A324025 - OEIS

%I #18 Mar 11 2023 07:55:42

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

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

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

%N Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

%C This square root of 6 in the 5-adic field ends with digit 1. The other, A324026, ends with digit 4.

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

%F a(n) = (A324023(n+1) - A324023(n))/5^n.

%F For n > 0, a(n) = 4 - A324026(n).

%F Equals A210850*A324030 = A210851*A324029, where each A-number represents a 5-adic number.

%e The solution to x^2 == 6 (mod 5^4) such that x == 1 (mod 5) is x == 516 (mod 5^4), and 516 is written as 4031 in quinary, so the first four terms are 1, 3, 0 and 4.

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

%Y Cf. A324023, A324024.

%Y Digits of 5-adic square roots:

%Y A324029, A324030 (sqrt(-6));

%Y A269591, A269592 (sqrt(-4));

%Y A210850, A210851 (sqrt(-1));

%Y this sequence, A324026 (sqrt(6)).

%K nonn,base

%O 0,2

%A _Jianing Song_, Sep 07 2019