A269592 Digits of one of the two 5-adic integers sqrt(-4). Here the ones related to A269590.

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, 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

Offset: 0

Author

Wolfdieter Lang, Mar 02 2016

Keywords

• nonn
• easy

Comments

This is the scaled first difference sequence of A269590.

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.

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.

Examples

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

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000
• BCMATH Congruence Programs, Finding a p-adic square root of a quadratic residue (mod p), p an odd prime..

Crossrefs

Cf. A269590, A269591 (companion), A269594.

Programs

PARI

a(n) = truncate(-sqrt(-4+O(5^(n+1))))\5^n; \\ Michel Marcus, Mar 04 2016

Formula

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

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.

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

a(n) = 4 - A269591(n) if n > 0 and a(0) = 5 - A269591(0) = 5-4 = 1. - Michel Marcus, Mar 31 2016