A question about sequence of positive integers related to their density in the reals Let $a_1 < a_2 < a_3 \ldots < a_n \dots$ be an increasing
sequence of positive integers. Let's say such a
sequence has:
Property A if, when consecutive terms of
the sequence are concatenated to form the
decimal $0.a_1a_2a_3a_4 \ldots $, the resulting
decimal is irrational. Examples of such sequences
are powers of positive
integers $d^n$,  and the sequence of primes.
Property B if, given an arbirary
positive integer $c$, there is a term of the
sequence whose decimal expansion begins with the digits of $c$.
Again, examples of such sequences are powers of positive integers
$d^n$, but with $d \ne 10^m$ , and the sequence of primes.
Property C if for any positive real numbers $a$, $b$ with
$a < b$ there exists terms $a_n$ and $a_m$ with
$a < a_n/a_m < b$.  The sequence of
primes is again an example, but the power sequence
of course does not have this property.
If
$\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = 1$
then all three properties hold. This is
proved in various places in the literature,
and on this forum.
Also, Property B implies Property A, since
if Property B holds there are terms of the
sequence which contain
arbitrary long strings of zeros.
Property C seems to be the strongest, so my question
is: Does Property C imply the other properties,
or is there a counterexample?
 A: Property C does not imply B. For instance, "All integers that do not begin with $9$" has property C, but clearly doesn't have property B. If we strengthen B to "contains the digits of $c$ consecutively and in order", rather than "begins with the digits of $c$", my gut tells me that "All integers that do not contain a $9$" still has property C, but this needs further investigation.
I strongly suspect that C does imply A. To work towards an answer, I propose to look at the contrapositive of this implication: No sequence that makes a rational number (as in the description of property A) can have property C.
The digits of the rational number $r = 0.a_1a_2a_3\ldots$ must eventually start to repeat in a periodic pattern. Let $i\in \Bbb N$ be such that this repetition starts before $a_i$. Now let the natural number $j\geq i$ be such that $a_j$ has strictly more digits than the length of the minimal repeating period of $r$.
We will now consider four cases:

*

*$m, n < j$: There are only finitely many ratios possible

*$m < j \leq n$: There are only finitely many ratios possible below, say, $10$

*$n < j \leq m$: There are only finitely many ratios possible above, say, $1$

*$j\leq m, n$: Let $P$ be the set of all possible minimal periods of $r$. Take the collection of all intervals of the form $[\frac p{q+1}\cdot 10^k, \frac{p+1}q\cdot 10^k]$ for $k\in \Bbb Z$ and $p, q\in P$. The ratio $\frac{a_n}{a_m}$ must lie in one of those intervals. Only finitely many of these intervals intersect with $[1, 10]$ (for each choice of $p$ and $q$, at most two values of $k$ makes the interval intersect $[1, 10]$), and they cannot cover all of it.

This last emphasized claim is where I'm currently stuck. It seems like most of these intervals are pretty small, and there aren't that many of them. (If the fundamental period is $k$ long, then there are on the order of $k^2$ intervals, and they are of the order of $10^{-k}$ wide, so it ought to work, but there are subtleties at play, especially with small $k$ and also periods with many zeroes. I haven't gone into the nitty-gritty to actually calculate functioning bounds.)
But once that's proven, it is easy to find (or, at least, prove that there exists) $1<a<b<10$ such that none of the finitely many ratios from the first three points are in $[a, b]$, and also such that $[a, b]$ doesn't intersect with any of the intervals from the fourth point.
