# Find generating function

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?

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Judging from the code, I think you want $${n \choose s} \sum_{i=s}^{n-1} \frac{a_i}{i \choose s}$$ – Robert Israel Apr 18 '12 at 18:50
Sorry. Yes, of cource. – Melania Apr 18 '12 at 19:39
Can you do the case $s=0$ ... $b_n = \sum_{i=0}^{n-1}a_i$ ?? How about the case $s=1$ ... $b_n = n\sum_{i=1}^{n-1} a_i/i$ ?? – Gerald Edgar Apr 18 '12 at 20:14
@Gerald Yes, for the first case $s=0$ the generation function is ${\frac {z}{ \left( 1-z \right) ^{2}}}$. For the case s=1 the is not elementary function and for the case $s=2$ I have got $${\frac {z(-1+3\,z)}{ \left( 1-z \right) ^{3}}}.$$ – Melania Apr 18 '12 at 20:45
I think Gerald was talking about general $a_n$. The ordinary generating function of $b_n$ in the case $s=1$ is, I think, $$b(x) = \frac{x}{(1-x)^2} \int_0^x \frac{a(t)}{t}\ dt + \frac{x a(x)}{1-x}$$ – Robert Israel Apr 18 '12 at 22:49

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 $$B_0(x)= A_s(x) \frac{x}{1-x}$$ $$B_{k+1}(x)=\frac{d}{dx} B_k(x)$$ $$\sum_{n=s}^{\infty} b_n x^n = B_s(x) x^s$$
Thanks. I have got similar result $$\sum_{n=s}^{\infty} b_n x^n=x^s \frac{d^s}{dx^s}\left( \frac{x}{1-x} \underbrace{\int_0^x \cdots \int_0^x}_{s \text{ times}} \frac{a(x)}{x^s} \underbrace{dx \cdots dx}_{s \text{ times}} \right).$$ – Melania Apr 21 '12 at 15:10