Diophantine triples and reduced quadruples with the Lucas sequence of recurrence un=Aun-1-un-2

In 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 {un}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

Lucas numbers of the form (2t/k)

Let Lm denote the mth Lucas number. We show that the solutions to the diophantine equation (2t/k) = Lm, in non-negative integers t, k ≤ 2t−1, and m, are (t, k, m) = (1, 1, 0), (2, 1, 3), and (a, 0, 1) with non-negative integers a

On k-periodic binary recurrences

We apply a new approach, namely the fundamental theorem of homo- geneous 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. Keywords: linear recurrences, k-periodic binary recurrence

Balancing with balancing powers

WOS: 000338514500018In this paper, the Diophantine equation B-1(k) + B-2(k) + ... + B-n-1(k) = B-n+1(k) + B-n+2(k) + ... + B-n+r(l) + for the positive integer unknowns n >= 2, k, l and r is studied in certain cases, where B-n denotes the nth term of the balancin

Factorial-like values in the balancing sequence

In this paper, we solve a few diophantine equations linked to balancing numbers and factorials. The basic problem (B_y=x!) has only the one nontrivial solution (B_2=6=3!), but it is a direct consequence of a theorem of F. Luca. The more difficult problem (B_y=x_2!/x_1!) is still open, but we solve it under different conditions. Two related problems are also studied

Tribonacci numbers with indices in arithmetic progression and their sums

WOS: 000328081500012In this paper, we give a recurrence relation for the Tribonacci numbers with indices in aritmetics progression, {Trn+s} for 0 <= s < n We find sums of {T-rn} g for arbitrary integer r via matrix methods

Schreier Multisets and the ss-step Fibonacci Sequences

Inspired by the surprising relationship (due to A. Bird) between Schreier sets and the Fibonacci sequence, we introduce Schreier multisets and connect these multisets with the $s$-step Fibonacci sequences, defined, for each $s\geqslant 2$, as: $F^{(s)}_{2-s} = \cdots = F^{(s)}_0 = 0$, $F^{(s)}_1 = 1$, and F^{(s)}_{n} = F^{(s)}_{n-1} + \cdots + F^{(s)}_{n-s}, \mbox{ for } n\geqslant 2. Next, we use Schreier-type conditions on multisets to retrieve a family of sequences which satisfy a recurrence of the form $a(n) = a(n-1) + a(n-u)$, with $a(n) = 1$ for $n = 1,\ldots, u$. Finally, we study nonlinear Schreier conditions and show that these conditions are related to integer decompositions, each part of which is greater than the number of parts raised to some power.Comment: 11 pages. To appear in Proceedings of the Integers Conference 202

SUM OF THE FIBONOMIAL COEFFICIENTS AT MOST ONE AWAY FROM FIBONACCI NUMBERS

WOS: 000389783800012In this note, we show that the equation Sigma(m)(k=0)[(2m+1)(k)](F) +/- 1 = F-n has no solution except that (m, n) = (1,3), (3,14) where [(m)(k)](F) is Fibonomial coefficient