Irmak, NurettinTogbe, Alain2024-11-072024-11-0720181310-51322367-8275https://doi.org/10.7546/nntdm.2018.24.3.95-102https://hdl.handle.net/11480/14549Let (L-n)(n >= 0 )be the Lucas sequence. D. Marques and A. Togbe [7] showed that if F-n . . . Fn+k-1 is a repdigit with at least two digits, then (k, n) = (1, 10), where (F-n)(>= 0) is the Fibonacci sequence. In this paper, we solve the equation L-n . . . Ln+k-1 = a (10(m) - 1/9) , where 1 <= a <= 9, n, k >= 2 and in are positive integers.eninfo:eu-repo/semantics/openAccessLucas numbersRepdigitsArticle2439510210.7546/nntdm.2018.24.3.95-102WOS:000448478500012N/A