Project Euler Forum

Someone posed this problem to me: Is there a number whose decimal expansion is equal to its continued fraction expansion? That is, if a,b,c,d,etc are digits, is there a number satisfying
a.bcde... = [a;b,c,d,e,...] ?

With some coding, I found a number that satisfies this criteria for the first four digits after the decimal point, but I haven't found a better one. Can you?

Can one also find a number that satisfies the criteria for all digits, or prove such a number cannot exist?

Just my thoughts:
If you use the strict definition of continued fractions, there is no more number with more than a(=any integer) and b=3.
But if you allow any of the digits of the continued fraction to be 0, then the search is much more complicated...