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Motivated by this question, I started to ask myself why it is not known any Smarandache prime.

Note that we can generate the first nine Smarandache numbers with the expression $$10^{n-1}+2*10^{n-2}+3*10^{n-3}+...+(n-1)10+n$$

Therefore, it comes naturally the question of for which values of $x\in \mathbb Z^+$ does the polynomial $$x^{n-1}+2*x^{n-2}+3*x^{n-3}+...+(n-1)x+n$$ generate prime numbers.

Surprisingly, I have not been able to find any prime number generated by this polynomial for $x=3k+1,$ $k\in \mathbb Z^+$. Numbers get big very fast, so it would be great if someone could share some checking program to continue searching, provide any theoretical reason of why the polynomial does not generate prime numbers for those values of $x$, or provide some counterexample (i.e., a prime number generated by the polynomial for $x=3k+1$).

Thanks in advance!

Barry Cipra
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Juan Moreno
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  • So, you are looking for $x$ of the form $3k+1$ and $n$ such that the above expression is prime, is that right ? How far have you searched for primes ? – Peter Sep 16 '21 at 09:29
  • That no Smarandache prime is known is probably a coincidence. In general, a sequence growing exponential and not giving primes for small inputs can easily have the property that , if it contains a prime number at all , it might be too large to find it. There are no "forced factors" , so there is no reason that there is no Smarandache prime. – Peter Sep 16 '21 at 09:34
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    What about $n=7$ and $x=208=3\cdot 69+1$ ? – Peter Sep 16 '21 at 09:45
  • @Peter thanks for your useful comments, and the counterexample! Can I ask you which kind of program have you used to get the result? – Juan Moreno Sep 16 '21 at 10:13
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    I used PARI/GP , a calculator being able to handle very large numbers and high precision maniplulations. A useful tool for such problems. – Peter Sep 16 '21 at 10:42

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