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A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.

%I A325490 #12 Mar 24 2023 02:45:46
%S A325490 2,4,0,3,0,2,0,4,0,3,0,2,2,2,1,2,1,4,0,3,4,2,1,4,1,1,2,0,0,3,0,1,1,3,
%T A325490 1,4,4,0,2,4,0,4,1,2,0,1,2,3,2,4,2,4,1,3,0,2,1,0,3,3,3,3,0,2,2,3,1,1,
%U A325490 4,1,1,0,1,4,0,3,3,3,0,3,0,0,4,0,3,2,3,1
%N A325490 Digits of one of the four 5-adic integers 6^(1/4) that is congruent to 2 mod 5.
%C A325490 One of the two square roots of A324026, where an A-number represents a 5-adic number. The other square root is A325491.
%C A325490 For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.
%H A325490 Robert Israel, <a href="/A325490/b325490.txt">Table of n, a(n) for n = 0..10000</a>
%H A325490 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A325490 Equals A325489*A210850 = A325492*A210851.
%F A325490 a(n) = (A325485(n+1) - A325485(n))/13^n.
%F A325490 For n > 0, a(n) = 4 - A325491(n).
%e A325490 The unique number k in [1, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 22 = (42)_5, so the first three terms are 2, 4 and 0.
%p A325490 S:= select(t -> op([1,3,1],t)=2, [padic:-rootp(_Z^4-6,5,100)]):
%p A325490 op([1,1,3],S); # _Robert Israel_, Mar 23 2023
%o A325490 (PARI) a(n) = lift(sqrtn(6+O(5^(n+1)), 4) * sqrt(-1+O(5^(n+1))))\5^n
%Y A325490 Cf. A210850, A210851, A324026, A325484, A325485, A325486, A325487.
%Y A325490 Digits of p-adic fourth-power roots:
%Y A325490 A325489, this sequence, A325491, A325492 (5-adic, 6^(1/4));
%Y A325490 A324085, A324086, A324087, A324153 (13-adic, 3^(1/4)).
%K A325490 nonn,base
%O A325490 0,1
%A A325490 _Jianing Song_, Sep 07 2019