On a problem of Pillai with k–generalized Fibonacci numbers and powers of 2

Abstract

For an integer k ≥ 2, let {F (k) n } n≥0 be the k-generalized Fibonacci sequence which starts with 0, . . . , 0, 1 (k terms) and each term afterwards is the sum of the k preceding terms. In this paper, we find all integers c having at least two representations as a difference between a k-generalized Fibonacci number and a powers of 2 for any fixed k 4. This paper extends previous work from [9] for the case k = 2 and [6] for the case k = 3.

Pillai’s problem with the Fibonacci andPadovan sequences

“…In the same way, ̸ = 1 and 1 ̸ = 1. With this conventions, our result is the following: (18,10,17,9), (18,11,8,6), (18,11,5,5), (18,11,0,4), (19,11,14,4), (19,12,7,9), (20,11,17,4), (20,12,14,8), (20,12,11,5), (20,12,10,3), (20,12,9,0), (20,13,9,11), (20,14,9,13), (21,…”

“…This version was started by Ddamulira, Luca and Rakotomalala in [8] where they considered U as being the Fibonacci sequence and V as being the sequence of powers of 2. Many other cases have been studied, see for example [3,6,7,10,12,13]. In [5], there is a general result, namely that if U and V satisfy some natural conditions, then equation (12) has only finitely many solutions which furthermore are all effectively computable.…”

Pillai's problem with Padovan numbers and powers of two

“…This version was started in [8] by Ddamulira, Luca and Rakotomalala where they take U being the Fibonacci sequence and V being the sequence of powers of 2. Many other cases have been studied, see for example [5,3,7,10]. In [6] it is proved that, under some natural conditions on U and V equation (2) has finitely many solutions and all of them are effectively computable.…”

“…This is A000931 sequence in [16]. Its first few terms are 0, 1, 1, 1, 2, 2, 3,4,5,7,9,12,16,21,28,37,49,65,86,114,151, . .…”

“…(3, 1, 0, 0), (5, 1, 3, 0), (5, 2, 0, 1), (6, 1, 5, 0), (6, 2, 0, 0), (6, 2, 3, 1), (7, 1, 6, 0), (7, 2, 3, 0), (7, 2, 5, 1), (7, 3, 0, 2), (8, 1, 7, 0), (8, 2, 5, 0), (8, 2, 6, 1), (8, 3, 3, 2), (9, 2, 7, 0), (9, 2, 8, 1), (9, 3, 0, 0), (9, 3, 3, 1), (9, 3, 6, 2), (10, 2, 9, 1), (10, 3, 5, 0), (10, 3, 6, 1), (10,3,8,2), (10,4,3,3), (11, 2, 10, 0), (11,3,8,0), (11,4, 0, 2), (11,4,7,3), (12, 3, 10, 0), (12,3,11,2), (12,4,3,0), (12,4,5,1), (12,4,7,2), (12,5,0,4), (13,4,9,1), (13,4,10,2), (13,5,8,4),…”