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A319199 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).

%I A319199 #54 Aug 29 2019 11:40:11
%S A319199 0,6,34,83,769,3170,36784,36784,3330956,26390160,187804588,470279837,
%T A319199 470279837,83518003043,180407013450,180407013450,23918214563165,
%U A319199 90384075702367,1020906131651195,7534560523292991,53130141264785563,212714673860009565,1888352266109861586
%N A319199 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).
%C A319199 For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 6 mod 7 such that k^3 - 6 is divisible by 7^n.
%C A319199 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.
%H A319199 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A319199 a(n) = A319097(n)*(A210852(n)-1) mod 7^n = A319097(n)*A210852(n)^2 mod 7^n.
%F A319199 a(n) = A319098(n)*(A212153(n)-1) mod 7^n = A319098(n)*A212153(n)^2 mod 7^n.
%e A319199 The unique number k in [1, 7^2] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 34, so a(2) = 24.
%e A319199 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, so a(3) = 122.
%o A319199 (PARI) a(n) = lift(sqrtn(6+O(7^n), 3))
%Y A319199 Cf. A319297, A319305, A319555.
%Y A319199 Approximations of p-adic cubic roots:
%Y A319199 A290567 (5-adic, 2^(1/3));
%Y A319199 A290568 (5-adic, 3^(1/3));
%Y A319199 A309444 (5-adic, 4^(1/3));
%Y A319199 A319097, A319098, this sequence (7-adic, 6^(1/3));
%Y A319199 A320914, A320915, A321105 (13-adic, 5^(1/3)).
%K A319199 nonn
%O A319199 0,2
%A A319199 _Jianing Song_, Aug 27 2019