cp's OEIS Frontend

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.

A321105 One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 11 (mod 13) case (except for n = 0).

This page as a plain text file.
%I A321105 #14 Aug 29 2019 10:21:07
%S A321105 0,11,154,999,25166,82288,82288,43523569,43523569,4937907895,
%T A321105 121587400998,1362313827639,12115276191861,175201872049228,
%U A321105 2901077831379505,10775830602778083,471448867729594896,6460198350378213465,23761030189140889331,361127251045013068718,4746888122171351400749
%N A321105 One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 11 (mod 13) case (except for n = 0).
%C A321105 For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 11 mod 13 such that k^3 - 5 is divisible by 13^n.
%C A321105 For k not divisible by 13, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.
%H A321105 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%e A321105 The unique number k in [1, 13^2] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 154, so a(2) = 154.
%e A321105 The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 999, so a(3) = 999.
%o A321105 (PARI) a(n) = lift(sqrtn(5+O(13^n), 3) * (-1-sqrt(-3+O(13^n)))/2)
%Y A321105 Cf. A320914, A320915, A321106, A321107, A321108.
%Y A321105 For 5-adic cubic roots, see A290567, A290568, A309444.
%K A321105 nonn
%O A321105 0,2
%A A321105 _Jianing Song_, Aug 27 2019