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Let's define an infinite sequence of positive integers as :

$a_n=k^2+(2n-1)k+2n-1 $ , where $ k,n \in \mathbf{Z^{+}}$

Suppose that one can prove that this sequence contains infinitely many prime numbers for any particular $k$. The first term of the sequence for any $k$ is of the form :

$a_1=k^2+k+1$

My question is : what is sufficient condition to prove that this polynomial produces primes for infinitely many values of $k$ ?

Note that there is strong experimental evidence that $k^2+k+1$ is prime for infinitely many values of $k$ and that this polynomial satisfies conditions of Bunyakovsky's conjecture.

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    I assumed that you knew that *no* quadratic polynomial is known that represents infinitely many primes. So I thought the question could not be about fixed $n$. The primality of $k^2+k+1$ for infinitely many $k$ is a well-known open problem. I am therefore deleting the answer, which dealt with fixed $k$ and variable $n$.2011-11-05
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    @Andre,I know that it isn't known whether such polynomial exist, that's why I have involved this sequence in my question..2011-11-05
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    Supposing isn't necessary; Andre answered that fixing $k$ allows infinitely primes in $a_n$ due to Dirichlet's Theorem (his answer is now deleted). Now would you mind explaining what $n$ and $a_n$ have to do with anything if your question is about $k^2+k+1$? Why the irrelevant, misleading build-up?2011-11-05
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    @anon,I think that fact that $k^2+k+1$ is the first term of sequence above may be interpreted as necessary condition2011-11-05
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    @pedja: I see, you are hoping that the additional parameter can help in settling the $k^2+k+1$ problem. Unfortunately each value of $n$ leads to another open problem. The fact that for each $k$, there are infinitely many $n$ for which we get a prime gives no information about $n=1$, or any other value of $n$. Each "row" can have infinitely many primes, while each "column" has only finitely many. It is not hard to come up with examples of such situations, though I certainly cannot for your two-parameter family.2011-11-05
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    @Andre,Do you think that we may claim that there exist at least one "column" with infinitely many primes ?2011-11-05
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    @pedja: My earlier comment was to the effect that in *general* situations, knowing that each row has infinitely many primes does not show that any column does. But one cannot rule out the possibility that for *this family* one could push such an argument through. I am not optimistic, but would love to be wrong.2011-11-05
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    I think if we could make such a claim, someone would have done so some time in the last 200 years. As no one has....2011-11-05
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    @Andre,thanks for your comments..2011-11-05
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    Related: https://math.stackexchange.com/questions/12849632018-11-30

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