I came across this post [Coefficients of the Inflated Eulerian Polynomial](https://arxiv.org/pdf/1504.01089.pdf) by AULI-GRAHAM-SAVAGE. In particular, the polynomials related to **descents** interested me
$$P\_n(x)=\sum\_{\pi\in\mathfrak{S}\_n}x^{n\cdot\text{des}(\pi)+\pi\_n}.$$
Here $\pi=\pi\_1\pi\_2\cdots\pi\_n\in\mathfrak{S}\_n$ is a permutation on the letters $[n]=\{1,2,\dots,n\}$.


Let's introduce the following variant with respect to [**major index**](https://en.wikipedia.org/wiki/Major_index) or [**inversion number**](https://en.wikipedia.org/wiki/Inversion_(discrete_mathematics)) statistics
$$Q\_{n,a}(x)=\sum\_{\pi\in\mathfrak{S}\_n}x^{a\cdot\text{maj}(\pi)+\pi\_n}.$$



> 
> **QUESTION.** Is this true?
> $$
> Q\_{n,a}(x)
> =x^n\left(1+x^{a-1}+\cdots+x^{(a-1)(n-1)}\right)\prod\_{j=0}^{n-2}\left(1+x^a+\cdots+x^{aj}\right).$$
> 
> 
> 



A close reative of "Inflated" Eulerian polynomials
Yes, this is true. First of all, we denote $x^a=t$, then the relation to prove reads as $$\sum\_{\pi} t^{{\rm maj}(\pi)}x^{\pi\_n}=(x^n+x^{n-1}t+\ldots+xt^{n-1})\prod\_{j=0}^{n-2}(1+t+\ldots+t^{j}).\quad\quad\quad (1)$$
Taking the coefficients of the fixed power of $x$, say $x^s$, in both sides of (1), we reduce (1) to the identity
$$
\sum\_{\pi:\pi\_n=s} t^{{\rm maj}(\pi)}=t^{n-s}\prod\_{j=0}^{n-2}(1+t+\ldots+t^{j}). \quad\quad\quad (2)
$$
We prove that (2) holds for all $n$ by induction on $n$. The base case $n=1$ is trivial. Assume that (2) holds for $n-1$ and prove it for $n$. Partition the set $\Omega(n,s)$ of all $(n-1)!$
permutations of $(\pi\_1,\ldots,\pi\_{n-1})$ of $\{1,\ldots,n\}\setminus \{s\}$ onto $n-1$ disjoints subsets $\Omega(n,s,j)=\{\pi:\pi\_{n-1}=j\}$.


For $j<s$ the element $n-1$ does not contribute to the major index of $\pi$ and by induction proposition the corresponding sum over $\Omega(n,s,t)$ equals $t^{n-1-j}\prod\_{j=0}^{n-3}(1+t+\ldots+t^{j})$.


For $j>s$ the element $n-1$ does contribute to the major index of $\pi$ and by induction proposition the corresponding sum over $\Omega(n,s,t)$ equals $t^{2n-2-(j-1)}\prod\_{j=0}^{n-3}(1+t+\ldots+t^{j})$. Totally we get
$$
\prod\_{j=0}^{n-3}(1+t+\ldots+t^{j})\cdot \left((t^{n-s}+\ldots+t^{n-2})+(t^{n-1}+\ldots+t^{2n-2-s})\right),
$$
that's RHS of (2).


