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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 ?

Arnold C
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    http://en.wikipedia.org/wiki/Bertrand%27s_postulate. – Martín-Blas Pérez Pinilla Apr 23 '15 at 09:46
  • It is a theorem that there is always a prime between consecutive cubes, a conjecture (but one that everyone believes) that there's always a prime between consecutive squares. – Gerry Myerson Apr 23 '15 at 11:06
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    @GerryMyerson: Not quite. It's been proved that there is always a prime between consecutive cubes for large enough cubes, but we still haven't (!) ruled out the possibility that there are some integers $n>0$ with $\pi((n+1)^3)=\pi(n^3).$ – Charles Apr 23 '15 at 15:22
  • @Charles, you're right, of course. I may have been thinking of a result that's conditional on the Riemann Hypothesis. – Gerry Myerson Apr 23 '15 at 23:56

2 Answers2

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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 $2$.

egreg
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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.

P Vanchinathan
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