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A322562 Digits of one of the two 17-adic integers sqrt(2) that is related to A322560.

%I A322562 #14 Aug 29 2019 11:17:17
%S A322562 11,2,2,8,11,12,2,2,9,14,1,1,5,11,10,9,14,2,10,2,1,0,13,8,2,11,4,0,16,
%T A322562 12,9,16,8,6,14,0,0,1,7,9,4,7,2,2,11,4,13,12,9,7,7,14,14,2,11,7,4,10,
%U A322562 14,6,11,16,6,6,5,5,14,13,2,6,5,14,10,4,16,12
%N A322562 Digits of one of the two 17-adic integers sqrt(2) that is related to A322560.
%C A322562 This square root of 2 in the 17-adic field ends with digit 11 (B when written as a 17-adic number). The other, A322561, ends with digit 6.
%H A322562 Seiichi Manyama, <a href="/A322562/b322562.txt">Table of n, a(n) for n = 0..10000</a>
%H A322562 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A322562 a(n) = (A322560(n+1) - A322560(n))/17^n.
%F A322562 For n > 0, a(n) = 16 - A322561(n).
%F A322562 Equals A309989*A322565 = A309990*A322566.
%e A322562 The solution to x^2 == 2 (mod 17^4) such that x == 11 (mod 17) is x == 39927 (mod 17^4), and 39927 is written as 822B in heptadecimal, so the first four terms are 11, 2, 2 and 8.
%o A322562 (PARI) a(n) = truncate(-sqrt(2+O(17^(n+1))))\17^n
%Y A322562 Cf. A322559, A322560.
%Y A322562 Digits of 17-adic square roots:
%Y A322562 A309989, A309990 (sqrt(-1));
%Y A322562 A322561, this sequence (sqrt(2));
%Y A322562 A322565, A322566 (sqrt(-2)).
%K A322562 nonn,base
%O A322562 0,1
%A A322562 _Jianing Song_, Aug 29 2019