This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A319305 #48 Nov 13 2024 08:29:29 %S A319305 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 A319305 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 A319305 2,5,1,1,0,4,0,0,5,6,4,1,5,2,4,3,0,1,2,5 %N A319305 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098. %C A319305 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 A319305 Seiichi Manyama, <a href="/A319305/b319305.txt">Table of n, a(n) for n = 0..10000</a> %H A319305 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a> %F A319305 Equals A319297*(A212155-1) = A319297*A212155^2, where each A-number represents a 7-adic number. %F A319305 Equals A319555*(A212152-1) = A319555*A212152^2. %e A319305 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 A319305 (PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1-sqrt(-3+O(7^(n+1))))/2)\7^n %Y A319305 Cf. A319097, A319098, A319199. %Y A319305 Digits of p-adic cubic roots: %Y A319305 A290566 (5-adic, 2^(1/3)); %Y A319305 A290563 (5-adic, 3^(1/3)); %Y A319305 A309443 (5-adic, 4^(1/3)); %Y A319305 A319297, this sequence, A319555 (7-adic, 6^(1/3)); %Y A319305 A321106, A321107, A321108 (13-adic, 5^(1/3)). %K A319305 nonn,base %O A319305 0,1 %A A319305 _Jianing Song_, Aug 27 2019