# Is this pleasing polynomial irreducible?

Let:

$f(x)=x^n+2x^{n-1}+3x^{n-2}+4x^{n-3}+\ldots + (n-1)x^2+nx+(n+1)$.

Is $f(x)$ irreducible?

In light of the answers to this question, I now know that this is true when $n+1$ is prime. What about when $n+1$ is composite? I have checked a lot of cases and it seems to be true.

(Unnecessary background information: this is linked to this recent question of mine. At the point where I say:

"$g_n(x,−n)=x^2h(x)$,"

the $h(x)$ in question has the property that substituting $x-1$ for $x$ puts it in the form of $f(x)$ above. If I can show that $h(x)$ is irreducible, then I will have shown that the galois group of the original polynomial is doubly transitive. Not that I am trying to draw attention back to my original question!)

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Not an answer, but: it was observed a few years ago that for $n=6$ this polynomial $x^6 + 2x^5 + 3x^4 + 4x^3 + 5x^2 + 6x + 7$, though irreducible, has non-generic Galois group: it's the transitive copy of $S_5$ in $S_6$! Anybody remember who first noted this? – Noam D. Elkies Sep 9 '11 at 18:35

It is a conjecture that for any $k < n$ one has $$\frac{d^k}{dx^k}(1+x+\cdots+x^n)$$ is irreducible in $\mathbb Z[x]$. Your question is the case $k=1$. It is known that the set of integers $n\in [0,t]$ for which the polynomial $nx^{n-1}+(n-1)x^{n-2}+\cdots+1$ is reducible has size at most $O(t^{1/3+\epsilon})$, see the paper