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A076453
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a(n+2) = abs(a(n+1)) - a(n), a(0)=1, a(1)=0.
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0
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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
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OFFSET
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0,5
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COMMENTS
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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.
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LINKS
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 07 2002
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STATUS
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approved
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