For integers k >= 2 and n not equal 0, let v(k)(n) denote the greatest nonnegative integer e such that k(e) divides n. Moreover, let (u(n))(n >= 0) be a nondegenerate Lucas sequence satisfying u(0) = 0, u(1) = 1, and u(n+2) = au(n) (+1) + bu(n), for some integers a and b. Shu and Yao showed that for any prime number p the sequence v(p) (u(n) +1 )n>0 is p-regular, while Medina and Rowland found the rank of v(p) (F-n+1)(n >= 0), where F-n is the n-th Fibonacci number. We prove that if k and b are relatively prime then v(k)(u(n)+1)n(>= 0) is a k-regular sequence, and for k a prime number we also determine its rank. Furthermore, as an intermediate result, we give explicit formulas for v(k) (u(n)), generalizing a previous theorem of Sanna concerning p-adic valuations of Lucas sequences.
On the k-regularity of the k-adic valuation of Lucas sequences / Murru, Nadir; Sanna, Carlo. - In: JOURNAL DE THÉORIE DES NOMBRES DE BORDEAUX. - ISSN 1246-7405. - 30:1(2018), pp. 227-237.
On the k-regularity of the k-adic valuation of Lucas sequences
Murru, Nadir;
2018-01-01
Abstract
For integers k >= 2 and n not equal 0, let v(k)(n) denote the greatest nonnegative integer e such that k(e) divides n. Moreover, let (u(n))(n >= 0) be a nondegenerate Lucas sequence satisfying u(0) = 0, u(1) = 1, and u(n+2) = au(n) (+1) + bu(n), for some integers a and b. Shu and Yao showed that for any prime number p the sequence v(p) (u(n) +1 )n>0 is p-regular, while Medina and Rowland found the rank of v(p) (F-n+1)(n >= 0), where F-n is the n-th Fibonacci number. We prove that if k and b are relatively prime then v(k)(u(n)+1)n(>= 0) is a k-regular sequence, and for k a prime number we also determine its rank. Furthermore, as an intermediate result, we give explicit formulas for v(k) (u(n)), generalizing a previous theorem of Sanna concerning p-adic valuations of Lucas sequences.File | Dimensione | Formato | |
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