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A322565 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322563.

%I A322565 #37 Aug 29 2019 11:29:11
%S A322565 7,1,12,0,9,0,16,8,5,16,14,0,1,15,11,16,15,8,13,15,11,5,11,3,9,16,16,
%T A322565 15,3,3,0,15,7,15,16,3,14,9,12,5,2,2,4,12,12,11,11,0,9,15,12,2,9,14,2,
%U A322565 10,6,0,8,5,15,6,6,14,9,2,10,1,7,2,13,12,3,13,6,16
%N A322565 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322563.
%C A322565 This square root of -2 in the 17-adic field ends with digit 7. The other, A322566, ends with digit 10 (A when written as a 17-adic number).
%H A322565 Seiichi Manyama, <a href="/A322565/b322565.txt">Table of n, a(n) for n = 0..10000</a>
%H A322565 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322565 a(n) = (A322563(n+1) - A322563(n))/17^n.
%F A322565 For n > 0, a(n) = 16 - A322566(n).
%F A322565 Equals A309989*A322561 = A309990*A322562.
%e A322565 The solution to x^2 == -2 (mod 17^4) such that x == 7 (mod 17) is x == 3492 (mod 17^4), and 3492 is written as C17 in heptadecimal, so the first four terms are 7, 1, 12 and 0.
%o A322565 (PARI) a(n) = truncate(sqrt(-2+O(17^(n+1))))\17^n
%Y A322565 Cf. A322563, A322564.
%Y A322565 Digits of 17-adic square roots:
%Y A322565 A309989, A309990 (sqrt(-1));
%Y A322565 A322561, A322562 (sqrt(2));
%Y A322565 this sequence, A322566 (sqrt(-2)).
%K A322565 nonn,base
%O A322565 0,1
%A A322565 _Jianing Song_, Aug 29 2019