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Copeland-Erdős theorem states that if $a_{1}, a_{2}, a_{3}, \ldots$ is an increasing sequence of integers such that for every $\theta < 1$ the number of $a$'s up to $N$ exceeds $N^{\theta}$ provided $N$ is sufficiently large, then the infinite decimal number $0.a_{1}a_{2}a_{3}\ldots$ is normal with respect to the base $b$ in which these integers are expressed. Thus, for instance, the number $0.23571113\dots$ formed by concatenating the primes for the fractional part is normal, since the number of primes up to $N$ grows like $N/\log{N}$ for large $N$.

I was wondering whether there is a converse to this theorem, namely, if $a_{1}, a_{2}, a_{3}, \ldots$ is an increasing sequence of integers such that there exists a $\theta < 1$ such that the number of $a$'s up to $N$ does not exceed $N^{\theta}$ provided $N$ is sufficiently large, then the infinite decimal number $0.a_{1}a_{2}a_{3}\ldots$ is not normal with respect to the base $b$ in which these integers are expressed. Maybe under some further restricting conditions...

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Consider the number $0.235711...$ that you mentioned, let $a_1=2$, $a_2=35$ and in general $a_n$ has at least one more digit than $a_{n-1}$ (with the caveat that you don't allow any of them to start with $0$, so if $a_{n+1}$ would start with zero, you add them to $a_n$ while still having $a_{n+1}$ have at least one more digit than $a_n$). Now, $0.a_1a_2\cdots$ is normal, as it's the number we considered at the start, but the number of $a_i\leq n$ is about $\log_{10}n<n^\theta$ for every fixed $\theta>0$

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  • $\begingroup$ Very well, this serves as a counterexample (which, incidentally, reminds me a bit of the idea behind the construction of Liouville numbers). Now the question is: can we characterize the non-normality of numbers constructed by concatenating series with specific growth properties? $\endgroup$ Commented 10 hours ago
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    $\begingroup$ I doubt growth properties will get you very far. One could take that same number, and group into integers such that each has at least twice as many digits as the preceding. You could make them grow much faster than that, and still concatenate to $0.235711\dotsc$. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ I agree with Gerry. Remember that if you found my answer satisfying, you may accept it by clicking the tick symbol just below the downvote button! $\endgroup$ Commented 9 hours ago

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