An infinite number of primes in the sequence

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Does the sequence $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contain an infinite number of primes?




I tried to find some theorems on this matter, but apparently the problem in general form is not solved. I will be happy with any results or suggestions, thank you.




sequences-and-series number-theory prime-numbers




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  • Not known for any polynomial in one variable.
    – Will Jagy
    yesterday










  • Those. there is not a single polynomial for which the problem was solved?
    – Vladislav Kharlamov
    yesterday










  • Not one. There are easy negatives, $3x^2 + 6$ or the like,
    – Will Jagy
    yesterday






  • 1




    To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
    – Hurkyl
    yesterday











  • This is understandable, I had in mind, of course, not trivial cases.
    – Vladislav Kharlamov
    yesterday














up vote
3
down vote

favorite













Does the sequence $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contain an infinite number of primes?




I tried to find some theorems on this matter, but apparently the problem in general form is not solved. I will be happy with any results or suggestions, thank you.




sequences-and-series number-theory prime-numbers




share|cite|improve this question





















  • Not known for any polynomial in one variable.
    – Will Jagy
    yesterday










  • Those. there is not a single polynomial for which the problem was solved?
    – Vladislav Kharlamov
    yesterday










  • Not one. There are easy negatives, $3x^2 + 6$ or the like,
    – Will Jagy
    yesterday






  • 1




    To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
    – Hurkyl
    yesterday











  • This is understandable, I had in mind, of course, not trivial cases.
    – Vladislav Kharlamov
    yesterday












up vote
3
down vote

favorite









up vote
3
down vote

favorite












Does the sequence $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contain an infinite number of primes?




I tried to find some theorems on this matter, but apparently the problem in general form is not solved. I will be happy with any results or suggestions, thank you.




sequences-and-series number-theory prime-numbers




share|cite|improve this question














Does the sequence $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contain an infinite number of primes?




I tried to find some theorems on this matter, but apparently the problem in general form is not solved. I will be happy with any results or suggestions, thank you.





sequences-and-series number-theory prime-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









William Elliot

5,0722414




5,0722414









asked yesterday









Vladislav Kharlamov

552115




552115











  • Not known for any polynomial in one variable.
    – Will Jagy
    yesterday










  • Those. there is not a single polynomial for which the problem was solved?
    – Vladislav Kharlamov
    yesterday










  • Not one. There are easy negatives, $3x^2 + 6$ or the like,
    – Will Jagy
    yesterday






  • 1




    To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
    – Hurkyl
    yesterday











  • This is understandable, I had in mind, of course, not trivial cases.
    – Vladislav Kharlamov
    yesterday
















  • Not known for any polynomial in one variable.
    – Will Jagy
    yesterday










  • Those. there is not a single polynomial for which the problem was solved?
    – Vladislav Kharlamov
    yesterday










  • Not one. There are easy negatives, $3x^2 + 6$ or the like,
    – Will Jagy
    yesterday






  • 1




    To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
    – Hurkyl
    yesterday











  • This is understandable, I had in mind, of course, not trivial cases.
    – Vladislav Kharlamov
    yesterday















Not known for any polynomial in one variable.
– Will Jagy
yesterday




Not known for any polynomial in one variable.
– Will Jagy
yesterday












Those. there is not a single polynomial for which the problem was solved?
– Vladislav Kharlamov
yesterday




Those. there is not a single polynomial for which the problem was solved?
– Vladislav Kharlamov
yesterday












Not one. There are easy negatives, $3x^2 + 6$ or the like,
– Will Jagy
yesterday




Not one. There are easy negatives, $3x^2 + 6$ or the like,
– Will Jagy
yesterday




1




1




To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
– Hurkyl
yesterday





To the best of my understanding, current mathematical knowledge is "almost certainly yes, but we can't prove it". I don't know if there are any special methods that might apply.
– Hurkyl
yesterday













This is understandable, I had in mind, of course, not trivial cases.
– Vladislav Kharlamov
yesterday




This is understandable, I had in mind, of course, not trivial cases.
– Vladislav Kharlamov
yesterday










1 Answer
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Bunyakovsky's conjecture says that if an irreducible polynomial $f(x)inmathbbZ[x]$ satisfies $1=mathrmgcdf(1), f(2), f(3), f(4), dots$ then $f(n)$ is prime for infinitely many $n$. It has not been proven for any polynomial of degree greater than $1$. According to this we conjecture that $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contains an infinite number of primes.






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    Bunyakovsky's conjecture says that if an irreducible polynomial $f(x)inmathbbZ[x]$ satisfies $1=mathrmgcdf(1), f(2), f(3), f(4), dots$ then $f(n)$ is prime for infinitely many $n$. It has not been proven for any polynomial of degree greater than $1$. According to this we conjecture that $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contains an infinite number of primes.






    share|cite|improve this answer

























      up vote
      2
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      Bunyakovsky's conjecture says that if an irreducible polynomial $f(x)inmathbbZ[x]$ satisfies $1=mathrmgcdf(1), f(2), f(3), f(4), dots$ then $f(n)$ is prime for infinitely many $n$. It has not been proven for any polynomial of degree greater than $1$. According to this we conjecture that $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contains an infinite number of primes.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Bunyakovsky's conjecture says that if an irreducible polynomial $f(x)inmathbbZ[x]$ satisfies $1=mathrmgcdf(1), f(2), f(3), f(4), dots$ then $f(n)$ is prime for infinitely many $n$. It has not been proven for any polynomial of degree greater than $1$. According to this we conjecture that $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contains an infinite number of primes.






        share|cite|improve this answer













        Bunyakovsky's conjecture says that if an irreducible polynomial $f(x)inmathbbZ[x]$ satisfies $1=mathrmgcdf(1), f(2), f(3), f(4), dots$ then $f(n)$ is prime for infinitely many $n$. It has not been proven for any polynomial of degree greater than $1$. According to this we conjecture that $ a_n = left|-fracn^46+frac3n^32-frac13n^23+6n-1right|$ contains an infinite number of primes.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered yesterday









        Dietrich Burde

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