Let us start with $$f(x)=4x+6$$ formed by the first two composite numbers. To get the next polynomial, take the next composite number (let us call it $\ s\ $) and replace $\ f\ $ by $\ fx+s\ $
This gives the polynomials $\ 4x+6,4x^2+6x+8,4x^3+6x^2+8x+9\ $ and so on. The polynomials are just formed by using the first $\ n+1\ $ composite numbers as coefficients.
Do we ever get a polynomial reducible over $\mathbb Q$ , or are all poylnomials in this sequence irreducible over $\mathbb Q$ ? From the third on, the polynomials are primitive, so that this is equivalent to irreducible in $\mathbb Z$ from here on.
The first $\ 600\ $ polynomials are irreducible. Any ideas how to prove that they are all irreducible ?
If this is too difficult, partial results (like that there cannot be a linear factor , or just that the polynomials are all squarefree) are welcome as well.