 2, 8, 14}, D a = 5.

4 The Discriminator Example For a = {1, 2, 8, 14}, D a = 5. For a = {1 2, 2 2, 3 2, 4 2, 5 2 }, D a = 10.

5 The Discriminator Problem Give an easy description or characterization of the Discriminator.

6 The Discriminator Problem Give an easy description or characterization of the Discriminator. Remark: D a (n) n.

7 Motivation: A problem in computer simulation Let S = {1 2, 2 2,..., n 2 } be a set of integers. Compute the square roots of its elements.

8 Motivation: A problem in computer simulation Let S = {1 2, 2 2,..., n 2 } be a set of integers. Compute the square roots of its elements. Solution: Take A be a 1 n 2 array with A(x) = x 1/2. For any value s S, A(s) is the square root of s.

9 Motivation: A problem in computer simulation Let S = {1 2, 2 2,..., n 2 } be a set of integers. Compute the square roots of its elements. Solution: Take A be a 1 n 2 array with A(x) = x 1/2. For any value s S, A(s) is the square root of s. modulo function: if 1 2, 2 2,..., n 2 are distinct modulo k, then letting A(r) = x 1/2 with r x mod k and 1 r k allow the same look up procedure to be performed.

10 Motivation: A problem in computer simulation Let S = {1 2, 2 2,..., n 2 } be a set of integers. Compute the square roots of its elements. Solution: Take A be a 1 n 2 array with A(x) = x 1/2. For any value s S, A(s) is the square root of s. modulo function: if 1 2, 2 2,..., n 2 are distinct modulo k, then letting A(r) = x 1/2 with r x mod k and 1 r k allow the same look up procedure to be performed.

11 History Theorem (Arnold, Benkoski, McCabe (1985)) If n > 4, and a = {1 2, 2 2,..., n 2 } D a (n) = min{m 2n : m = p or m = 2p with p an odd prime}.

12 History Bremser, Schumer and Washington (1990): for a cycle polynomial f = x d and d is odd, D f (n) = min{k n : f : Z/kZ Z/kZ is a permutation}

13 History Bremser, Schumer and Washington (1990): for a cycle polynomial f = x d and d is odd, D f (n) = min{k n : f : Z/kZ Z/kZ is a permutation} Moree and Mullen (1996): when f is a Dickson polynomial of degree coprime to 6.

14 On function taking only prime values Theorem (Zhi-Wei Sun, 2013) For n Z + let S(n) denote the smallest integer m > 1 such that those 2k(k 1) modulo m for k = 1,..., n are pairwise distinct. Then S(n) is the least prime greater than 2n 2.

15 On function taking only prime values Theorem (Zhi-Wei Sun, 2013) For n Z + let S(n) denote the smallest integer m > 1 such that those 2k(k 1) modulo m for k = 1,..., n are pairwise distinct. Then S(n) is the least prime greater than 2n 2. Fact S(n) is exactly the set of all prime numbers!

16 On function taking only prime values Theorem (Zhi-Wei Sun, 2013) For n Z + let S(n) denote the smallest integer m > 1 such that those 2k(k 1) modulo m for k = 1,..., n are pairwise distinct. Then S(n) is the least prime greater than 2n 2. Fact S(n) is exactly the set of all prime numbers! Remark The way to generate all primes via this theorem is simple in concept, but it has no advantage in algorithm. Nevertheless, it is of certain theoretical interest since it provides a surprising new characterization of primes.

17 Discriminator for non polynomial sequences Moree and Zumalacárregui(2016) : u(j) = 3j 5( 1) j 4, j = 1, 2, 3,... D u (n) := min{2 e, 5 f }, where e = log 2 (n) and f = log 5 (5n/4).

