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A324077 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0).

%I A324077 #31 Sep 07 2019 19:08:41
%S A324077 0,2,28,1211,3408,346140,2573898,60495606,311489674,6837335442,
%T A324077 70464331680,208322823529,18129926763899,111322267253823,
%U A324077 1928572906807341,29490207606702364,438977351719428420,7093143443551226830,15743559362932564763,1365208442786421282311
%N A324077 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0).
%C A324077 For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 2 mod 13 such that k^4 - 3 is divisible by 13^n.
%C A324077 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 A324077 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A324077 a(n) = A324082(n)*A286840(n) mod 13^n = A324083(n)*A286841(n) mod 13^n.
%F A324077 For n > 0, a(n) = 13^n - A324084(n).
%F A324077 a(n)^2 == A322085(n) (mod 13^n).
%e A324077 The unique number k in [1, 13^2] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 28, so a(2) = 28.
%e A324077 The unique number k in [1, 13^3] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1211, so a(3) = 1211.
%o A324077 (PARI) a(n) = lift(sqrtn(3+O(13^n), 4) * sqrt(-1+O(13^n)))
%Y A324077 Cf. A286840, A286841, A322085, A324082, A324083, A324084, A324085, A324086, A324087, A324153.
%K A324077 nonn
%O A324077 0,2
%A A324077 _Jianing Song_, Sep 01 2019