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Let $a_n = 10^n+n$. Then, $a_n$ is prime only when $n \in \{1, 9, 69, 313, 451\}$ for any $n < 10101$.

I would like to know what is the sixth value of $n$ so that $a_n$ is a prime number (I checked that $n > 10101$), and whether there exists an infinite number of prime numbers in the sequence $\{a_n\}_{n=0,1,\ldots}$.

G. Gare
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Travis Wang
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  • Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Oct 04 '22 at 22:11
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    Your first question was answered by Peter and Henrik in comments that have since been moved to chat. As for infinitude, see https://math.stackexchange.com/questions/4523193/are-there-infinitely-many-primes-of-the-form-x-we-probably-dont-know. – Ravi Fernando Oct 05 '22 at 00:39
  • $$N\equiv 1,3\pmod 6$$ $$N\not\equiv 1 \pmod 7\iff N\equiv 3\pmod 6$$ $$N\not\equiv 4 \pmod 7\iff N\equiv 1\pmod 6$$ $$N\not\equiv 8 \pmod 9$$ $$N\equiv 1,3,7,9\pmod {10}$$ – Roddy MacPhee Oct 07 '22 at 20:55
  • @RoddyMacPhee Anyone who will actually search for futher primes sieves out the small factors anyway. There are however plenty of candidates , so a primality search will take long unless some helpers check distinct ranges. – Peter Oct 08 '22 at 08:46
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    If someone wants to continue the search, upto $51\ 000$ , there are no furhter exponents (except those in OEIS) leading to a prime. Based on the candidates, the expected number of primes upto $n=500\ 000$ is about $1$. So, it can be a long way to the next such prime. – Peter Oct 09 '22 at 07:53
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    I arrived at $60\ 000$ without any further PRP beyond $n=23\ 769$ – Peter Oct 09 '22 at 13:00

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