|
%I #42 Aug 29 2019 12:12:07
%S 5,5,2,2,1,5,6,0,5,4,4,3,0,0,3,3,2,4,5,2,1,1,6,1,4,6,2,2,2,6,2,4,1,0,
%T 0,2,0,3,4,2,2,1,5,2,6,1,0,4,1,6,0,6,0,1,3,0,3,4,1,1,1,0,6,5,2,4,0,2,
%U 2,5,1,1,0,4,0,0,5,6,4,1,5,2,4,3,0,1,2,5
%N Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.
%C For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.
%H Seiichi Manyama, <a href="/A319305/b319305.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 A319297*(A212155-1) = A319297*A212155^2, where each A-number represents a 7-adic number.
%F Equals A319555*(A212152-1) = A319555*A212152^2.
%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 = (255)_7, so the first three terms are 5, 5 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 A319297, this sequence, 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
|
