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Do there exist $2013$ distinct positive integers $k_1,k_2,\dotsc,k_{2013}$, such that for any $n\geq 2012$, there is at least one prime number in

$$\{k_1 \cdot 2^{a_1(n)}+1, k_2\cdot2^{a_2(n)}+1, \dotsc, k_{2013} \cdot 2^{a_{2013}(n)}+1\}$$

where $a_1(n)=n, a_{i+1}(n)=2^{a_i(n)}, i=1,2,\dotsc,2012$.

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Almost surely not. That would imply, for example, that there exists $j$ such that $k_j \cdot 2^{a_j(n)} + 1$ is prime a positive proportion of the time, which is farfetched to say the least (possibly even provably false). –  Greg Martin Sep 7 '13 at 6:45
...I am convinced, due to the presence of 2013, that this is a contest question. –  Fernando Pessoa Sep 8 '13 at 5:08

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