18 Discriminator for non polynomial sequences Moree and Zumalacárregui(2016) : u(j) = 3j 5( 1) j 4, j = 1, 2, 3,... D u (n) := min{2 e, 5 f }, where e = log 2 (n) and f = log 5 (5n/4). Ciolan and Moree(Preprint arxiv 2017): For every prime q 7, the discriminator of the family u q (j) = 3j q( 1) j+(q 1)/2, j = 1, 2, 3,... 4

19 Shallit s Conjecture In May 2016, Jeffrey Shallit posed the following conjecture. Conjecture: Given k 1, consider the recurrence with numbers determined by u k (n + 2) = (4k + 2)u k (n + 1) u k (n), u k (0) = 0, u k (1) = 1.

20 Shallit s Conjecture In May 2016, Jeffrey Shallit posed the following conjecture. Conjecture: Given k 1, consider the recurrence with numbers determined by u k (n + 2) = (4k + 2)u k (n + 1) u k (n), u k (0) = 0, u k (1) = 1. Then, for k 6, the discriminator D k (n) of the sequence u k (n) is the smallest number m n taken from the sets D k defined as.

21 Shallit s Conjecture In May 2016, Jeffrey Shallit posed the following conjecture. Conjecture: Given k 1, consider the recurrence with numbers determined by u k (n + 2) = (4k + 2)u k (n + 1) u k (n), u k (0) = 0, u k (1) = 1. Then, for k 6, the discriminator D k (n) of the sequence u k (n) is the smallest number m n taken from the sets D k defined as. D 1 = {2 i, i }. D 2 = {2 i, 3 2 i }. D 3 = {2 i 3 j }. D 4 = {2 i 5 j+1 }. D 5 = {2 i+1 3 j 5 l }. D 6 = {2 i+1 3 j 7 l }.

22 Case k = 1 Theorem (Moree, Luca, F. (2017)) Let v n be the smallest power of two such that v n n. Let w n be the smallest integer of the form 2 a 5 b satisfying 2 a 5 b 5n/3 with a, b 1. Then D 1 (n) = min{v n, w n }.

23 Case k = 1 Theorem (Moree, Luca, F. (2017)) Let v n be the smallest power of two such that v n n. Let w n be the smallest integer of the form 2 a 5 b satisfying 2 a 5 b 5n/3 with a, b 1. Then D 1 (n) = min{v n, w n }. Let M = { { m 1 : m log 5 } 1 log(6/5) } = {3, 6, 9, 12, 15,...}. log 2 log 2

24 Case k = 1 Theorem (Moree, Luca, F. (2017)) Let v n be the smallest power of two such that v n n. Let w n be the smallest integer of the form 2 a 5 b satisfying 2 a 5 b 5n/3 with a, b 1. Then D 1 (n) = min{v n, w n }. Let M = We have { { m 1 : m log 5 } 1 log(6/5) } = {3, 6, 9, 12, 15,...}. log 2 log 2 {D 1 (2), D 1 (3), D 1 (4),...} = {2 a 5 b : a 1, b M {0}}.

25 Case k = 2 Theorem (Moree, Luca, F. (2017)) Let e 0 be the smallest integer such that 2 e n and f 1 the smallest integer such that 3 2 f n. Then D 2 (n) = min{2 e, 3 2 f }.

26 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9),

27 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = and B k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9), { {m even : if p m, then p k(k + 1)} if k 2 (mod 9); {m even, 9 m : if p m, then p k(k + 1)} if k 2 (mod 9).

28 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = and B k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9), { {m even : if p m, then p k(k + 1)} if k 2 (mod 9); {m even, 9 m : if p m, then p k(k + 1)} if k 2 (mod 9). Let k > 2. We have D k (n) min{m n : m A k B k }, with equality if the interval [n, 3n/2) contains an integer m A k B k and with at most finitely many n for which strict inequality holds.

