Conjecture about the digits of $\pi$

Question

Consider irrational numbers between 1 and 9.

Lets call a specific one $a$. Let $n>0$ be an integer.

In decimal consider the first $n$ digits of $a$. Call that string $A(a,n)$. Now consider the next $n$ digits of $a$ after the first $n$. Call that string $B(a,n)$.

Define $a$ has the repeat digits property (rdp) iff :

$$A(a,n) = B(a,n) $$

For some $n$. Also $rdp(a) = $ true.

Now it is tempting to think statistically about this. What is the probability that $rdp(a) =$ true ? And probably that probability is equal to

$$ 10^{-1} + 10^{-2} + 10^{-3} + ... = 1/9 $$

Or close to it. Is that correct ??

However i wonder about actual proofs rather than statistical reasoning.

So I make a conjecture

There exists NO $n$ such that

$$ A(\pi,n) = B(\pi,n)$$

Now i picked the number $\pi$ because we know alot about its digits. ( unlike say zeta(5) , euler gamma etc , in fact we are not even sure they are irrational ! ) For instance we can compute the 100000 th digit in base 16 without needing to store or compute all the previous ones.

. See https://en.m.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula

So how to handle this ??

Or is this one of the simplest undecidable problems ?

Or the simplest example of computational irreducibility ? ( see wolfram's book a new kind of science ).

Is there any hope of solving they things besides brute force Search and luck ?

Is this THE example of the ultimate halting problem ?

Answer 1

Not an answer but here are a couple thoughts that are too long for a comment:

Your calculation of the probability rdp(a) is true is not quite right. You are over counting numbers that satisfy the condition for multiple values of $n$. For example $.1111$ satisfies the condition for both $n=1$ and $n=2$. In any case what you computed is an upper bound, and $1/10$ is trivially a lower bound.

You can rewrite rdp(a)being true as $10^na-a - \lfloor10^na-a\rfloor < 10^{-n}$ for some $n$ so $a - K/(10^n-1) < \frac{1}{10^n(10^n-1)}$ which is reminds me of diophantine approximation, perhaps something can be said in that direction.

Answer 2

I believe your conjecture is open, but here are some partial results:

Let $a$ be defined to have the infinitely-often-repeating-digits property (IORDP) if there are infinitely many values of $n$ such that $A(a,n) = B(a,n)$.

If $a$ has IORDP, it has irrationality measure at least $2$. That's because in the repeat-digits cases we can match the first $2 \cdot n$ digits of $a$ with a fraction whose denominator only has $n$ digits (all $9$'s). This isn't very helpful since every irrational number has irrationality measure at least $2$. (However, irrationality measure is defined as an infimum so maybe it is possible to strengthen this to a strict inequality?)

But, we can define the infinitely-often-twice-repeating-digits property (IOTRDP) to mean that there are infinitely many $n$ with $A(a,n) = B(a,n) = C(a,n)$ where $C$ is the third sequence of $n$ digits. If $a$ has IOTRDP, then it has irrationality measure at least $3$. So one thing we can say is that algebraic numbers like $\sqrt{2}$ do not have IOTRDP.

And since $\pi$ is known to have an irrationality measure less than $8$, we can be sure that only finitely often is it the case that the first $n$ digits of $\pi$ repeat $8$ times. But this is not enough to establish that any particular number of repetitions of the initial digit sequence of $\pi$ never occurs.

Also, there is a relationship between irrationality measure and the runtime analysis of digit-extraction algorithms like BBP. Although the expected number of summands for BBP to extract the $n^{th}$ nybble is $n+O(1)$, we can't rule out the existence of cases where it requires more than $7 \cdot n$ terms, i.e. the first $n$ nybbles of $\pi$ are followed by more than $6 \cdot n$ $0$'s or $f$'s. This doesn't affect the asymptotic runtime of BBP since it's a constant factor but it means maybe there are digits that take more than seven times as long as typical ones to extract.