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A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).

%I A322085 #15 Jun 13 2019 04:11:57
%S A322085 0,4,108,1122,18698,361430,1104016,5930825,570667478,7912243967,
%T A322085 113957237697,251815729546,11004778093768,104197118583692,
%U A322085 3132948184506222,26757206498701956,589802029653700283,7909384730668678534,85763128005100719931,648040162764887685576
%N A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).
%C A322085 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 4 modulo 13.
%C A322085 A322086 is the approximation (congruent to 9 mod 13) of another square root of 3 over the 13-adic field.
%H A322085 Robert Israel, <a href="/A322085/b322085.txt">Table of n, a(n) for n = 0..896</a>
%H A322085 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322085 For n > 0, a(n) = 13^n - A322086(n).
%F A322085 a(n) = Sum_{i=0..n-1} A322087(i)*13^i.
%F A322085 a(n) = A286840(n)*A322089(n) mod 13^n = A286841(n)*A322090(n) mod 13^n.
%e A322085 4^2 = 16 = 1*13 + 3.
%e A322085 108^2 = 11664 = 69*13^2 + 3.
%e A322085 1122^2 = 1258884 = 573*13^3 + 3.
%p A322085 S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
%p A322085 S4:= op(select(t -> t[1]=4, S)):
%p A322085 seq(add(S4[i]*13^(i-1),i=1..n-1),n=1..31); # _Robert Israel_, Jun 13 2019
%o A322085 (PARI) a(n) = truncate(sqrt(3+O(13^n)))
%Y A322085 Cf. A286840, A286841, A322086, A322087, A322089, A322090.
%K A322085 nonn
%O A322085 0,2
%A A322085 _Jianing Song_, Nov 26 2018