9

Is there a prime which is the form of $10^n + 1$ except $2, 11, 101$?

I confirm there isn't such prime for $n < 64$.

TOM
  • 1,563
  • 3
    We know if $n$ has an odd prime factor, the result is composite. So the only possibilities are for $n$ a power of two. These grow rapidly enough that heuristic arguments suggest an "expected" number of these primes will be small. – hardmath Jan 22 '17 at 00:24
  • 1
    These are also a special case of generalized Fermat primes and you can find a little more information about them under that name. – Steven Stadnicki Jan 22 '17 at 00:25
  • That $10^{2^{10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}}}+1$ is not prime is almost certain but it's likely to be unprovable. – Count Iblis Jan 22 '17 at 01:03

2 Answers2

7

Note that $(10^{n}+1)|10^{n(2k+1)}+1$ for $n$, $k\in \mathbb{N}$.

According to the Table $\boldsymbol{1}$ (page $24$ or $\frac{30}{55}$ of the pdf) from this journal:

$10^{2^{n}}+1$ has no (known?) prime factors for $n=13$, $14$, $21$, $23$, $24$, $25$, $\ldots $

That means $10^m+1$ is composite continually for $3\le m \le 8195$.

Ng Chung Tak
  • 19,693
2

Based on the previous answer by Ng Chung Tak, I intuitively estimate the probability of the existence of primes of the form $10^m + 1$. The prior answer indicated that such primes can only be of the form $10^{2^n} + 1$, and it has been proven that $10^{2^n} + 1$ is not prime for $2 \leq n \leq 12$.

If we assume the probability of a large integer $N$ being prime approaches $1/\ln(N)$, we can estimate that the probability of finding a large prime of the form $10^m + 1$ is that of finding a prime of the form $10^{2^n} + 1$ for $n \geq 13$. This probability does not exceed

$$ \sum_{n \geq 13} \frac{1}{\ln(10^{2^n})} \approx \frac{1}{\ln(10) \cdot 2^{12}} \approx 0.0001, $$

which is extremely small. Therefore, intuitively, it is almost impossible to find a prime of the form $10^m + 1$ for $m \geq 3$.

log2
  • 512