Primes of the form $2^{m_0}p_1^{m_1}\ldots p_r^{m_r}+1$

Is it known any example of a set of primes $\{p_1,\ldots,p_r\}$ with the following property: there are infinitely many $(m_0,\ldots,m_r)\in\mathbb N^{r+1}$ such that $2^{m_0}p_1^{m_1}\ldots p_r^{m_r}+1$ is prime?

Otherwise, is there any conjecture about that? Can anybody provide any reference for such conjecture in literature?

Thanks a lot!

• No, there is no known such set of primes. If that were the case, it would follow that there is a polynomial $p(x)$ of degree $> 1$ for which $p(n)$ is prime infinitely often, and this is open. On the other hand, there are probably infinitely many primes of the form $2^x \cdot 3^{y} + 1$, and probably infinitely many such even with $y = 1$. For example, $3 \cdot 2^{10829346} + 1$ is prime. – Jeremy Rouse Feb 1 '16 at 21:45
• @JeremyRouse how do you get the polynomial? – user9072 Feb 1 '16 at 22:56
• @quid: For each such set of $m_i$, rewrite all the $m_i$ as $2k_i+l_i$ with $l_i \in \{0,1\}$. Then over all primes of the given form, there is some $(l_0, l_1,...)$ which occurs infinitely often, hence the polynomial $(2^{l_0}p_1^{l_1}\ldots p_r^{l_r})*x^2+1$ assumes infinitely many prime values. This generalizes mod any number, not just 2, so we can actually make the polynomial be a binomial of as high a degree as we like. – ARupinski Feb 2 '16 at 0:51