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Digits of one of the two 17-adic integers sqrt(2) that is related to A322560.
6

%I #14 Aug 29 2019 11:17:17

%S 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 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 14,6,11,16,6,6,5,5,14,13,2,6,5,14,10,4,16,12

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

%C 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 Seiichi Manyama, <a href="/A322562/b322562.txt">Table of n, a(n) for n = 0..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%F a(n) = (A322560(n+1) - A322560(n))/17^n.

%F For n > 0, a(n) = 16 - A322561(n).

%F Equals A309989*A322565 = A309990*A322566.

%e 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 (PARI) a(n) = truncate(-sqrt(2+O(17^(n+1))))\17^n

%Y Cf. A322559, A322560.

%Y Digits of 17-adic square roots:

%Y A309989, A309990 (sqrt(-1));

%Y A322561, this sequence (sqrt(2));

%Y A322565, A322566 (sqrt(-2)).

%K nonn,base

%O 0,1

%A _Jianing Song_, Aug 29 2019