A271223 Digits of one of the two 3-adic integers sqrt(-2).

1, 1, 2, 0, 0, 2, 0, 1, 0, 0, 0, 2, 1, 2, 0, 1, 0, 1, 1, 0, 2, 2, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 2, 1, 0, 1, 2, 0, 2, 2, 0, 2, 0, 1, 2, 0, 1, 2, 2, 2, 1, 0, 2, 0, 1, 2, 0, 2, 0, 0, 1, 1, 2, 1, 0, 1, 2, 1, 1, 2, 0

Offset: 0

Author

Wolfdieter Lang, Apr 05 2016

Keywords

• nonn
• base
• easy

Comments

This is the scaled first difference sequence of A268924. See the formula.

The digits of the other 3-adic integer sqrt(-2), are given in A271224. See also A268924 for the two 3-adic numbers sqrt(-2), called there u and -u.

a(n) is the unique solution of the linear congruence 2*A268924(n)*a(n) + A271225(n) == 0 (mod 3), n>=1. Therefore only the values 0, 1, and 2 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.

For details see the Wolfdieter Lang link under A268992.

The first k digits in the base 3 representation of Lucas(3^n) give the first k terms of the sequence. For example, the base 3 representation of Lucas(3^5) = 84722519070079276 begins 1 + 1*3 + 2*(3^2) + 0*(3^3) + 0*(3^4) + ... so the sequence begins [1, 1, 2, 0, 0, ...]. - Peter Bala, Nov 15 2022

Examples

			a(4) = 0 because 2*22*3 + 6 = 138 == 0 (mod 3).
a(4) = - 6*(2*22) (mod 3) = -0*(2*1) (mod 3) = 0.
A268924(5) =  22 = 1*3^0 + 1*3^1 + 2*3^2 + 0*3^3 + 0*3^4.
		

References

• Trygve Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964, pp. 86 and 77-78.

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. A006267, A268924, A268992, A271224, A271225.

Programs

Maple

# uses properties of the numbers Lucas(3^n) = A006267(n)
a := proc(n) option remember; if n = 1 then 1 else irem(a(n-1)^3 + 3*a(n-1), 3^n) end if; end proc:
convert(a(70), base, 3); # Peter Bala, Nov 15 2022

PARI

a(n) = truncate(sqrt(-2+O(3^(n+1))))\3^n; \\ Michel Marcus, Apr 09 2016

Formula

a(n) = (b(n+1) - b(n))/3^n, n >= 0, with b(n) = A268924(n), n >= 0.

a(n) = - A271225(n)*2*A268924(n) (mod 3), n >= 1. Solution of the linear congruence given above in a comment. See, e.g., Nagell, Theorem 38 pp. 77-78.

A268924(n+1) = sum(a(k)*3^k, k=0..n), n >= 0.