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A322564 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-2). This is the 10 (mod 17) case (except for n = 0).

%I A322564 #14 Mar 26 2022 17:46:37
%S A322564 0,10,265,1421,80029,664676,23382388,23382388,3306091772,80039423623,
%T A322564 80039423623,4112027224521,552462368146649,9291795926593064,
%U A322564 19196373959499001,861085506756503646,861085506756503646,49522277382423372127,6667444372473117485543,48856697728676292458570,287929133413827617305723
%N A322564 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-2). This is the 10 (mod 17) case (except for n = 0).
%C A322564 For n > 0, a(n) is the unique solution to x^2 == 2 (mod 17^n) in the range [0, 17^n - 1] and congruent to 10 modulo 17.
%C A322564 A322563 is the approximation (congruent to 7 mod 17) of another square root of -2 over the 17-adic field.
%H A322564 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322564 For n > 0, a(n) = 17^n - A322563(n).
%F A322564 a(n) = Sum_{i=0..n-1} A322566(i)*17^i.
%F A322564 a(n) = A286877(n)*A322560(n) mod 17^n = A286878(n)*A322559(n) mod 17^n.
%e A322564 10^2 = 100 = 6*17 - 2;
%e A322564 265^2 = 70225 = 243*17^2 - 2;
%e A322564 1421^2 = 2019241 = 411*17^3 - 2.
%o A322564 (PARI) a(n) = truncate(-sqrt(-2+O(17^n)))
%Y A322564 Cf. A322565, A322566.
%Y A322564 Approximations of 17-adic square roots:
%Y A322564 A286877, A286878 (sqrt(-1));
%Y A322564 A322559, A322560 (sqrt(2));
%Y A322564 A322563, this sequence (sqrt(-2)).
%K A322564 nonn
%O A322564 0,2
%A A322564 _Jianing Song_, Aug 29 2019