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A321076 One of the two successive approximations up to 11^n for 11-adic integer sqrt(5). Here the 4 (mod 11) case (except for n = 0).

%I A321076 #11 Nov 14 2018 08:46:21
%S A321076 0,4,48,1258,6582,6582,1456041,10313846,166211214,1666723381,
%T A321076 18172357218,95984631021,2663789666520,24632788303567,162723636879291,
%U A321076 542473470462532,33960458825787740,493457757461509350,3020692899957978205,58619866034880293015,547892589622196663343
%N A321076 One of the two successive approximations up to 11^n for 11-adic integer sqrt(5). Here the 4 (mod 11) case (except for n = 0).
%C A321076 For n > 0, a(n) is the unique solution to x^2 == 5 (mod 11^n) in the range [0, 11^n - 1] and congruent to 4 modulo 11.
%C A321076 A321077 is the approximation (congruent to 7 mod 11) of another square root of 5 over the 11-adic field.
%H A321076 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A321076 For n > 0, a(n) = 11^n - A321077(n).
%F A321076 a(n) = Sum_{i=0..n-1} A321078(i)*11^i.
%e A321076 4^2 = 16 = 5 + 1*11.
%e A321076 48^2 = 2304 = 5 + 19*11^2.
%e A321076 1258^2 = 1582564 = 5 + 1189*11^3.
%o A321076 (PARI) a(n) = truncate(sqrt(5+O(11^n)))
%Y A321076 Cf. A321077, A321078.
%K A321076 nonn
%O A321076 0,2
%A A321076 _Jianing Song_, Oct 27 2018