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A319305 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098. 10
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 (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

Seiichi Manyama, Table of n, a(n) for n = 0..10000

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

Cf. A319097, A319098, A319199.

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

Adjacent sequences:  A319302 A319303 A319304 * A319306 A319307 A319308

KEYWORD

nonn,base

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

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Last modified August 4 09:55 EDT 2021. Contains 346446 sequences. (Running on oeis4.)