29 Main Tools The main tools used for the proof of Shallit s conjecture are: properties of a binary recurrent sequence {u n } n 0,

30 Main Tools The main tools used for the proof of Shallit s conjecture are: properties of a binary recurrent sequence {u n } n 0, order of appearance z(m) of an integer m is the sequence u n, U 0, U 1,..., U z(m),...,

31 Main Tools The main tools used for the proof of Shallit s conjecture are: properties of a binary recurrent sequence {u n } n 0, order of appearance z(m) of an integer m is the sequence u n, U 0, U 1,..., U z(m),..., U i 0 (mod m), i [0, z(m) 1] but U z(m) 0 (mod m).

32 Method of the Proof Find an interval for the D u (n) : for example if 2 e is a discriminator, then 2 e n, and, D u (n) [n, 2n].

33 Method of the Proof Find an interval for the D u (n) : for example if 2 e is a discriminator, then 2 e n, and, D u (n) [n, 2n]. Find a form for an eligible or ineligible value of the discriminator: For example, we consider a 0, a 1,..., a n 1, with a 0 = 0. For an integer k to be a discriminant, z(k) n.

34 Method of the Proof Find a form for an eligible or ineligible value of the discriminator: For example, we consider a 0, a 1,..., a n 1, with a 0 = 0. For an integer k to be a discriminant, z(k) n.

35 Method of the Proof Find a form for an eligible or ineligible value of the discriminator: For example, we consider a 0, a 1,..., a n 1, with a 0 = 0. For an integer k to be a discriminant, z(k) n. Study the congruence relation U i U j (mod k).

36 Sketch of the Proofs The congruence U i U j (mod k) Consider the sequence U 0, U 1,..., U n 1.

37 Sketch of the Proofs The congruence U i U j (mod k) Consider the sequence U 0, U 1,..., U n 1. If k is a discriminator, then U i U j (mod k) for 0 i, j n 1

38 Sketch of the Proofs The congruence U i U j (mod k) Consider the sequence U 0, U 1,..., U n 1. If k is a discriminator, then U i U j (mod k) for 0 i, j n 1 Otherwise, (i, j) [0, n 1] such that U i U j (mod k). = k U i U j.

39 Some properties of Lucas recurrent sequences We consider the Lucas sequence {u n } n 0, with u 0 = 0, u 1 = 1 and u n+2 = ru n+1 + su n for all n 0, (1) where s = 1 and r := 4k + 2 are integers. Put = r 2 4 and assume that 0.

40 Some properties of Lucas recurrent sequences We consider the Lucas sequence {u n } n 0, with u 0 = 0, u 1 = 1 and u n+2 = ru n+1 + su n for all n 0, (1) where s = 1 and r := 4k + 2 are integers. Put = r 2 4 and assume that 0. Let (α, β) be the roots of the characteristic equation x 2 rx + 1 = 0 of the binary sequence {u n } n 0, then the so-called Binet formula u n = αn β n α β holds for all n 0. (2)

41 We let {v n } n 0 for the companion Lucas sequence of {u n } n 0 given by v 0 = 2, v 1 = r and v n+2 = rv n+1 v n. Its Binet formula is v n = α n + β n for all n 0. (3)

42 We let {v n } n 0 for the companion Lucas sequence of {u n } n 0 given by v 0 = 2, v 1 = r and v n+2 = rv n+1 v n. Its Binet formula is v n = α n + β n for all n 0. (3) Lemma Let i and j be two integers with the same parity i.e i j (mod 2). Then u i u j = u (i j)/2 v (i+j)/2.

43 We let {v n } n 0 for the companion Lucas sequence of {u n } n 0 given by v 0 = 2, v 1 = r and v n+2 = rv n+1 v n. Its Binet formula is v n = α n + β n for all n 0. (3) Lemma Let i and j be two integers with the same parity i.e i j (mod 2). Then u i u j = u (i j)/2 v (i+j)/2. Lemma (Bertrand s Postulate(1845)) For every natural number n > 3, there is a prime p satisfying n < p < 2n.

