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A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.

%I A324025 #18 Mar 11 2023 07:55:42
%S A324025 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 A324025 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 A324025 1,1,0,1,4,2,3,4,4,4,4,0,3,3,1,2,3,2,3,1
%N A324025 Digits of one of the two 5-adic integers sqrt(6) that is related to A324023.
%C A324025 This square root of 6 in the 5-adic field ends with digit 1. The other, A324026, ends with digit 4.
%H A324025 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A324025 a(n) = (A324023(n+1) - A324023(n))/5^n.
%F A324025 For n > 0, a(n) = 4 - A324026(n).
%F A324025 Equals A210850*A324030 = A210851*A324029, where each A-number represents a 5-adic number.
%e A324025 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 A324025 (PARI) a(n) = truncate(sqrt(6+O(5^(n+1))))\5^n
%Y A324025 Cf. A324023, A324024.
%Y A324025 Digits of 5-adic square roots:
%Y A324025 A324029, A324030 (sqrt(-6));
%Y A324025 A269591, A269592 (sqrt(-4));
%Y A324025 A210850, A210851 (sqrt(-1));
%Y A324025 this sequence, A324026 (sqrt(6)).
%K A324025 nonn,base
%O A324025 0,2
%A A324025 _Jianing Song_, Sep 07 2019