A324030 - OEIS

%I #19 Mar 11 2023 07:56:17

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

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

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

%N Digits of one of the two 5-adic integers sqrt(-6) that is related to A324028.

%C This square root of -6 in the 5-adic field ends with digit 3. The other, A324029, ends with digit 2.

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

%F a(n) = (A324028(n+1) - A324028(n))/5^n.

%F For n > 0, a(n) = 4 - A324029(n).

%F Equals A210850*A324025 = A210851*A324026, where each A-number represents a 5-adic number.

%e The solution to x^2 == -6 (mod 5^4) such that x == 3 (mod 5) is x == 463 (mod 5^4), and 463 is written as 3323 in quinary, so the first four terms are 3, 2, 3 and 3.

%o (PARI) a(n) = truncate(-sqrt(-6+O(5^(n+1))))\5^n

%Y Cf. A324027, A324028.

%Y Digits of 5-adic square roots:

%Y A324029, sequence (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 07 2019