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

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

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    $\begingroup$ 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. $\endgroup$
    – hardmath
    Jan 22, 2017 at 0:24
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    $\begingroup$ These are also a special case of generalized Fermat primes and you can find a little more information about them under that name. $\endgroup$ Jan 22, 2017 at 0:25
  • $\begingroup$ 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. $\endgroup$ Jan 22, 2017 at 1:03

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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 $3\le m \le 8195$.

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  • $\begingroup$ @TransfiniteNumbers : n = 2 :) since $10^4+1=73\times 37$ $\endgroup$
    – Adren
    Nov 15, 2019 at 11:10
  • $\begingroup$ @Adren Do you mean 10^4+1 = 73 x 137? $\endgroup$
    – user4414
    Nov 15, 2019 at 19:10
  • $\begingroup$ @TransfiniteNumbers ooops , yes of course $\endgroup$
    – Adren
    Nov 16, 2019 at 6:43

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