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A319297
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Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.
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10
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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
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OFFSET
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0,1
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COMMENTS
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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.
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LINKS
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Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number
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FORMULA
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Equals A319305*(A212152-1) = A319305*A212152^2, where each A-number represents a 7-adic number.
Equals A319555*(A212155-1) = A319555*A212155^2.
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EXAMPLE
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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.
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PROG
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(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1+sqrt(-3+O(7^(n+1))))/2)\7^n
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CROSSREFS
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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
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KEYWORD
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nonn,base
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AUTHOR
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Jianing Song, Aug 27 2019
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STATUS
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approved
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