Define a transformation $T\_s$ of integer sequence $\{ a\_n \}$ by 
$$
b\_n=T\_s(a\_n)={n \choose s} \sum\_{i=s}^{n-1} \frac{a\_i}{{i \choose s}}, 
$$
for a fixed $s \in \mathbb{N}.$


For instance, if we aplly the transformation $T\_2$ to the sequence $a\_n=1$ then we get the sequence $b\_n=n(n-2).$
Maple code for the sample



```
    T:=(a,s)->factor(simplify(product(n-i,i=0..s-1)*sum(a(k)/product(k-i,i=0..s-1),k=s..n-1)));
a:=n->1:T(a,2);
                                 n(n-2)

```

**Question.** Suppose that $a\_n$ has a generating function (ordinary or exponential or another one) $a(x).$ What is the generating function of transformed sequence? 



Find generating function
This can be done step by step.
First note that $\binom{n}{s}/\binom{i}{s}$ can be written as $n(n-1)\cdots(n-s+1)/i(i-1)\cdots(i-s+1)$
Since we have the generating function (with assuming $a\_i=0$ for $i< s$)
$$a(x)x^{-s}=\sum\_{i=s}^{\infty} a\_ix^{i-s}$$
We obtain the following by integrating $s$ times.
Let $A\_0(x)=a(x)x^{-s}$, and $A\_{k+1}(x)= \int\_0^x A\_k(t)dt$.
Then 
$$A\_s(x)=\sum\_{i=s}^{\infty} a\_i\frac{x^i}{i(i-1)\cdots (i-s+1)}$$
The generating function for $b\_n/ n(n-1)\cdots (n-s+1)$ can be obtained from the product
$$\left(\sum\_{i=s}^{\infty} a\_i\frac{x^i}{i(i-1)\cdots (i-s+1)}\right)\left(\sum\_{j=1}^{\infty} x^j\right)$$


Now the generating function for $b\_n$ follows from differentiating s times
Again assuming $b\_i =0$ for $i< s$, we have
$$B\_0(x)= A\_s(x) \frac{x}{1-x}$$
$$B\_{k+1}(x)=\frac{d}{dx} B\_k(x)$$
$$\sum\_{n=s}^{\infty} b\_n x^n = B\_s(x) x^s$$


