A319297 - OEIS

%I #52 Nov 13 2024 08:29:24

%S 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,

%T 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,

%U 5,6,4,4,0,1,5,2,1,1,5,0,4,1,6,5,5,5,0,4

%N Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.

%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. [Typo corrected by _Keyang Li_, Nov 04 2024]

%H Seiichi Manyama, <a href="/A319297/b319297.txt">Table of n, a(n) for n = 0..10000</a>

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

%F Equals A319305*(A212152-1) = A319305*A212152^2, where each A-number represents a 7-adic number.

%F Equals A319555*(A212155-1) = A319555*A212155^2.

%e 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.

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

%Y Cf. A319097, A319098, A319199.

%Y Digits of p-adic cubic roots:

%Y A290566 (5-adic, 2^(1/3));

%Y A290563 (5-adic, 3^(1/3));

%Y A309443 (5-adic, 4^(1/3));

%Y this sequence, A319305, A319555 (7-adic, 6^(1/3));

%Y A321106, A321107, A321108 (13-adic, 5^(1/3)).

%K nonn,base

%O 0,1

%A _Jianing Song_, Aug 27 2019