# Find $a_n$ in terms of $b_n$ given $b_n = \sum_{k=0}^{n} {n \choose k} a_k$

Given sequences $$a_n$$ and $$b_n$$ satifying $$b_n = \sum_{k=0}^{n} {n \choose k} a_k$$

I am required to find $$a_n$$ in terms of $$b_n$$

My attempt:

The generating fuction for $$b_n$$ will be \begin{align} B(x) &= \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} {n \choose k} a_k \right)\\ &= \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} {n \choose n-k} a_k \right) \end{align} This looks like the product of two generating functions $$A(x)$$ ( for $$a_n$$ ) and $$C(x)$$. Hence the given sequence $$b_n$$ is an convolution of two $$a_n$$ and some $$c_n$$.

If now I can find $$c_n$$, and a closed form for $$C(x)$$ (which I believe exists), the sequence $$a_n$$ can be found since $$A(x) = \frac{B(x)}{C(x)}$$

My question:

I am unable to find the sequence $$c_n$$. I tried using $$c_k = {n \choose k}$$ but I am quite sure that it is incorrect.

• Hint: Try to find $a_n$ for small $n$, for example $n=1, 2, 3$. See the pattern that is occurring – Jakobian Oct 7 '18 at 10:03
• @Jakobian Thanks for the hint. I've obtained that $a_n = \sum_{k=0}^{n} (-1)^k {n \choose k} b_{n-k}$. Proving by induction will make sure that the answer is correct. Can the question be solved by generating functions? – sc_ Oct 7 '18 at 10:15
• Yes, another way would be to consider exponential generating functions i. e. $B(x) = \sum \frac{b_nx^n}{n!}$ and $A(x) = \sum \frac{a_nx^n}{n!}$, then $B(x) = e^xA(x)$ – Jakobian Oct 7 '18 at 10:15

Here we are looking for so-called binomial inverse pairs. To show the relationship we multiply exponential generating functions (egfs). Let $$A(x)=\sum_{n\ge0}a_{n}\frac{x^n}{n!}$$ and $$B(x)=\sum_{n\ge0}b_{n}\frac{x^n}{n!}$$ egfs with $$B(x)=A(x)e^x$$. Comparing coefficients gives the following
Binomial inverse pair \begin{align*} B(x)&=A(x)e^x&A(x)&=B(x)e^{-x}\\ b_n&=\sum_{k=0}^{n}\binom{n}{k}a_k&a_n&=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}b_k \end{align*}