cp's OEIS Frontend

A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.

%I A324153 #14 Sep 09 2019 04:31:42
%S A324153 11,10,5,11,0,6,0,8,4,6,11,2,8,6,5,4,2,11,0,3,3,5,12,0,9,6,8,7,1,0,9,
%T A324153 1,3,7,4,8,8,10,5,8,1,4,8,2,11,12,10,11,8,9,1,5,9,6,9,10,6,5,9,6,11,
%U A324153 12,9,12,1,4,1,6,1,12,9,7,8,5,3,2,0,6,1,7,11
%N A324153 Digits of one of the four 13-adic integers 3^(1/4) that is congruent to 11 mod 13.
%C A324153 One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324085.
%C A324153 For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.
%H A324153 Robert Israel, <a href="/A324153/b324153.txt">Table of n, a(n) for n = 0..10000</a>
%H A324153 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A324153 Equals A324086*A286839 = A324087*A286838.
%F A324153 a(n) = (A324084(n+1) - A324084(n))/13^n.
%F A324153 For n > 0, a(n) = 12 - A324085(n).
%e A324153 The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986 = (5AB)_13, so the first three terms are 11, 10 and 5.
%p A324153 R:= select(t -> op([1,3,1],t)=11, [padic:-rootp(x^4-3, 13,101)]):
%p A324153 op([1,1,3],R); # _Robert Israel_, Sep 08 2019
%o A324153 (PARI) a(n) = lift(-sqrtn(3+O(13^(n+1)), 4) * sqrt(-1+O(13^(n+1))))\13^n
%Y A324153 Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324087.
%K A324153 nonn,base
%O A324153 0,1
%A A324153 _Jianing Song_, Sep 01 2019