AoPS - 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 $m$ .

hxy09

01.06.2009 08:23

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

JBL

16.06.2009 04:10

This is sequence A007538 in the OEIS.