Conjecture about the digits of $\pi$ 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 ?
 A: 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.
A: 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.
