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A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097. 12
3, 3, 2, 2, 4, 6, 6, 1, 4, 4, 0, 3, 0, 5, 3, 5, 1, 5, 3, 6, 2, 2, 6, 4, 3, 3, 2, 0, 2, 1, 2, 3, 3, 4, 5, 6, 1, 5, 3, 0, 0, 3, 2, 6, 6, 0, 3, 5, 0, 6, 5, 1, 0, 3, 6, 4, 6, 6, 2, 4, 0, 4, 3, 3, 1, 4, 1, 5, 5, 6, 4, 4, 0, 1, 5, 2, 1, 1, 5, 0, 4, 1, 6, 5, 5, 5, 0, 4 (list; graph; refs; listen; history; text; internal format)
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 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.
LINKS
Wikipedia, p-adic number
FORMULA
Equals A319305*(A212152-1) = A319305*A212152^2, where each A-number represents a 7-adic number.
Equals A319555*(A212155-1) = A319555*A212155^2.
EXAMPLE
The unique number k in [1, 7^3] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 122 = (233)_7, so the first three terms are 3, 3 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));
this sequence, A319305, A319555 (7-adic, 6^(1/3));
A321106, A321107, A321108 (13-adic, 5^(1/3)).
Sequence in context: A007515 A352455 A293729 * A309569 A356528 A292600
KEYWORD
nonn,base
AUTHOR
Jianing Song, Aug 27 2019
STATUS
approved

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Last modified April 25 12:53 EDT 2024. Contains 371969 sequences. (Running on oeis4.)