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A322090 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 7 (mod 13) case (except for n = 0).

%I A322090 #12 Dec 05 2022 08:18:10
%S A322090 0,7,124,124,13306,70428,1926893,40541365,542529501,2989721664,
%T A322090 45407719156,458983194703,18380587135073,111572927624997,
%U A322090 2231698673770768,2231698673770768,462904735800587581,5120821000082846468,74324148355133549932,1423789031778622267480,10195310774031298931542
%N A322090 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 7 (mod 13) case (except for n = 0).
%C A322090 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 7 modulo 13.
%C A322090 A322089 is the approximation (congruent to 6 mod 13) of another square root of -3 over the 13-adic field.
%H A322090 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322090 For n > 0, a(n) = 13^n - A322089(n).
%F A322090 a(n) = Sum_{i=0..n-1} A322092(i)*13^i.
%F A322090 a(n) = A286840(n)*A322085(n) mod 13^n = A286841(n)*A322086(n) mod 13^n.
%F A322090 a(n) == L(13^n,7) (mod 13^n) == ((7 + sqrt(53))/2)^(13^n) + ((7 - sqrt(53))/2)^(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 A322090 7^2 = 49 = 4*13 - 3.
%e A322090 124^2 = 15376 = 91*13^2 - 3 = 7*13^3 - 3.
%e A322090 13306^2 = 177049636 = 6199*13^4 - 3.
%o A322090 (PARI) a(n) = truncate(-sqrt(-3+O(13^n)))
%Y A322090 Cf. A114525, A286840, A286841, A322085, A322086, A322089, A322092.
%K A322090 nonn,easy
%O A322090 0,2
%A A322090 _Jianing Song_, Nov 26 2018