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A319555 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199. 10
6, 4, 1, 2, 1, 2, 0, 4, 4, 4, 1, 0, 6, 1, 0, 5, 2, 4, 4, 4, 2, 3, 1, 0, 6, 3, 1, 4, 2, 6, 1, 6, 1, 2, 1, 5, 4, 5, 5, 3, 4, 2, 6, 4, 0, 4, 3, 4, 4, 1, 0, 6, 5, 2, 4, 1, 4, 2, 2, 1, 5, 2, 4, 4, 2, 5, 4, 6, 5, 1, 0, 1, 6, 1, 1, 4, 0, 6, 3, 4, 4, 2, 3, 4, 0, 0, 4, 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

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

Wikipedia, p-adic number

FORMULA

Equals A319297*(A212152-1) = A319297*A212152^2, where each A-number represents a 7-adic number.

Equals A319305*(A212155-1) = A319305*A212155^2.

EXAMPLE

The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83 = (146)_7, so the first three terms are 6, 4 and 1.

PROG

(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3))\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, A319305, this sequence (7-adic, 6^(1/3));

A321106, A321107, A321108 (13-adic, 5^(1/3)).

Sequence in context: A106333 A104748 A117335 * A244980 A021863 A259620

Adjacent sequences:  A319552 A319553 A319554 * A319556 A319557 A319558

KEYWORD

nonn,base

AUTHOR

Jianing Song, Aug 27 2019

STATUS

approved

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Last modified June 14 20:21 EDT 2021. Contains 345038 sequences. (Running on oeis4.)