Eulerian polynomial

The Eulerian polynomials $A_{n} (x)$ are defined to be:

f_{n} (x) = \sum_{i \geq 0} (1 + x)^{n} x^{i} = \frac{A_{n} (x)}{(1 - x)^{n + 1}}

Examples

\begin{aligned} 1 + x + x^{2} + x^{3} + \dots & = \frac{1}{1 - x} \\ 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots & = {(x \cdot \frac{1}{1 - x})}^{'} = \frac{1}{(1 - x)^{2}} \\ 1 + 2^{2} x + 3^{2} x^{2} + 4^{2} x^{3} + \dots & = (x \cdot \frac{1}{(1 - x)^{2}}) = \frac{1 + x}{(1 - x)^{3}} \\ f_{3} (x) & = \frac{1 + 4 x + x^{2}}{(1 - x)^{4}} \\ f_{4} (x) & = \frac{1 + 11 x + 11 x^{2} + x^{3}}{(1 - x)^{5}} \\ f_{5} (x) & = \frac{1 + 26 x + 66 x^{2} + 26 x^{4} + x^{4}}{(1 - x)^{6}} \end{aligned}

The $f_{i}$ 's follow the following recurrence relation:

f_{n + 1} (x) = (x \cdot f_{n} (x))^{'}

for $n \geq 0$ , and $f_{0} (x) = \frac{1}{1 - x}$ .

Plugging this in, we have

\begin{aligned} \frac{A_{n + 1} (x)}{(1 - x)^{n + 2}} & = {(\frac{x A_{n} (x)}{(1 - x)^{n + 1}})}^{'} \\ = \frac{(A_{n} (x) + x A_{n}^{'} (x)) (1 - x)^{n + 1} + (n + 1) (1 - x)^{n} x A_{n} (x)}{(1 - x)^{2 n + 2}} \\ = \frac{(A_{n} (x) + x A_{n}^{'} (x)) (1 - x) + (n + 1) x A_{n} (x)}{(1 - x)^{n + 2}} \\ A_{n + 1} (x) & = (1 + n x) A_{n} (x) + x (1 - x) A_{n}^{'} (x) \end{aligned} $ $ f o r $ n \geq 0 $, a n d $ A_{0} (x) = 1 $ .