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A324084 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 11 (mod 13) case (except for n = 0).

%I A324084 #9 Sep 07 2019 19:09:18
%S A324084 0,11,141,986,25153,25153,2252911,2252911,504241047,3767163931,
%T A324084 67394160169,1583837570508,5168158358582,191552839338430,
%U A324084 2008803478891948,21695685407388393,226439257463751421,1557272475830111103,96711847589024828366,96711847589024828366
%N A324084 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 11 (mod 13) case (except for n = 0).
%C A324084 For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 11 mod 13 such that k^4 - 3 is divisible by 13^n.
%C A324084 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 A324084 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A324084 a(n) = A324082(n)*A286841(n) mod 13^n = A324083(n)*A286840(n) mod 13^n.
%F A324084 For n > 0, a(n) = 13^n - A324077(n).
%F A324084 a(n)^2 == A322085(n) (mod 13^n).
%e A324084 The unique number k in [1, 13^2] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 141, so a(2) = 141.
%e A324084 The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986, so a(3) = 986.
%o A324084 (PARI) a(n) = lift(-sqrtn(3+O(13^n), 4) * sqrt(-1+O(13^n)))
%Y A324084 Cf. A286840, A286841, A322085, A324077, A324082, A324083, A324085, A324086, A324087, A324153.
%K A324084 nonn
%O A324084 0,2
%A A324084 _Jianing Song_, Sep 01 2019