Bariz - Problem

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Tags: floor function, induction, More Sequences

Peter

25.05.2007 03:25

The infinite sequence of 2's and 3's \[\begin{array}{l}2,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3, \\ 3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,\cdots \end{array}\] has the property that, if one forms a second sequence that records the number of 3's between successive 2's, the result is identical to the given sequence. Show that there exists a real number $r$ such that, for any $n$, the $n$th term of the sequence is 2 if and only if $n = 1+\lfloor rm \rfloor$ for some nonnegative integer $m$.

We claim that $ m$ th $ 2$ appears as $ [(2 + \sqrt {3})(m - 1)] + 1$th term is this sequence Prove by induction,the foundation is trivial,and for every positive integer $ \le m$ is true. Let's take a look at thr case of $ m + 1$ $ \exists$ $ r,r\in N$ s.t. $ [(2 + \sqrt {3})(r - 1)] + 1\le m < [(2 + \sqrt {3})r] + 1$ After some simple calculation.We obtain $ r = [\frac {m}{2 + \sqrt {3}}] + 1$ This yields the number of $ 2$ $ 3$ in the first $ m$ terms are $ r$ and $ m - r$ respectively So the position of $ m + 1$th $ 2$ is $ 2r + 3(m - r) + m + 1 = 4m - r + 1 = [(2 + \sqrt {3})m] + 1$ which complete yhe induction. QED