A076453 a(n+2) = abs(a(n+1)) - a(n), a(0)=1, a(1)=0.

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

Offset

0,5

Comments

For any initial values a(0), a(1), the sequence of real numbers {a(n)} satisfying the relation a(n+2) = |a(n+1)| - a(n) is periodic with period 9.

Links

Table of n, a(n) for n=0..99.

M. Brown and J. F. Slifker, Solution to Problem 6439, Amer. Math. Monthly, 92 (1985), p. 218.

M. Crampin, Piecewise linear recurrence relations, Math. Gaz., November 1992, p. 355.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 1).

Mathematica

LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1}, {1, 0, -1, 1, 2, 1, -1, 0, 1}, 100] (* Ray Chandler, Aug 26 2015 *)

Crossrefs

Sequence in context: A145865 A341281 A076452 * A263657 A261769 A005590

Adjacent sequences: A076450 A076451 A076452 * A076454 A076455 A076456

Keyword

sign

Author

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 07 2002

Status

approved