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A322566 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322564.

%I A322566 #26 Aug 29 2019 11:29:26
%S A322566 10,15,4,16,7,16,0,8,11,0,2,16,15,1,5,0,1,8,3,1,5,11,5,13,7,0,0,1,13,
%T A322566 13,16,1,9,1,0,13,2,7,4,11,14,14,12,4,4,5,5,16,7,1,4,14,7,2,14,6,10,
%U A322566 16,8,11,1,10,10,2,7,14,6,15,9,14,3,4,13,3,10,0
%N A322566 Digits of one of the two 17-adic integers sqrt(-2) that is related to A322564.
%C A322566 This square root of -2 in the 17-adic field ends with digit 10 (A when written as a 17-adic number). The other, A322565, ends with digit 7.
%H A322566 Seiichi Manyama, <a href="/A322566/b322566.txt">Table of n, a(n) for n = 0..10000</a>
%H A322566 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322566 a(n) = (A322564(n+1) - A322564(n))/17^n.
%F A322566 For n > 0, a(n) = 16 - A322565(n).
%F A322566 Equals A309989*A322562 = A309990*A322561.
%e A322566 The solution to x^2 == -2 (mod 17^4) such that x == 10 (mod 17) is x == 80029 (mod 17^4), and 80029 is written as G4FA in heptadecimal, so the first four terms are 10, 15, 4 and 16.
%o A322566 (PARI) a(n) = truncate(-sqrt(-2+O(17^(n+1))))\17^n
%Y A322566 Cf. A322563, A322564.
%Y A322566 Digits of 17-adic square roots:
%Y A322566 A309989, A309990 (sqrt(-1));
%Y A322566 A322561, A322562 (sqrt(2));
%Y A322566 A322565, this sequence (sqrt(-2)).
%K A322566 nonn,base
%O A322566 0,1
%A A322566 _Jianing Song_, Aug 29 2019