A322561 - OEIS

%I #14 Aug 29 2019 11:30:01

%S 6,14,14,8,5,4,14,14,7,2,15,15,11,5,6,7,2,14,6,14,15,16,3,8,14,5,12,

%T 16,0,4,7,0,8,10,2,16,16,15,9,7,12,9,14,14,5,12,3,4,7,9,9,2,2,14,5,9,

%U 12,6,2,10,5,0,10,10,11,11,2,3,14,10,11,2,6,12,0,4

%N Digits of one of the two 17-adic integers sqrt(2) that is related to A322559.

%C This square root of 2 in the 17-adic field ends with digit 6. The other, A322562, ends with digit 11 (B when written as a 17-adic number).

%H Seiichi Manyama, <a href="/A322561/b322561.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) = (A322559(n+1) - A322559(n))/17^n.

%F For n > 0, a(n) = 16 - A322562(n).

%F Equals A309989*A322566 = A309990*A322565.

%e The solution to x^2 == 2 (mod 17^4) such that x == 6 (mod 17) is x == 43594 (mod 17^4), and 43594 is written as 8EE6 in heptadecimal, so the first four terms are 6, 14, 14 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 this sequence, A322562 (sqrt(2));

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

%K nonn,base

%O 0,1

%A _Jianing Song_, Aug 29 2019