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Öğe Balancing Diophantine triples with distance 1(SPRINGER, 2015) Alp, Murat; Irmak, Nurettin; Szalay, LaszloFor a positive real number let the Balancing distance be the distance from to the closest Balancing number. The Balancing sequence is defined by the initial values , and by the binary recurrence relation , . In this paper, we show that there exist only one positive integer triple such that the Balancing distances , and all are exactly 1.Öğe DIOPHANTINE TRIPLES AND REDUCED QUADRUPLES WITH THE LUCAS SEQUENCE OF RECURRENCE(CROATIAN MATHEMATICAL SOC, 2014) Irmak, Nurettin; Szalay, LaszloIn this study, we show that there is no positive integer triple {a, b, c} such that all of ab+1, ac+1 and bc+1 are in the sequence {u(n)}n= 0 satisfies the recurrence un=Aun-1-un-2 with the initial values u0=0, u1=1. Further, we investigate the analogous question for the quadruples {a,b,c,d} with abc+1=ux, bcd+1=uy, cda+1=uz and dab+1=ut, and deduce the non-existence of such quadruples.Öğe Factorial-like values in the balancing sequence(Univ Osijek, Dept Mathematics, 2018) Irmak, Nurettin; Liptai, Kalman; Szalay, LaszloIn this paper, we solve a few Diophantine equations linked to balancing numbers and factorials. The basic problem consists of solving the equation B-y = x! in positive integers x, y, which has only one nontrivial solution B-2 = 6 = 3!, as a direct consequence of the theorem of F. Luca [5]. A more difficult problem is to solve B-y = x(2)!/x(1)!, but we were able to handle it under some conditions. Two related problems are also studied.Öğe Lucas numbers of the form (k2t)(Univ Tartu Press, 2019) Irmak, Nurettin; Szalay, LaszloLet L-m denote the m(th) Lucas number. We show that the solutions to the diophantine equation ((2t)(k)) = L-m, in non-negative integers t, k <= 2(t-1), and m, are (t, k, m) = (1, 1, 0), (2,1, 3), and (a, 0,1) with non-negative integers a.Öğe On k-periodic binary recurrences(E K F Liceum Kiado, 2012) Irmak, Nurettin; Szalay, LaszloWe apply a new approach, namely the fundamental theorem of homogeneous linear recursive sequences, to k-periodic binary recurrences which allows us to determine Binet's formula of the sequence if k is given. The method is illustrated in the cases k = 2 and k = 3 for arbitrary parameters. Thus we generalize and complete the results of Edson-Yayenie, and Yayenie linked to k = 2 hence they gave restrictions either on the coefficients or on the initial values. At the end of the paper we solve completely the constant sequence problem of 2-periodic sequences posed by Yayenie.Öğe ONLY FINITELY MANY TRIBONACCI DIOPHANTINE TRIPLES EXIST(Walter De Gruyter Gmbh, 2017) Fuchs, Clemens; Hutle, Christoph; Irmak, Nurettin; Luca, Florian; Szalay, LaszloDiophantine triples taking values in recurrence sequences have recently been studied quite a lot. In particular the question was raised whether or not there are finitely many Diophantine triples in the Tribonacci sequence. We answer this question here in the affirmative. We prove that there are only finitely many triples of integers 1 <= u < v < w such that uv + 1, uw 1, vw + 1 are Tribonacci numbers. The proof depends on the Subspace theorem.Öğe Reduced diophantine quadruples with the binary recurrence G(n) = AG(n-1) - G(n-2)(OVIDIUS UNIV PRESS, 2015) Alp, Murat; Irmak, Nurettin; Szalay, LaszloGiven a positive integer A not equal 2. In this paper, we show that there do not exist two positive integer pairs {a, b} not equal {c, d} such that the values of ac + 1, ad + 1 and bc + 1, bd + 1 are the terms of the sequence {G(n)}(n >= 0) which satisfies the recurrence relation G(n) = AG(n-1) - G(n-2) with the initial values G(0) = 0, G(1) = 1.Öğe TRIBONACCI NUMBERS CLOSE TO THE SUM 2(a)+3(b)+5(c)(MATEMATISK INST, 2016) Irmak, Nurettin; Szalay, LaszloWe show that there are exactly 22 solutions to the inequalities 0 <= T-n - 2(a) - 3(b) - 5(c) <= 10, where T-n denotes the nth term (n >= 0) of the Tribonacci sequence, and 0 <= a, b <= c are integers. All the solutions are explicitly determined.