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 A319297 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097. 10
 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 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: A113780 A007515 A293729 * A309569 A292600 A014967 Adjacent sequences:  A319294 A319295 A319296 * A319298 A319299 A319300 KEYWORD nonn,base AUTHOR Jianing Song, Aug 27 2019 STATUS approved

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Last modified April 13 15:23 EDT 2021. Contains 342936 sequences. (Running on oeis4.)