Does every positive integer appear in the digits of $2\cdot 0.1234567891011…$?

Let $C = 0.1234567891011121314…$ the Champernowne constant. My question is :

Does the real number $2 \cdot C \simeq 0.24691357820222426283032343638404244464850525456586062646668707274...$ contain every positive integer in its digits?

For instance, $2022$ appears here : $0.246913578\underbrace{2022}...$

Obviously this is true for $C$, and also for $0.246810121416...$. These number are known to be normal. My question is to determine whether $2C$ is at least a disjunctive number (in base $10$).

More generally, I would like to know if a non-zero multiple $n \cdot x \; (n \in \Bbb Z)$ of a disjunctive number is also disjunctive (true if $n=10^k,k\in \Bbb N$).

I looked at some theorems about disjunctive/normal numbers (for instance, if $f$ is a non-constant polynomial with real coefficients which is always positive, then the "concatenation" of the integer parts $x=0.[f(1)][f(2)][f(3)]...$ is normal), but I wasn't able to conclude.

Any comment would be helpful.

• Yes. Every non-negative integer. Write $5n=a_1a_2\dots a_k$ in base $10$. Then when $10a_1a_2\dots a_k0$ occurs in $0.123\dots$ then $n$ occurs in $2(0.123\dots)$. – Thomas Andrews Feb 9 '16 at 14:40
• @ThomasAndrews : thank you! This was easier than I thought. Your argument also works e.g. for $5C$. But I don't see if it is possible to generalize your argument e.g. for $3C$. What do you think? – Watson Feb 9 '16 at 15:01
• For $3C$ you could argue as follows: Write $n = 3n' + j$, $j \in \{0,1,2\}$. If now $n' = a_k \ldots a_0$ in base $10$, we can look for $10a_k \ldots a_0 0$ in $C$ to find $3n'$. If $j = 1$, look for $10 a_k \ldots a_0 4 0$ in $C$, if $j = 2$, look for $10 a_k \ldots a_0 7 0$ in $C$ – martini Feb 9 '16 at 15:15
• Yes, $3$ is harder. For example, if you want to find $73$ in $3C$, first take $M=10002/3 = 3334$. Then compute $73\cdot 3334 = 243382$. Then find $102344820$ in $C$. Then $3C$ contains $30730146$ which contains $73$. It can be done, but it is in a more case-by-case basis, where the number we multiply by ($5$ in my original example) varies with $N$. – Thomas Andrews Feb 9 '16 at 15:17
• In general, if you want to show that every $m$ occurs in $nC$, then write $m/n$ is decimal to a very large amount of digits, then round up at the end. Then when that sequence of digits occurs in $C$ surrounded by "enough" zeros, then $m$ occurs in the $nC$. Again, for $m=73$, $n=6$ then $m/n=12.16666\dots$ so we pick the digits $12167$. Then where $10121670$ occurs in $C$, $6073002$ occurs in $6C$. – Thomas Andrews Feb 9 '16 at 15:27

1 Answer

In general, if you want to show that an $m$ occurs in $nC$, then write $m/n$ in decimal to a large amount of digits, then round up at the end. Then when that sequence of digits occurs in $C$ surrounded by "enough" zeros, then $m$ occurs in the same location of $nC$.

(When writing $m/n$, you'll have to keep the zeros that start the quotient in the case $n>m$.)

For example, given $m=73$, $n=6$ then $m/n=12.16666\dots$ so we pick the digits $12167$. Then where $10121670$ occurs in $C$, $6073002$ occurs in $6C$.

This is particularly easy for $n=2^k$ or $n=5^k$ because the decimals for $m/n$ terminate in these cases, so there is no question of "how many digits." I think, more generally, you are safe if you take more digits of $m/n$ after the decimal than there are digits of $n$, but I'm not 100% sure of that.

You have to start with $1$ then add as many digits of $0$s as digits of $n$, then add your sequence of digits from $m/n$, then again as many $0$s to the end as digits of $n$.

So if $m=84,n=13$ then $m/n \approx 6.461538462$ so we'll take the digits $10064700$ and find them in $C$. Where these digits occur, we'll get $308411xx$ in the same location of $13C$.

The only property of $C$ we are using is that every finite sequence of digits occurs in it.