If a number with more than one digit occurs in the fraction, it should be expanded to as many digits in the expansion. I will be even more impressed, however, if the fraction consists entirely of 1-digit terms. No integers allowed.
There is no solution with single digits. You clearly have to start with a $3$ to get $0.3 \approx [0;3]=\frac 13$ Now whatever you put next the CF is too small.
You have the amusing $\frac 13=0.3333333\ldots=[0;3,333333333\ldots]$ In a bit of playing, this "attractive fixed point" is hard to avoid. You need a big second number to get anywhere close. As you work at it, the $3$'s keep coming. I haven't made it into a proof that there isn't one.
I presume you don't like $1=[1;]$ or any other integer.