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problem 57 · 11 lines

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<p>It is possible to show that the square root of two can be expressed as an infinite continued fraction.</p>
<p class="center">$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$</p>
<p>By expanding this for the first four iterations, we get:</p>
<p>$1 + \frac 1 2 = \frac  32 = 1.5$<br/>
$1 + \frac 1 {2 + \frac 1 2} = \frac 7 5 = 1.4$<br/>
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 2}} = \frac {17}{12} = 1.41666 \dots$<br/>
$1 + \frac 1 {2 + \frac 1 {2+\frac 1 {2+\frac 1 2}}} = \frac {41}{29} = 1.41379 \dots$<br/></p>
<p>The next three expansions are $\frac {99}{70}$, $\frac {239}{169}$, and $\frac {577}{408}$, but the eighth expansion, $\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.</p>
<p>In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?</p>
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