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Friend of mine gave me this problem :

Consider the following array of natural numbers :

$\begin{array}{ccccccccc} 1 & 2 & 4 & 7 & 11 & 16 & 22 & 29 & \ldots \\ 3 & 5 & 8 & 12 & 17 & 23 & 30 & 38 & \ldots \\ 6 & 9 & 13 & 18 & 24 & 31 & 39 & 48 & \ldots \\ 10 & 14 & 19 & 25 & 32 & 40 & 49 & 59 & \ldots \\ 15 & 20 & 26 & 33 & 41 & 50 & 60 & 71 & \ldots \\ 21 & 27 & 34 & 42 & 51 & 61 & 72 & 84 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \end{array}$

Question : Are there infinitely many primes in every row of this array ?

My attempt :

The nth term of the first row is given by :

$a_n=\frac{1}{2}(n^2-n+2)$

for $n=2k$ we have :

$a_{2k}=P(k)=2k^2-k+1$

and for $n=2k-1$ we have :

$a_{2k-1}=Q(k)=2k^2-3k+2$

Note that both $P(k)$ and $Q(k)$ are irreducible over integers .

Also note that : $\gcd(P(1),P(2),\ldots)=1$ and $\gcd(Q(1),Q(2),\ldots)=1$

So according to Bunyakowsky conjecture both $P(k)$ and $Q(k)$ generates for natural arguments infinitely many prime numbers . Therefore , if Bunyakowsky conjecture is true first row contains infinitely many primes . One can draw same conclusion for all other rows .

Is my reasoning correct ? Is there some other approach to this problem ?

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    @lewist : please edit this question and add content of your comment on the top replacing "a friend of mine ...", also do the same with previous question.2013-05-21

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