| x [ n ]| = x [ n –1] + x [ n +1] has period 9

$$ | x _ {n} | = x _ {n - 1} + x _ {n + 1} \mathrm {h a s p e r i o d} g. $$

This morning I heard the at first sight surprising theorem that the sequence of real numbers $x_{n}$ $(-\infty \leq n\leq +\infty)$ such that $|x_{n}| = x_{n - 1} + x_{n + 1}$ has a period of length 8. Here is my proof.

Since $x_{n - 1} + x_{n + 1}\geq 0$ , there exists an element $x_{i}$ such that $x_{i}\geq 0$ . Since $x_{i + 1} + x_{i + 1}\geq 0$ $x_{i - 1}\geq 0$ v $x_{i + 1}\geq 0$ , i.e. the sequence contains to successive elements p and q , such that p20 A q20 Without loss of generality we may choose p49 . But this means that the sequence contains 3 consecutive nonnegative elements p q q-p or, after renaming p p+ r for nonnegative p and r. Let r5p and let us extend the sequence in the direction of p -since the relation

$$ x _ {0} = p \quad (x _ {0} \geq 0) $$

$$ x _ {1} = p + r \quad (x _ {1} \geq 0) $$

$$ x _ {2} = r \quad (x _ {2} \geq 0) $$

$$ x _ {3} = - p \quad (x _ {3} \leq 0) $$

$$ x _ {4} = p - r \quad (x _ {4} \geq 0) $$

$$ x _ {5} = 2 \cdot p - r (x _ {5} \geq 0) $$

$$ x _ {6} = p \quad (x _ {6} > 0) $$

$$ x _ {7} = r - p \quad (x _ {7} \leq 0) $$

$$ x _ {8} = - r \quad (x _ {8} \leq 0) $$

$$ x _ {g} = p \quad (x _ {g} > 0) $$

$$ x _ {1 0} = p + r \quad (x _ {1 0} \geqslant 0) $$

For $x_{n}$ is symmetric in $x_{n + 1}$ and $x_{n + 1}$ , the direction of indexing is (relevant)- . With $x_0 = x_0$ and $X_{1} = X_{10}$ , the theorem has been proved without case analysis!

2g August 1983 prof. dr. Edsger W. Dgksha Burroughs Research Fellow

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