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Let us suppose that $\{\alpha_{n}\}_{n \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers and that the number obtained by concatenating the decimal representations of the elements of $\{\alpha_{n}\}_{n \in \mathbb{N}}$ after the decimal point, i.e.,

$0.\alpha_{1}\alpha_{2}\alpha_{3}\ldots$

has period $s$ (e.g., $0.12 \mathbf{12} \mathrm{121212}...$ has period 2).

If $a_{k}$ denotes the number of elements in $\{\alpha_{n}\}_{n \in \mathbb{N}}$ with exactly $k$ digits in their decimal representation, does the inequality

$a_{k} \leq s$

always hold?

What would be, in your opinion, the right way to approach this question? I've tried a proof by exhaustion without much success. I'd really appreciate any (self-contained) hints you can provide me with.

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From your question it is not fully clear whether you know if the answer is yes, or no, or if you want to prove that $\alpha_k \leq s$ actually holds. If you don't know, the best idea would seem to be to look for a counterexample. Also, it may be more accurate to say that the concatenated sequence has length $\leq s $ –  gary Jul 27 '11 at 15:59
    
Also, if understood the problem well, it would seem you can get a counterexample by using s=1, and fiding sequences that generate irrational numbers; I think concatenating the elements of an arithmetic progression would do. –  gary Jul 27 '11 at 16:06
    
@gary: I think you misunderstood the question; see the two answers given. –  joriki Jul 27 '11 at 16:28
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2 Answers 2

All numbers with exactly $k$ digits are consecutive in $\{\alpha_{n}\}_{n \in \mathbb{N}}$. If $a_k>s$, then string the first $s$ numbers with $k$ digits together. The resulting string has $ks$ digits, which is exactly $k$ periods, so you're now at the same point in the period as you were at the beginning of the first number with $k$ digits. By peridocity, the next number with $k$ digits would have to be the same as the first, which isn't possible since the sequence strictly increases.

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If the period is $s$ then there are essentially $s$ starting places in the recurring decimal for a $k$-digit integer - begin at the first digit of the decimal, the second etc - beyond $s$ you get the same numbers coming round again. If you had $a_k > s$ then two of your $\alpha_n$ with $k$ digits would be the same by the pigeonhole principle.

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