More haskelling. This problem's nice - it's very easy to express recursively. It's easy to express iteratively too, of course. Actually, I find the iterative solution a lot clearer, which probably means my Haskell sucks. :(

euler57

//
//  HASKELL
//

matches i (num, den)    | i == 0 = 0
                        | g (num' + den') > g den' = 1 + f
                        | otherwise = f
                        where   num'    = den
                                den'    = num + (2 * den)
                                f       = matches (i - 1) (num', den')
                                g       = length . show

solution = matches 1000 (1, 2)

main = print $ solution

//
//  PYTHON
//

num = 1
den = 2

matches = 0

for i in xrange(1000):    
    num, den = den, (num + 2 * den)
    if len(str(num + den)) > len(str(den)):
        matches = matches + 1

print matches