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Details and Motivation:

Let $x\in(0,1)$. Call the following property $(*)$:

Let $b\in\Bbb N$. Suppose $x=0.x_0x_1\dots$ is the expansion of $x$ in base $b$. For all $n\in\Bbb N$ with expansion $n=n_0\dots n_m$ in base $b$, we have some $t\in\Bbb N$ such that $$x=0.x_0\dots x_{t-1}(n_0 \dots n_m)x_{m+t}\dots$$

This property is equivalent to saying every finite sequence of numbers $u=u_0\dots u_v$ (for $0\le u_i\le b-1$) appears somewhere in the base $b$ expansion of $x$.

For simplicity, fix $b$.

It is not known whether $\pi$ is a normal number. It is always mentioned when asking whether $\pi-3$ has property $(*)$ (in base $b=10$).

The Question:

Is it possible for a number to have property $(*)$ but not be (simply) normal?

Thoughts:

This is certainly not a question I think I can answer myself.

As far as I can tell, there's not much known about normal numbers. Almost all real numbers are normal but there is only a handful of families we can demonstrate to be.

Perhaps I'm asking too much of MSE . . .

My guess: maybe! Each such finite sequence $u$ would appear the same amount of times (countably many), given that the concatenation

$$\underbrace{u\| \dots \| u}_{w\text{ times}}$$

appears in a number of property $(*)$ by definition, for all $w\in\Bbb N$. My issue is that it does not follow that the natural density of such concatenations is the same per $u$.

Shaun
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    Is it assumed only for one base b or for every base? – coffeemath May 27 '23 at 10:14
  • I wrote it up deliberately so that both could be an answer, @coffeemath; hence "(simply) normal". Should I edit it to pick one? – Shaun May 27 '23 at 10:18
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    If you pick one it will determine what question you are asking and preclude the other. But if you seek both answers, you could just mention that in the post. [Though some users here may respond that you should only ask one quetion per post.] – coffeemath May 27 '23 at 10:21
  • I have edited it accordingly, @coffeemath. Please leave your comments for context. (I link to one.) Thank you. – Shaun May 27 '23 at 10:23

1 Answers1

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Create a number like the Champernowne constant but before every integer, add a sequence of zeros which is of equal length to the length of that integer's digits. Since the Champernowne constant is normal, this number cannot be, since zero occurs much more frequently. However, the constructed number satisfies property $(*)$.

To be more clear, in base 10, this number's decimal expansion would start

$0.0102030405060708090010001100120013001400150016...$

Of course, the choice of how long the zero sequences should be, compared to the lengths of the integer representations is arbitrary as long as they are at least of the same size. One could think of "going Liouville" and skewing the distributions much further towards zeros.

Other variations on this theme could probably be used to produce numbers satisfying property $(*)$ and having different combinations of properties. For example, to create a number which is "simply normal" in base 10 but not "normal" in base 10, my guess is that using a different repeated digit in the prefix each time would work (unproven).

Ron Kaminsky
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