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A325486 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 3 (mod 5) case (except for n = 0).

%I A325486 #7 Sep 11 2019 20:34:59
%S A325486 0,3,3,103,228,2728,8978,71478,71478,1633978,3587103,42649603,
%T A325486 140305853,628587103,3069993353,21380540228,82415696478,540179368353,
%U A325486 540179368353,15798968430853,34872454758978,34872454758978,988546771165228,8141104144212103,8141104144212103
%N A325486 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 3 (mod 5) case (except for n = 0).
%C A325486 For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 3 mod 5 such that k^4 - 6 is divisible by 5^n.
%C A325486 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 A325486 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A325486 a(n) = A325484(n)*A048899(n) mod 13^n = A325485(n)*A048898(n) mod 13^n.
%F A325486 For n > 0, a(n) = 5^n - A325485(n).
%F A325486 a(n)^2 == A324024(n) (mod 5^n).
%e A325486 The unique number k in [1, 5^2] and congruent to 3 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 3, so a(2) = 3.
%e A325486 The unique number k in [1, 5^3] and congruent to 3 modulo 5 such that k^4 - 6 is divisible by 5^3 is k = 103, so a(3) = 103.
%o A325486 (PARI) a(n) = lift(-sqrtn(6+O(5^n), 4) * sqrt(-1+O(5^n)))
%Y A325486 Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.
%Y A325486 Approximations of p-adic fourth-power roots:
%Y A325486 A325484, A325485, this sequence, A325487 (5-adic, 6^(1/4));
%Y A325486 A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).
%K A325486 nonn
%O A325486 0,2
%A A325486 _Jianing Song_, Sep 07 2019