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%I #9 Sep 07 2019 19:09:02
%S 0,10,101,1622,14804,214731,214731,9868349,637353519,637353519,
%T 637353519,552071320915,552071320915,23850156443396,1538225689404661,
%U 48786742317796129,560645672458703699,5218561936740962586,13868977856122300519,126324384808079693648
%N One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (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 10 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)*A286840(n) mod 13^n = A324084(n)*A286841(n) mod 13^n.
%F For n > 0, a(n) = 13^n - A324082(n).
%F a(n)^2 == A322086(n) (mod 13^n).
%e The unique number k in [1, 13^2] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 101, so a(2) = 101.
%e The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622, so a(3) = 1622.
%o (PARI) a(n) = lift(-sqrtn(3+O(13^n), 4))
%Y Cf. A286840, A286841, A322085, A324077, A324082, A324084, A324085, A324086, A324087, A324153.
%K nonn
%O 0,2
%A _Jianing Song_, Sep 01 2019
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