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A319305
Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.
12
5, 5, 2, 2, 1, 5, 6, 0, 5, 4, 4, 3, 0, 0, 3, 3, 2, 4, 5, 2, 1, 1, 6, 1, 4, 6, 2, 2, 2, 6, 2, 4, 1, 0, 0, 2, 0, 3, 4, 2, 2, 1, 5, 2, 6, 1, 0, 4, 1, 6, 0, 6, 0, 1, 3, 0, 3, 4, 1, 1, 1, 0, 6, 5, 2, 4, 0, 2, 2, 5, 1, 1, 0, 4, 0, 0, 5, 6, 4, 1, 5, 2, 4, 3, 0, 1, 2, 5
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
Wikipedia, p-adic number
FORMULA
Equals A319297*(A212155-1) = A319297*A212155^2, where each A-number represents a 7-adic number.
Equals A319555*(A212152-1) = A319555*A212152^2.
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
Digits of p-adic cubic roots:
A290566 (5-adic, 2^(1/3));
A290563 (5-adic, 3^(1/3));
A309443 (5-adic, 4^(1/3));
A319297, this sequence, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).
Sequence in context: A280868 A229160 A011501 * A196614 A319905 A319593
KEYWORD
nonn,base,changed
AUTHOR
Jianing Song, Aug 27 2019
STATUS
approved