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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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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.
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REFERENCES
| M. Crampin, Piecewise linear recurrence relations, Math. Gaz., November 1992, p. 355.
J. F. Slifker, Solution to Problem 6439 proposed by M. Brown, Amer. Math. Monthly, 92 (1985), p. 218.
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FORMULA
| a(n)=(1/81)*{n mod 9 - 8*[(n + 1) mod 9] - 8*[(n + 2) mod 9] + 19*[(n + 3) mod 9] + 10*[(n + 4) mod 9]-8*[(n + 5) mod 9]-17*[(n + 6) mod 9] + 10*[(n + 7) mod 9] + 10*[(n + 8) mod 9]} with n>=0 - Paolo P. Lava (paoloplava(AT)gmail.com), Nov 21 2006
a(n)=1/81*{(n mod 9)-8*[(n+1) mod 9]-8*[(n+2) mod 9]+19*[(n+3) mod 9]+10*[(n+4) mod 9]-8*[(n+5) mod 9]-17*[(n+6) mod 9]+10*[(n+7) mod 9]+10*[(n+8) mod 9]} with n>=0 - Paolo P. Lava (paoloplava(AT)gmail.com), Nov 27 2006
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CROSSREFS
| Sequence in context: A101808 A145865 A076452 * A005590 A142598 A037800
Adjacent sequences: A076450 A076451 A076452 * A076454 A076455 A076456
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KEYWORD
| sign
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AUTHOR
| Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 07 2002
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