Regularity in normal numbers

Question

Define $f(m,n,x)$ such; let $a_j$ be the $j$'th digit in the decimal expansion of x, then $f(m,n,x)=0.a_ma_{m+n}a_{m+2n}a_{m+3n}...$

Then, if $x$ is normal, is there for every computable number $0<y<1$, integers $m,n$, such that $y=f(m,n,x)$ ?

Answer 1

This is impossible. For example, take $y=1/3=0.333\ldots$, and suppose $m$ and $n$ are as stated. Then after the $m$th digit of $x$, every $n$th digit is a $3$. But then the decimal expansion of $x$ can't contain $m+n$ $2$s in a row, so $x$ is not normal.