OFFSET
0,1
COMMENTS
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots. [Typo corrected by Keyang Li, Nov 04 2024]
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number
FORMULA
EXAMPLE
The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138 = (255)_7, so the first three terms are 5, 5 and 2.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1-sqrt(-3+O(7^(n+1))))/2)\7^n
CROSSREFS
KEYWORD
nonn,base,changed
AUTHOR
Jianing Song, Aug 27 2019
STATUS
approved