44 Index of appearance z(m) Lemma (Bilu, Hanrot, Voutier (2001).) The index of appearance z of the sequence U(k) has the following properties. i) If p (k), then z(p) = p. ii) If p (k), then z(p) p e, where e = ( (k) p ). iii) Let c = ν p (U z(p) (k)). Then z(p b ) = p max{b c,0} z(p). iv) If p U m (k), then z(p) m. v) If n = m 1 m s with m 1,..., m s pairwise coprime, then z(m 1 m s ) = lcm[z(m 1 ),..., z(m s )].

45 Steps of the Proof for k = 1 In this case, we consider the binary recurrent sequence {u n } n 0 given by u 0 = 0, u 1 = 1 and u n+1 = 6u n u n 1 for all n 0. Its first terms are 0, 1, 6, 35, 204, 1189, 6930, 40391, , , ,....

46 Steps of the Proof for k = 1 In this case, we consider the binary recurrent sequence {u n } n 0 given by u 0 = 0, u 1 = 1 and u n+1 = 6u n u n 1 for all n 0. Its first terms are 0, 1, 6, 35, 204, 1189, 6930, 40391, , , ,.... Result: D u (n) = min{2 e, 2 a 5 b }, with 2 a 5 b 5n/3.

47 Step 1: Structure of D 1 (n) Lemma Let m = D 1 (n) for some n > 1. Then i) m has at most one odd prime divisor. ii) If m is divisible by exactly one odd prime p, then e = ( 2 p ) = 1 and z(p) = (p + 1)/2. iii) If m is not a power of 2, then m can be written as 2 a p b with a, b 1 and p 5 (mod 8).

48 Step 1: Structure of D 1 (n) Lemma Let m = D 1 (n) for some n > 1. Then i) m has at most one odd prime divisor. ii) If m is divisible by exactly one odd prime p, then e = ( 2 p ) = 1 and z(p) = (p + 1)/2. iii) If m is not a power of 2, then m can be written as 2 a p b with a, b 1 and p 5 (mod 8). Lemma Assume that m = 2 a p b 1 1 is such that a 1, p 1 5 (mod 8) and z(p 1 ) = (p 1 + 1)/2. Then U i U j (mod m) holds if and only if i j (mod z(m)).

49 Proof Assume that D 1 (n) = m and write it as m = 2 a p b 1 1 pbr r, where the p i are distinct odd primes.

50 Proof Assume that D 1 (n) = m and write it as m = 2 a p b 1 1 pbr r, where the p i are distinct odd primes. If r 2, we obtain the inequality ( ) ( ) z(m) 2 a p b p b p1 + 1 pr + 1 k 1 r < m 2 2 2, (4)

51 Proof Assume that D 1 (n) = m and write it as m = 2 a p b 1 1 pbr r, where the p i are distinct odd primes. If r 2, we obtain the inequality ( ) ( ) z(m) 2 a p b p b p1 + 1 pr + 1 k 1 r < m 2 2 2, (4) it follows that the interval [z(m), 2z(m)) contains a power of 2, say 2 b < 2z(m) < m. But then since 2 b z(m) n, we get a contradiction.

52 Proof Assume that D 1 (n) = m and write it as m = 2 a p b 1 1 pbr r, where the p i are distinct odd primes. If r 2, we obtain the inequality ( ) ( ) z(m) 2 a p b p b p1 + 1 pr + 1 k 1 r < m 2 2 2, (4) it follows that the interval [z(m), 2z(m)) contains a power of 2, say 2 b < 2z(m) < m. But then since 2 b z(m) n, we get a contradiction. If r = 1 and e 1 = ( 2 p 1 ) = 1, then a contradiction z(m) = z(2 a p b 1 1 ) 2a p b (p 1 1)/2 < m/2,

53 Assume now that e 1 = 1 and that z(p 1 ) is a proper divisor of (p + 1)/2. Then z(m) 2 a p b z(p 1 ) 2 a p b (p 1 + 1)/4 < m/2, again the same contradiction.

