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?

## Decimal expansion = Continued fraction

### Re: Decimal expansion = Continued fraction

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...

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...