A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.

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

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

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

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));

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

Adjacent sequences: A319294 A319295 A319296 * A319298 A319299 A319300

Keyword

nonn,base

Author

Jianing Song, Aug 27 2019

Status

approved