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A319097 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 3 (mod 7) case (except for n = 0).

%I A319097 #50 Aug 29 2019 11:39:23
%S A319097 0,3,24,122,808,10412,111254,817148,1640691,24699895,186114323,
%T A319097 186114323,6118094552,6118094552,490563146587,2525232365134,
%U A319097 26263039914849,59495970484450,1222648540420485,6107889334151832,74501260446390690,234085793041614692,1351177521208182706,24810103812706111000,134285093173029776372
%N A319097 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 3 (mod 7) case (except for n = 0).
%C A319097 For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 3 mod 7 such that k^3 - 6 is divisible by 7^n.
%C A319097 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 A319097 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A319097 a(n) = A319098(n)*(A210852(n)-1) mod 7^n = A319098(n)*A210852(n)^2 mod 7^n.
%F A319097 a(n) = A319199(n)*(A212153(n)-1) mod 7^n = A319199(n)*A212153(n)^2 mod 7^n.
%e A319097 The unique number k in [1, 7^2] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 24, so a(2) = 24.
%e A319097 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, so a(3) = 122.
%o A319097 (PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1+sqrt(-3+O(7^n)))/2)
%Y A319097 Cf. A319297, A319305, A319555.
%Y A319097 Approximations of p-adic cubic roots:
%Y A319097 A290567 (5-adic, 2^(1/3));
%Y A319097 A290568 (5-adic, 3^(1/3));
%Y A319097 A309444 (5-adic, 4^(1/3));
%Y A319097 this sequence, A319098, A319199 (7-adic, 6^(1/3));
%Y A319097 A320914, A320915, A321105 (13-adic, 5^(1/3)).
%K A319097 nonn
%O A319097 0,2
%A A319097 _Jianing Song_, Aug 27 2019