%I #16 Jul 31 2017 12:56:58

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

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

%N Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

%C This is the scaled first difference sequence of A269590.

%C The digits of the other 5-adic integer sqrt(-4), are given in A269591. See also A268922 for the two 5-adic numbers -u and u.

%C a(n) is the unique solution of the linear congruence 2*A269590(n)*a(n) + A269594(n) == 0 (mod 5), n>=1. Therefore only the values 0, 1, 2, 3 and 4 appear. See the Nagell reference given in A268922, eq. (6) on p. 86, adapted to this case. a(0) = 4 follows from the formula given below.

%H Seiichi Manyama, <a href="/A269592/b269592.txt">Table of n, a(n) for n = 0..10000</a>

%H BCMATH Congruence Programs, <a href="http://www.numbertheory.org/php/p-adic.html">Finding a p-adic square root of a quadratic residue (mod p), p an odd prime.</a>.

%F a(n) = (b(n+1) - b(n))/5^n, n>=0, with b(n) = A269590(n), n >= 0.

%F a(n) = -A269594(n)*(2*A269590(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.

%F A269590(n+1) = sum(a(k)*5^k, k=0..n), n>=0.

%F a(n) = 4 - A269591(n) if n > 0 and a(0) = 5 - A269591(0) = 5-4 = 1. - _Michel Marcus_, Mar 31 2016

%e a(4) = -212*(2*364)^3 (mod 5) = -2*(2*(-1))^3 (mod 5) = 1.

%o (PARI) a(n) = truncate(-sqrt(-4+O(5^(n+1))))\5^n; \\ _Michel Marcus_, Mar 04 2016

%Y Cf. A269590, A269591 (companion), A269594.

%K nonn,easy

%O 0,1

%A _Wolfdieter Lang_, Mar 02 2016