If $p_n$ is the $n$'th prime, let $A_n(x) = x^n + p_1x^{n1}+\cdots + p_{n1}x+p_n$. Is $A_n$ then irreducible in $\mathbb{Z}[x]$ for any natural number $n$? I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible. I have thought about this for a long time now, and asked many others, with no answer yet.

I will prove that $A_n$ is irreducible for all $n$, but most of the credit goes to Qiaochu. We have $$(x1)A_n = b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x  p_n$$ for some positive integers $b_{n+1},\ldots,b_1$ summing to $p_n$. If $x \le 1$, then $$b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x \le b_{n+1}+\cdots+b_1 = p_n$$ with equality if and only $x=1$, so the only zero of $(x1)A_n$ inside or on the unit circle is $x=1$. Moreover, $A_n(1)>0$, so $x=1$ is not a zero of $A_n$, so every zero of $A_n$ has absolute value greater than $1$. If $A_n$ factors as $B C$, then $B(0) C(0) = A_n(0) = p_n$, so either $B(0)$ or $C(0)$ is $\pm 1$. Suppose that it is $B(0)$ that is $\pm 1$. On the other hand, $\pm B(0)$ is the product of the zeros of $B$, which are complex numbers of absolute value greater than $1$, so it must be an empty product, i.e., $\deg B=0$. Thus the factorization is trivial. Hence $A_n$ is irreducible. 


Since I don't have enough "reputation" to comment on Bjorn's answer, I will write this in an answer. The remark about the location of the zeros of $A_n$ goes back at least to Kakeya, On the Limits of the Roots of an Algebraic Equation with Positive Coefficients, Tôhoku Math. J., vol. 2, 140142, 1912. It also appears in the nice book by E. Landau, Darstellung und Begründung einiger neuerer Ergebnisse der Funktionentheorie, Springer 1916, (Hilfssatz p.20). 

