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%I #14 Sep 20 2019 12:32:02
%S 1,4,1,4,4,3,3,0,2,4,2,2,1,2,0,0,3,3,2,2,1,4,2,2,0,2,3,0,3,0,4,4,4,2,
%T 0,3,3,1,3,3,4,0,3,2,3,2,2,3,3,2,4,4,1,3,2,4,0,2,4,1,0,0,4,4,4,4,3,0,
%U 4,1,0,4,3,0,0,1,1,4,2,1,2,1,1,1,3,0,2,0
%N Digits of one of the two 5-adic integers sqrt(-9) that is related to A327302.
%C This is the 5-adic solution to x^2 = -9 that ends in 1. A327305 gives the other solution that ends in 4.
%H G. P. Michon, <a href="http://www.numericana.com/answer/p-adic.htm#integers">Introduction to p-adic integers</a>, Numericana.
%F For n > 0, a(n) is the unique m in {0, 1, 2, 3, 4} such that (A327302(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).
%F a(n) = (A327302(n+1) - A327302(n))/5^n.
%F For n > 0, a(n) = 4 - A327305(n).
%e Equals ...3313302444030320224122330021224203344141.
%o (PARI) a(n) = truncate(-sqrt(-9+O(5^(n+1))))\5^n
%Y Cf. A327302, A327303.
%Y Digits of 5-adic square roots:
%Y this sequence, A327305 (sqrt(-9));
%Y A324029, A324030 (sqrt(-6));
%Y A269591, A269592 (sqrt(-4));
%Y A210850, A210851 (sqrt(-1));
%Y A324025, A324026 (sqrt(6)).
%K nonn,base
%O 0,2
%A _Jianing Song_, Sep 16 2019
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