54 Assume now that e 1 = 1 and that z(p 1 ) is a proper divisor of (p + 1)/2. Then z(m) 2 a p b z(p 1 ) 2 a p b (p 1 + 1)/4 < m/2, again the same contradiction. We write m = 2 a p b 1 1. We know that a 1 and e = 1. Thus, p ±3 (mod 8). If p 3 (mod 8), then z(m) = lcm[z(2 a ), z(p b )] 2 a p b 1 (p + 1)/4. In particular, z(m) < m/2, and we get again a contradiction. Thus, p 5 (mod 8).

55 Step 2 Lemma For n the interval [5n/3, 37n/19) contains a number of the form 2 a 5 b with a 1 and b 0.

56 Step 2 Lemma For n the interval [5n/3, 37n/19) contains a number of the form 2 a 5 b with a 1 and b 0. Corollary Suppose that m = 2 a p b, p > 5, a, b 1. If m , then m is not a discriminator value.

57 Step 2 Lemma For n the interval [5n/3, 37n/19) contains a number of the form 2 a 5 b with a 1 and b 0. Corollary Suppose that m = 2 a p b, p > 5, a, b 1. If m , then m is not a discriminator value. Suppose that D 1 (n) = m, then we must have z(m) = 2 a p b 1 (p + 1)/(2) 19m/37 n, that is m 37n/19; a contradiction

58 Step 3 Lemma We say that m discriminates U 0,..., U n 1 if these integers are pairwise distinct modulo m. i) The integer m = 2 a discriminates U 0,..., U n 1 if and only if m n.

59 Step 3 Lemma We say that m discriminates U 0,..., U n 1 if these integers are pairwise distinct modulo m. i) The integer m = 2 a discriminates U 0,..., U n 1 if and only if m n. ii) The integer m = 2 a 5 b with a, b 1 discriminates U 0,..., U n 1 if and only if m 5n/3.

60 Step 3 Lemma We say that m discriminates U 0,..., U n 1 if these integers are pairwise distinct modulo m. i) The integer m = 2 a discriminates U 0,..., U n 1 if and only if m n. ii) The integer m = 2 a 5 b with a, b 1 discriminates U 0,..., U n 1 if and only if m 5n/3. Proof. m discriminates U 0,..., U z(m) 1, but not U 0,..., U z(m).

61 Step 3 Lemma We say that m discriminates U 0,..., U n 1 if these integers are pairwise distinct modulo m. i) The integer m = 2 a discriminates U 0,..., U n 1 if and only if m n. ii) The integer m = 2 a 5 b with a, b 1 discriminates U 0,..., U n 1 if and only if m 5n/3. Proof. m discriminates U 0,..., U z(m) 1, but not U 0,..., U z(m). then m discriminates U 0,..., U n 1 iff n z(m).

62 Step 3 Lemma We say that m discriminates U 0,..., U n 1 if these integers are pairwise distinct modulo m. i) The integer m = 2 a discriminates U 0,..., U n 1 if and only if m n. ii) The integer m = 2 a 5 b with a, b 1 discriminates U 0,..., U n 1 if and only if m 5n/3. Proof. m discriminates U 0,..., U z(m) 1, but not U 0,..., U z(m). then m discriminates U 0,..., U n 1 iff n z(m). As it is easily seen that z(m) = 3m/5, the result follows.

63 Case k = 2, D 2 (n) = min{2 e, 3 2 f }. if z(m) = m, then m 3 2 a for some a 0.

64 Case k = 2, D 2 (n) = min{2 e, 3 2 f }. if z(m) = m, then m 3 2 a for some a 0. or z(m) 3m/5.

65 Case k = 2, D 2 (n) = min{2 e, 3 2 f }. if z(m) = m, then m 3 2 a for some a 0. or z(m) 3m/5. if m is a discriminator then m 5n/3.

