For $n \ge 1,$ if $2^n$ in base ten has $d$ digits, $k$ of which are odd, define $f(n)=k/d.$ I'm wondering whether $f(n)$ can get arbitrarily close to $1,$ or on the other hand does $f(n)$ have a least upper bound which is less than $1.$ [If the latter can we find that lub?]
We have $f(9)=2/3,$ since $2^9=512.$ we can't have $f(n)=1$ since the units digit is even. I'd be interested in some specific cases where $f(n)$ is "fairly near" $1$ [whatever that means].