A327305 - OEIS

%I #18 Aug 31 2020 03:44:28

%S 4,0,3,0,0,1,1,4,2,0,2,2,3,2,4,4,1,1,2,2,3,0,2,2,4,2,1,4,1,4,0,0,0,2,

%T 4,1,1,3,1,1,0,4,1,2,1,2,2,1,1,2,0,0,3,1,2,0,4,2,0,3,4,4,0,0,0,0,1,4,

%U 0,3,4,0,1,4,4,3,3,0,2,3,2,3,3,3,1,4,2,4

%N Digits of one of the two 5-adic integers sqrt(-9) that is related to A327303.

%C This is the 5-adic solution to x^2 = -9 that ends in 4. A327304 gives the other solution that ends in 1.

%H Robert Israel, <a href="/A327305/b327305.txt">Table of n, a(n) for n = 0..10000</a>

%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 (A327303(n) + m*5^n)^2 + 9 is divisible by 5^(n+1).

%F a(n) = (A327303(n+1) - A327303(n))/5^n.

%F For n > 0, a(n) = 4 - A327304(n).

%e Equals ...1131142000414124220322114423220241100304.

%p op([1,1,3], select(t -> padic:-ratvaluep(t,1)=4, [padic:-rootp(x^2+9,5,100)])); # _Robert Israel_, Aug 31 2020

%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 A327304, this sequence (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 0,1

%A _Jianing Song_, Sep 16 2019