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%I #9 Sep 07 2019 19:08:54
%S 0,3,68,575,13757,156562,4612078,52880168,178377202,9967145854,
%T 137221138330,1240089073122,22746013801566,279024950148857,
%U 2399150696294628,2399150696294628,104770936724476142,3431853982640375347,98586429095835092610,1335595905567366417029
%N One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 3 (mod 13) case (except for n = 0).
%C For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 3 mod 13 such that k^4 - 3 is divisible by 13^n.
%C 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F a(n) = A324077(n)*A286841(n) mod 13^n = A324084(n)*A286840(n) mod 13^n.
%F For n > 0, a(n) = 13^n - A324083(n).
%F a(n)^2 == A322086(n) (mod 13^n).
%e The unique number k in [1, 13^2] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 68, so a(2) = 68.
%e The unique number k in [1, 13^3] and congruent to 3 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 575, so a(3) = 575.
%o (PARI) a(n) = lift(sqrtn(3+O(13^n), 4))
%Y Cf. A286840, A286841, A322085, A324077, A324083, A324084, A324085, A324086, A324087, A324153.
%K nonn
%O 0,2
%A _Jianing Song_, Sep 01 2019
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