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It is possible to show that the square root of two can be expressed as an infinite continued fraction.

sqrt(2) = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2                         = 3 /2  = 1.5
1 + 1/(2 + 1/2)                 = 7 /5  = 1.4
1 + 1/(2 + 1/(2 + 1/2))         = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?