66 Case k = 2, D 2 (n) = min{2 e, 3 2 f }. if z(m) = m, then m 3 2 a for some a 0. or z(m) 3m/5. if m is a discriminator then m 5n/3. for n 2, there is a power of two or a number of the form 3 2 a in the interval [n, 5n/3).

67 Case k = 2 Theorem (Moree, Luca, F. (2017)) Let e 0 be the smallest integer such that 2 e n and f 1 the smallest integer such that 3 2 f n. Then D 2 (n) = min{2 e, 3 2 f }.

68 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9),

69 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = and B k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9), { {m even : if p m, then p k(k + 1)} if k 2 (mod 9); {m even, 9 m : if p m, then p k(k + 1)} if k 2 (mod 9).

70 Case k > 2 Theorem (Moree, Luca, F. (2017)) Put A k = and B k = { {m odd : if p m, then p k} if k 6 (mod 9); {m odd, 9 m : if p m, then p k} if k 6 (mod 9), { {m even : if p m, then p k(k + 1)} if k 2 (mod 9); {m even, 9 m : if p m, then p k(k + 1)} if k 2 (mod 9). Let k > 2. We have D k (n) min{m n : m A k B k }, with equality if the interval [n, 3n/2) contains an integer m A k B k and with at most finitely many n for which strict inequality holds.

71 Case k > 2 Here, we have to consider two cases: p k(k + 1)

72 Case k > 2 Here, we have to consider two cases: p k(k + 1) p k(k + 1)

73 The congruence U i (k) U j (k) (mod m) u k (n + 2) = (4k + 2)u k (n + 1) u k (n). (k) = 16k(k + 1).

74 The congruence U i (k) U j (k) (mod m) u k (n + 2) = (4k + 2)u k (n + 1) u k (n). (k) = 16k(k + 1). k(k + 1) = du 2, and let K = Q[ d].

75 The congruence U i (k) U j (k) (mod m) u k (n + 2) = (4k + 2)u k (n + 1) u k (n). (k) = 16k(k + 1). k(k + 1) = du 2, and let K = Q[ d]. Let p a k and π be any prime ideal diving p and let e be such that π e p.

76 The congruence U i (k) U j (k) (mod m) u k (n + 2) = (4k + 2)u k (n + 1) u k (n). (k) = 16k(k + 1). k(k + 1) = du 2, and let K = Q[ d]. Let p a k and π be any prime ideal diving p and let e be such that π e p. We assume U i λ (mod p b ) then α i α i 4 k(k + 1)λ 0 (mod π eb+ae/2 ).

77 The congruence U i (k) U j (k) (mod m) u k (n + 2) = (4k + 2)u k (n + 1) u k (n). (k) = 16k(k + 1). k(k + 1) = du 2, and let K = Q[ d]. Let p a k and π be any prime ideal diving p and let e be such that π e p. We assume U i λ (mod p b ) then α i α i 4 k(k + 1)λ 0 (mod π eb+ae/2 ). then it satisfied the quadratic congruence x 2 4 k(k + 1)λx 1 = 0 (mod π eb+ae/2 ).

78 Taking their difference we get (α i α j )(α i + α j 4 k(k + 1))λ 0 (mod π be+ae/2 ).

79 Taking their difference we get (α i α j )(α i + α j 4 k(k + 1))λ 0 (mod π be+ae/2 ). If p k, we have that α = 2k k(k + 1) 1 (mod π). Then (α i + α j 4 k(k + 1)) 2 (mod π ae/2 )

80 Taking their difference we get (α i α j )(α i + α j 4 k(k + 1))λ 0 (mod π be+ae/2 ). If p k, we have that α = 2k k(k + 1) 1 (mod π). Then (α i + α j 4 k(k + 1)) 2 (mod π ae/2 ) Thus, α i α j (mod π be+ae/2 )

