A miraculous number N Of course we can talk about 1 digit prime numbers, 2 digit primes , 3 digit primes , and so on..., my question is : is there an N (N greater than zero) such that there are no N-digit prime numbers ? I realize that it will be truly miraculous if there exist such an N , in other words the existence of such an N would be highly highly improbable due to the statement that there exists at least one prime number between 2 consecutive cubic numbers or even 2 consecutive square numbers , but these statements have not yet been proven (rigorously?). So does there exist such an N ?
 A: By Bertrand's postulate, for any integer $n>3$ there is a prime $p$ such that $n<p<2n-2$.
If $b>1$ is the base of your numbering system, we have that $b^{n+1}-1>2b^{n}-2$, because $b^{n+1}-2b^n=b^n(b-2)\ge0>-1$. So no number $n>1$ can be miraculous in base $b$, because $b^n>3$ as soon as $n>1$ or $b>3$.
The case $b=3$ and $n=1$ is ruled out, because $2$ is one digit long in base $3$.
So the only miraculous number is $1$, with respect to base $2$.
A: The comment by Martin-Bras points to Bertrand's postulate proved in 19th century: that there exists a prime number between  $n$ and $2n$. When you double an $n$ digit number if it becomes an $n+1$ digit then that $n+1$ digit number will have $1$ as leading digit. So the prime could be starting with 1 as leading digit; or if it is $n$ digit long then its next double will lead to such one (etc). So one will be able find for each $n$ a prime that is $n$ digit long and starting with 1 as leading digit; this guarantees in 3 more doublings 3 more primes that are equally long.
