Repdigits base b as products of two Lucas numbers
dc.contributor.author | Erduvan, F. and Keskin, R. and Şiar, Z. | |
dc.date.accessioned | 2021-04-08T12:06:31Z | |
dc.date.available | 2021-04-08T12:06:31Z | |
dc.date.issued | 2020 | |
dc.identifier | 10.2989/16073606.2020.1787539 | |
dc.identifier.issn | 16073606 | |
dc.identifier.uri | https://www.scopus.com/inward/record.uri?eid=2-s2.0-85087830086&doi=10.2989%2f16073606.2020.1787539&partnerID=40&md5=54e48d48817bc8a28974a4c43a4618ce | |
dc.identifier.uri | http://acikerisim.bingol.edu.tr/handle/20.500.12898/3984 | |
dc.description.abstract | Let (Ln) be the sequence of Lucas numbers defined by L 0 = 2, L 1 = 1, and Ln = L n−1 + L n−2 for n ≥ 2. Let 0 ≤ m ≤ n and b = 2, 3, 4, 5, 6, 7, 8, 9. In this study, we show that if LmLn is a repdigit in the base b and has at least two digits, then LmLn ∈ {3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 28, 36, 54, 121, 228}. Furthermore, it is shown that if Ln is a repdigit in the base b and has at least two digits, then (n, b) = (2, 2), (3, 3), (4, 6), (4, 2), (6, 5), (6, 8). Namely, L 2 = 3 = (11)2, L 3 = 4 = (11)3, L 4 = 7 = (11)6 and L 4 = 7 = (111)2, L 6 = 18 = (33)5, L 6 = 18 = (22)8. © 2020, © 2020 NISC (Pty) Ltd. | |
dc.language.iso | English | |
dc.source | Quaestiones Mathematicae | |
dc.title | Repdigits base b as products of two Lucas numbers |
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