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A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).

%I A325485 #7 Sep 11 2019 20:34:16
%S A325485 0,2,22,22,397,397,6647,6647,319147,319147,6178522,6178522,103834772,
%T A325485 592116022,3033522272,9137037897,70172194147,222760084772,
%U A325485 3274517897272,3274517897272,60494976881647,441964703444147,1395639019850397,3779824810866022,51463540631178522
%N A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).
%C A325485 For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 2 mod 5 such that k^4 - 6 is divisible by 5^n.
%C A325485 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 A325485 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A325485 a(n) = A325484(n)*A048898(n) mod 13^n = A325485(n)*A048899(n) mod 13^n.
%F A325485 For n > 0, a(n) = 5^n - A325486(n).
%F A325485 a(n)^2 == A324024(n) (mod 5^n).
%e A325485 The unique number k in [1, 5^2] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 22, so a(2) = 22.
%e A325485 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 also k = 22, so a(3) is also 22.
%o A325485 (PARI) a(n) = lift(sqrtn(6+O(5^n), 4) * sqrt(-1+O(5^n)))
%Y A325485 Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.
%Y A325485 Approximations of p-adic fourth-power roots:
%Y A325485 A325484, this sequence, A325486, A325487 (5-adic, 6^(1/4));
%Y A325485 A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).
%K A325485 nonn
%O A325485 0,2
%A A325485 _Jianing Song_, Sep 07 2019