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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, integers $m,n$, such that $y=f(m,n,x)$ ?

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    I think its no, but is it no for every normal x?2012-10-21

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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.

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    @da_baws: That's another question, and should be asked separately.2012-10-21