A076452 a(n+2) = abs(a(n+1)) - a(n), a(0)=0, a(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, 0, -1, 1, 2, 1, -1, 0

Offset

0,7

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).

Formula

a(n) = a(n-9), with a(0)=0, a(1)=1, a(2)=1, a(3)=0, a(4)=-1, a(5)=1, a(6)=2, a(7)=1, a(8)=-1. - Harvey P. Dale, May 23 2014

Mathematica

RecurrenceTable[{a[0]==0, a[1]==1, a[n]==Abs[a[n-1]]-a[n-2]}, a, {n, 100}] (* or *) LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 1, 1, 0, -1, 1, 2, 1, -1}, 100] (* or *) PadRight[{}, 100, {0, 1, 1, 0, -1, 1, 2, 1, -1}] (* Harvey P. Dale, May 23 2014 *)

Crossrefs

Sequence in context: A101808 A145865 A341281 * A076453 A263657 A261769

Adjacent sequences: A076449 A076450 A076451 * A076453 A076454 A076455

Keyword

sign,easy

Author

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

Status

approved