cp's OEIS Frontend

A319098 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).

%I A319098 #53 Aug 29 2019 11:40:48
%S A319098 0,5,40,138,824,3225,87260,793154,793154,29617159,191031587,
%T A319098 1320932583,7252912812,7252912812,7252912812,2041922131359,
%U A319098 16284606661188,82750467800390,1013272523749218,9155340513301463,31953130884047749,111745397181659750,670291261264943757
%N A319098 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 A319098 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 A319098 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 A319098 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A319098 a(n) = A319097(n)*(A212153(n)-1) mod 7^n = A319097(n)*A212153(n)^2 mod 7^n.
%F A319098 a(n) = A319199(n)*(A210852(n)-1) mod 7^n = A319199(n)*A210852(n)^2 mod 7^n.
%e A319098 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 A319098 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 A319098 (PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)
%Y A319098 Cf. A319297, A319305, A319555.
%Y A319098 Approximations of p-adic cubic roots:
%Y A319098 A290567 (5-adic, 2^(1/3));
%Y A319098 A290568 (5-adic, 3^(1/3));
%Y A319098 A309444 (5-adic, 4^(1/3));
%Y A319098 A319097, this sequence, A319199 (7-adic, 6^(1/3));
%Y A319098 A320914, A320915, A321105 (13-adic, 5^(1/3)).
%K A319098 nonn
%O A319098 0,2
%A A319098 _Jianing Song_, Aug 27 2019