MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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?

share|cite|improve this question
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
up vote 3 down vote accepted

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 $$B_0(x)= A_s(x) \frac{x}{1-x}$$ $$B_{k+1}(x)=\frac{d}{dx} B_k(x)$$ $$\sum_{n=s}^{\infty} b_n x^n = B_s(x) x^s$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.