A binary linear recurrence sequence of composite numbers

Dubickas, Artūras; Novikas, Aivaras; Šiurys, Jonas

doi:10.1016/j.jnt.2010.03.015

Cited by 8 publications

(11 citation statements)

References 12 publications

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“…The known values for admissible a and b have decreased over the years through the work of others including D. Knuth [19], J. W. Nicol [23] and M. Vsemirnov [29], the latter giving the smallest known such a and b (but notably the same number of digits as the a and b in [23]). The result has also been generalized to other recursions; see A Dubickas, A. Novikas and J. Šiurys [4], D. Ismailescu, A. Ko, C. Lee and J. Y. Park [16] and I. Lunev [21].…”

Section: Does There Exist Any Prime Number Such That If Any Digit (In...mentioning

confidence: 99%

Consecutive primes which are widely digitally delicate

Filaseta¹,

Juillerat²

2021

Preprint

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Section: Does There Exist Any Prime Number Such That If Any Digit (In...mentioning

confidence: 99%

Consecutive primes which are widely digitally delicate

Filaseta¹,

Juillerat²

2021

Preprint

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“…In this paper we present a new proof of the following result of Dubickas, Novikas, and Siurys: Theorem 1.1. [1] Let (a, b) ∈ Z 2 and let (x n ) n≥0 be the sequence defined by some initial values x 0 and x 1 and the second order linear recurrence (1) x n+1 = ax n + bx n−1…”

Section: Introductionmentioning

confidence: 99%

On second order linear sequences of composite numbers

Ismailescu,

Ko,

Lee

et al. 2018

Preprint

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In this paper we present a new proof of the following 2010 result of Dubickas, Novikas, and Siurys:Theorem. Let (a, b) ∈ Z 2 and let (x n ) n≥0 be the sequence defined by some initial values x 0 and x 1 and the second order linear recurrencefor n ≥ 1. Suppose that b = 0 and (a, b) = (2, −1), (−2, −1). Then there exist two relatively prime positive integers x 0 , x 1 such that |x n | is a composite integer for all n ∈ N.The above theorem extends a result of Graham who solved the problem when (a, b) = (1, 1). Lemma 2.2. Consider the sequence (x n ) n≥0 given by (1). Suppose that 1 ≤ |x 0 | < |x 1 | and |a| > |b| ≥ 1. Then the sequence (|x n |) n≥0 is strictly increasing. Proof. We use induction on n. By hypothesis, the statement is true for n = 0. Suppose that |x n−1 | < |x n | for some n ≥ 1. We intend to prove that |x n | < |x n+1 |. Indeedby the induction hypothesis and since |a| − |b| ≥ 1 we finally have |x n+1 | > |x n |, which completes the induction. Lemma 2.3. Let n 1 , n 2 and n 3 be three positive integers such that no prime number p divides all of them. Then, there exists an integer k ≥ 2 such that n 1 and n 2 + kn 3 are relatively prime. Proof. Let d := gcd(n 2 , n 3 ). Note that gcd(n 1 , d) = 1 otherwise d divides n 1 , n 2 and n 3 . By Dirichlet's theorem on arithmetic progressions there exists a k such that n 2 /d + kn 3 /d is a prime number greater than n 1 . Then d (n 2 /d + k • n 3 /d) = n 2 + k n 3 is both greater and relatively prime to n 1 .3. Two simple special cases: (i) a = 0 and (ii) a 2 + 4b = 0.Case (i). Since a = 0, it can be easily proved that x 2n = b n x 0 and x 2n+1 = b n x 1 . It suffices to take x 0 = 4 and x 1 = 9 to obtain that x n is composite for all n ≥ 0.Case (ii). If a 2 + 4b = 0, then a must be even, a = 2c and therefore b = −a 2 /4 = −c 2 . We divide the proof into two cases: |b| ≥ 2 and b = −1.If |b| ≥ 2 we have |c| ≥ 2. Let us now take x 0 = 4c 2 − 1 and x 1 = 2c 3 . Then x 0 and x 1 are relative prime positive composite integers. One immediately obtains x 2 = c 2 thus x 2 is also composite. Also, it can be easily proved that x n = c n ((n − 1) − (2n − 4)c 2 ) for all n ≥ 3.Note that for any n ≥ 3 one cannot haveIf a = 2 then a simple induction shows that x n+1 = (n + 1)x 1 − nx 0 = x 1 +n(x 1 −x 0 ) for all n ≥ 0, that is, (x n ) n≥0 is an arithmetic sequence whose first term and common difference are relatively prime. By Dirichlet's theorem on primes in arithmetic progressions, it follows that |x n | is a prime number for infinitely many values of n. If a = −2 then one can show that x n+1 = (−1) n (x 1 + n(x 0 + x 1 )). In this case, (x n ) n≥0 is the union of two arithmetic sequences both of which have the first term and the common difference relatively prime. Again, for any choice of x 0 and x 1 relatively prime, |x n | is a prime for infinitely many n. This proves the necessity of the condition (a, b) = (±2, −1). The case |b| ≥ 2Based on the results from the previous section, from now on one can assume that (5)|a| ≥ 1 and a 2 + 4b = 0.We divide the proof into three...

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“…For related results and references we refer to the book of Guy [20], p. 17, and the papers of Graham [19], Knuth [27], Wilf [47] and Dubickas et al [17], and the references given there.…”

Section: Introductionmentioning

confidence: 99%

On the GCD-s of k consecutive terms of Lucas sequences

Hajdu

Szikszai

2012

Journal of Number Theory

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Let u = (u n ) ∞ n=0 be a Lucas sequence, that is a binary linear recurrence sequence of integers with initial terms u 0 = 0 and u 1 = 1. We show that if k is large enough then one can find k consecutive terms of u such that none of them is relatively prime to all the others. We even give the exact values g u and G u for each u such that the above property first holds with k = g u ; and that it holds for all k G u , respectively. We prove similar results for Lehmer sequences as well, and also a generalization for linear recurrence divisibility sequences of arbitrarily large order. On our way to prove our main results, we provide a positive answer to a question of Beukers from 1980, concerning the sums of the multiplicities of 1 and −1 values in non-degenerate Lucas sequences. Our results yield an extension of a problem of Pillai from integers to recurrence sequences, as well.

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A binary linear recurrence sequence of composite numbers

Cited by 8 publications

References 12 publications

Consecutive primes which are widely digitally delicate

Consecutive primes which are widely digitally delicate

On second order linear sequences of composite numbers

On the GCD-s of k consecutive terms of Lucas sequences

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