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Let $s(x)$ is the length of continued fraction expansion of $x$, and let $l(x)$ be the sum of partial quotients. I can prove that for any rational $\alpha$ ratios $\frac{s(\alpha x)}{s(x)}$ and $\frac{l(\alpha x)}{l(x)}$ (for all rational $x$) are bounded with some constants depending on $\alpha$ only.

Is this result new?

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I would be really surprised if it wasn't. After all, for l(x) this is just the length of the Euclid algorithm for x=p/q, so if we, say, multiply it by 2/3, then it is just the Euclid algorithm for 2p/3q... must be well known. Perhaps, even Euclid himself knew it. ;) –  Nikita Sidorov Oct 9 '10 at 10:49
    
Yes, it is more or less clear. It is not surprising. It is not hard. But was it already proved or not? –  Alexey Ustinov Oct 9 '10 at 12:07
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@Nikita It is not Euclid algorithm for 2p/3q at all :), and all the proofs I know are quite non-trivial, though some of them are respectively short. –  Fedor Petrov Oct 9 '10 at 13:10

1 Answer 1

up vote 7 down vote accepted

It definitely is not new for the length, and I am nearly sure that is not for height either.

See, for example,

Labhalla, Salah; Lombardi, Henri Transformation homographique appliqu´ee `a un d´eveloppement en fraction continue fini ou infini. (French) [Fractional linear transformations applied to finite and infinite continued fractions] Acta Arith. 73 (1995), no. 1, 29–41.

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Thank you Fedor. I'll be greatful for any other refferences concerning this question. –  Alexey Ustinov Oct 9 '10 at 22:29
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see also Knuth (Art of Computer Programming), solution of the ex. 4.5.3.15 By the way, I know all these references just because this was a problem proposed there (mathsoc.spb.ru/konkurs/contest10.pdf, how to make normal links here?) (by Elena Golubeva), and students not only proposed their original solutions, but one of them found a lot of references. –  Fedor Petrov Oct 10 '10 at 15:40
    
Do you have these solutions and references in electronic form? –  Alexey Ustinov Oct 10 '10 at 22:36
    
@Alexey: I sent you smth by e-mail –  Fedor Petrov Oct 11 '10 at 7:52

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