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A321073 One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 6 (mod 11) case (except for n = 0).

%I A321073 #17 Dec 05 2022 08:35:39
%S A321073 0,6,94,578,3240,135009,296060,2067621,118990647,1619502814,
%T A321073 3977450505,211476847313,782100188535,22751098825582,229887371689168,
%U A321073 609637205272409,38204870730013268,84154600593585429,3622283800088641826,42541704994534262193,654132609478679725103
%N A321073 One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 6 (mod 11) case (except for n = 0).
%C A321073 For n > 0, a(n) is the unique solution to x^2 == 3 (mod 11^n) in the range [0, 11^n - 1] and congruent to 6 modulo 11.
%C A321073 A321072 is the approximation (congruent to 5 mod 11) of another square root of 3 over the 11-adic field.
%H A321073 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A321073 For n > 0, a(n) = 11^n - A321072(n).
%F A321073 a(n) = Sum_{i=0..n-1} A321075(i)*11^i.
%F A321073 a(n) == (3 + sqrt(8))^(11^n) + (3 - sqrt(8))^(11^n) (mod 11^n). - _Peter Bala_, Dec 04 2022
%e A321073 6^2 = 36 = 3 + 3*11.
%e A321073 94^2 = 8836 = 3 + 73*11^2.
%e A321073 578^2 = 334084 = 3 + 251*11^3.
%o A321073 (PARI) a(n) = truncate(-sqrt(3+O(11^n)))
%Y A321073 Cf. A321072, A321075.
%K A321073 nonn,easy
%O A321073 0,2
%A A321073 _Jianing Song_, Oct 27 2018