81 Taking their difference we get (α i α j )(α i + α j 4 k(k + 1))λ 0 (mod π be+ae/2 ). If p k, we have that α = 2k k(k + 1) 1 (mod π). Then (α i + α j 4 k(k + 1)) 2 (mod π ae/2 ) Thus, α i α j (mod π be+ae/2 ) This give π eb U i j and. i j 0 (mod z(p b ))

82 Lemma Assume p k is odd. Then U i U j (mod z(p b )). (mod p b ) if and only if i j

83 If p (k + 1), we have that the factors (α i α j ) and (α i + α j 4 k(k + 1)) ( 1) i + ( 1) j (mod π ae/2 ).

84 If p (k + 1), we have that the factors (α i α j ) and (α i + α j 4 k(k + 1)) ( 1) i + ( 1) j (mod π ae/2 ). Thus, π never divides both factors, and π ae/2 divides α i α j in case i j (mod 2), and it divides α i + α j 4 k(k + 1)λ in case i j (mod 2).

85 Lemma Assume that p is odd and p (k + 1). Then U i U j equivalent to one of the following: i) If i j (mod 2), then i j (mod z(p b )). (mod p b ) is

86 Lemma Assume that p is odd and p (k + 1). Then U i U j equivalent to one of the following: i) If i j (mod 2), then i j (mod z(p b )). ii) If i j (mod 2), then i + j 0 (mod z(p b )). (mod p b ) is

87 Lemma We have i j (mod m) U i U j (mod m), (5) precisely when m A k B k.

88 Lemma We have i j (mod m) U i U j (mod m), (5) precisely when m A k B k. {m odd : z(m) = m and m P(k)}. {m even : z(m) = m}.

89 The set F k For k > 1 there is a finite set F k such that D k = A k B k F k. Lemma There are infinitely many k for the finite set F k is non-empty. It can have a cardinality larger than any given bound.

90 The set F k For k > 1 there is a finite set F k such that D k = A k B k F k. A 1 = {1}, B 1 = {2 e : e 1} F 1 = {2 a 5 m : a 1 and m M}.

91 The set F k For k > 1 there is a finite set F k such that D k = A k B k F k. A 1 = {1}, B 1 = {2 e : e 1} F 1 = {2 a 5 m : a 1 and m M}. F 2 is empty.

92 The set F k For k > 1 there is a finite set F k such that D k = A k B k F k. A 1 = {1}, B 1 = {2 e : e 1} F 1 = {2 a 5 m : a 1 and m M}. F 2 is empty. F k is finite for k > 1.

93 Example For k = 3, we have the following sequence u n+1 = 14u n u n 1 for all n 0. Its first terms are 0, 1, 14, 195, 2716, 37829, , The discrimator for n = 1..., 20. D(1) = 1 D(2) = 2 D(3) = 3 D(4) = 2 2 D(5) = 2 3 D(6) = 2 3 D(7) = 2 3 D(8) = 2 3 D(9) = 3 2 D(10) = D(11) = D(12) = D(13) = 2 4 D(14) = 2 4 D(15) = 2 4 D(16) = 2 4 D(17) = D(18) = D(19) = D(20) = 2 3 3

94 Open Problems Give a characterization of D u (n) for a given Lucas sequence u n : Fibonacci sequence, Prime numbers sequence, etc. D P (n) P n Give a classification of all recurrent sequence with D u (2 e ) = 2 e D u (2 e ) = 2 e = D u [n, 2n]

95 Open Problems Conjecture: Let {u n } n 0 be a binary sequence given by the recurrence u 0 = a u 1 = b u n+2 = ru n+1 + su n for n 0 where r, s are two integers such that r > 0 and (r, s) (2, 1), (1, 1). For all e 1, D u (2 e ) = 2 e ν 2 (r) = 1, s 3 (mod 4) and a, b have different parity.

96 I love mathematics for its own sake, because it allows for no hypocrisy and no vagueness. Stendhal THANKS FOR YOUR ATTENTION!