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A322086 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 9 (mod 13) case (except for n = 0).

%I A322086 #13 Jun 13 2019 04:12:07
%S A322086 0,9,61,1075,9863,9863,3722793,56817692,245063243,2692255406,
%T A322086 23901254152,1540344664491,12293307028713,198677988008561,
%U A322086 804428201193067,24428686515388801,75614579529479558,741031188712659399,26692278946856673198,813880127610558425101,11047322160238681199840
%N A322086 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 9 (mod 13) case (except for n = 0).
%C A322086 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 9 modulo 13.
%C A322086 A322085 is the approximation (congruent to 4 mod 13) of another square root of 3 over the 13-adic field.
%H A322086 Robert Israel, <a href="/A322086/b322086.txt">Table of n, a(n) for n = 0..896</a>
%H A322086 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322086 For n > 0, a(n) = 13^n - A322085(n).
%F A322086 a(n) = Sum_{i=0..n-1} A322088(i)*13^i.
%F A322086 a(n) = A286840(n)*A322090(n) mod 13^n = A286841(n)*A322089(n) mod 13^n.
%e A322086 9^2 = 81 = 6*13 + 3.
%e A322086 61^2 = 3721 = 22*13^2 + 3.
%e A322086 1075^2 = 1155625 = 526*13^3 + 3.
%p A322086 S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
%p A322086 S9:= op(select(t -> t[1]=9, S)):
%p A322086 seq(add(S9[i]*13^(i-1),i=1..n-1),n=1..31); # _Robert Israel_, Jun 13 2019
%o A322086 (PARI) a(n) = truncate(-sqrt(3+O(13^n)))
%Y A322086 Cf. A286840, A286841, A322085, A322088, A322089, A322090.
%K A322086 nonn
%O A322086 0,2
%A A322086 _Jianing Song_, Nov 26 2018