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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.)