I tested a few numbers of the form $10^n+1$ (i.e. $100000001$) besides $11$ and $101$ and they were all composite. Clearly some of them if we have an even number of $0$'s in between are multiples of $11$, but is it just a coincidence that the ones with odd number of $0$'s are also composite?
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1You seem to forget $+$ somewhere. – Asaf Karagila Aug 15 '13 at 03:17
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See also http://math.stackexchange.com/questions/242949/are-there-infinite-many-integer-n-ge-0-such-that-102n1-prime-numbers ...oh, and even more relevant, http://math.stackexchange.com/questions/34877/are-there-infinite-amount-of-primes-and-non-primes-in-110n-formed-numbers – Chris Culter Aug 15 '13 at 03:34
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As André Nicolas mentions, $10^n+1$ can be prime only if $n=2^m$. These are generalized Fermat numbers, and I believe it is open if infinitely many of them are prime. Here is some data for the special case that you are interested. In particular, the data shows that 11 and 101 are the only primes of the form $10^n+1$ for all $n$ up to $2^{23}$.
Tony Huynh
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The answer is not known. But the only candidate $n$ have the shape $n=2^m$. So you can confine your testing to these.
André Nicolas
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