A269591 - OEIS

A269591

Digits of one of the two 5-adic integers sqrt(-4).

1, 2, 0, 2, 3, 0, 4, 2, 3, 3, 4, 4, 3, 1, 1, 3, 4, 0, 3, 1, 2, 0, 3, 1, 1, 0, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 4, 3, 2, 2, 3, 2, 4, 4, 0, 3, 1, 4, 0, 3, 3, 1, 0, 1, 3, 3, 2, 3, 3, 3, 4, 4, 3, 1, 3, 1

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OFFSET

0,2

COMMENTS

This is the scaled first difference sequence of A268922.

The digits of the other 5-adic integer sqrt(-4), are given in A269592. See also A268922 for the two 5-adic numbers sqrt(-4), called u and -u.

a(n) is the unique solution of the linear congruence 2*A268922(n)*a(n) + A269593(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) = 1 follows from the formula given below.

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.

FORMULA

a(n) = (b(n+1) - b(n))/5^n, n >= 0, with b(n) = A268922(n), n >= 0.

a(n) = -A269593(n)*(2*A268922(n))^3 (mod 5), n >= 1. Solution of the linear congruence see, e.g., Nagell, Theorem 38 pp. 77-78.

A268922(n+1) = sum(a(k)*5^k, k=0..n), n >= 0.

EXAMPLE

a(4) = 3 because 2*261*3 + 109 = 1675 == 0 (mod 5).

a(4) = - 109*(2*261)^3 (mod 5) = -(-1)*(2*1)^3 (mod 5) = 8 (mod 5) = 3.

A268922(5) = 2136 = 1*5^0 + 2*5^1 + 0*5^2 + 2*5^3 + 3*5^4.

PROG