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One of the four successive approximations up to 13^n for the 13-adic integer 6^(1/4). This is the 4 (mod 5) case (except for n = 0).
9

%I #7 Sep 11 2019 20:35:35

%S 0,4,4,4,379,1004,10379,26004,104129,1276004,9088504,28619754,

%T 126276004,614557254,3055963504,27470026004,57987604129,57987604129,

%U 820927057254,16079716119754,16079716119754,206814579401004,1637326054010379,6405697636041629,30247555546197879

%N One of the four successive approximations up to 13^n for the 13-adic integer 6^(1/4). This is the 4 (mod 5) case (except for n = 0).

%C For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^4 - 6 is divisible by 5^n.

%C 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%F a(n) = A325485(n)*A048898(n) mod 5^n = A325486(n)*A048899(n) mod 5^n.

%F For n > 0, a(n) = 5^n - A325484(n).

%F a(n)^2 == A324023(n) (mod 5^n).

%e The unique number k in [1, 5^2] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 4, so a(2) = 4.

%e The unique number k in [1, 5^3] and congruent to 4 modulo 5 such that k^4 - 6 is divisible by 5^3 is also k = 4, so a(3) is also 4.

%o (PARI) a(n) = lift(-sqrtn(6+O(5^n), 4))

%Y Cf. A048898, A048899, A324023, A325489, A325490, A325491, A325492.

%Y Approximations of p-adic fourth-power roots:

%Y A325484, A325485, A325486, this sequence (5-adic, 6^(1/4));

%Y A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

%K nonn

%O 0,2

%A _Jianing Song_, Sep 07 2019