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A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).

%I A322089 #12 Dec 05 2022 08:18:04
%S A322089 0,6,45,2073,15255,300865,2899916,22207152,273201220,7614777709,
%T A322089 92450772693,1333177199334,4917497987408,191302178967256,
%U A322089 1705677711928521,48954194340319989,202511873382592260,3529594919298491465,38131258596823843197,38131258596823843197,8809653000849500507259
%N A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).
%C A322089 For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 6 modulo 13.
%C A322089 A322090 is the approximation (congruent to 7 mod 13) of another square root of -3 over the 13-adic field.
%H A322089 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322089 For n > 0, a(n) = 13^n - A322090(n).
%F A322089 a(n) = Sum_{i=0..n-1} A322091(i)*13^i.
%F A322089 a(n) = A286840(n)*A322086(n) mod 13^n = A286841(n)*A322085(n) mod 13^n.
%F A322089 a(n) == L(13^n,6) (mod 13^n) == (3 + sqrt(10))^(13^n) + (3 - sqrt(10))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - _Peter Bala_, Dec 05 2022
%e A322089 6^2 = 36 = 3*13 - 3.
%e A322089 45^2 = 2025 = 12*13^2 - 3.
%e A322089 2073^2 = 4297329 = 1956*13^3 - 3.
%o A322089 (PARI) a(n) = truncate(sqrt(-3+O(13^n)))
%Y A322089 Cf. A114525, A286840, A286841, A322085, A322086, A322090, A322091.
%K A322089 nonn,easy
%O A322089 0,2
%A A322089 _Jianing Song_, Nov 26 2018