OFFSET
0,1
COMMENTS
For k not divisible by 5, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number
EXAMPLE
The unique number k in [1, 13^3] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 177 = (108)_13, so the first three terms are 8, 0 and 1.
PROG
(PARI) a(n) = lift(sqrtn(5+O(13^(n+1)), 3))\13^n
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Jianing Song, Aug 27 2019
STATUS
approved