%I #53 Aug 29 2019 11:40:48

%S 0,5,40,138,824,3225,87260,793154,793154,29617159,191031587,

%T 1320932583,7252912812,7252912812,7252912812,2041922131359,

%U 16284606661188,82750467800390,1013272523749218,9155340513301463,31953130884047749,111745397181659750,670291261264943757

%N One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 5 (mod 7) case (except for n = 0).

%C For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 5 mod 7 such that k^3 - 6 is divisible by 7^n.

%C 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%F a(n) = A319097(n)*(A212153(n)-1) mod 7^n = A319097(n)*A212153(n)^2 mod 7^n.

%F a(n) = A319199(n)*(A210852(n)-1) mod 7^n = A319199(n)*A210852(n)^2 mod 7^n.

%e The unique number k in [1, 7^2] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 40, so a(2) = 40.

%e 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, so a(3) = 138.

%o (PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)

%Y Cf. A319297, A319305, A319555.

%Y Approximations of p-adic cubic roots:

%Y A290567 (5-adic, 2^(1/3));

%Y A290568 (5-adic, 3^(1/3));

%Y A309444 (5-adic, 4^(1/3));

%Y A319097, this sequence, A319199 (7-adic, 6^(1/3));

%Y A320914, A320915, A321105 (13-adic, 5^(1/3)).

%K nonn

%O 0,2

%A _Jianing Song_, Aug 27 2019