Need to find a sequence $ a_n $ with following properties: $ \sum_{k= 0}^{n}a_k*a_{n-k}=1 $ in context of generating Funtions

Question

Its in the Chapter for generating Functions but i am not sure, what is the purpose of the exercise, it looks completly random to me. Is there any use in:

$ a_n $ with $ \sum_{k= 0}^{n}a_k*a_{n-k}=1 $ or something like this $ a_n $ with $ \sum_{k= 0}^{n}a_k*a_{n-k}=2 $ or something like this $ a_n $ with $ \sum_{k= 0}^{n}a_k*a_{n-k}=3 $

creating such functions? They don not have to hold at the same time. So Basically i need to find 2 sequences: $ a_n$ with properties $\sum_{k= 0}^{n}a_k*a_{n-k}=1 $

example 1: $$a_k=(1,-1,1,-1,......)$$ $$a_{n-k}=(1,1,1,1,........)$$ The Product: $$a_k*a_{n-k}=(1,-1,1,-1,........)$$ And the Sum must be 1

example 2 $$(1,1,1,1,......,1,1)$$ $$(0,0,0,0,.......,0,1)$$ The Product: $$(0,0,0,0,........,0,1)$$

What is the meaning of this exercice? update: I think i need to find a product of two generating functions with a result sequence (0,...0,1).

Answer 1

From the binomial theorem, we know that $$ \sum_{n=0}^\infty\binom{2n}{n} x^n=\frac{1}{\sqrt{1-4x}} $$ whence $$ \sum_{k=0}^n\binom{2k}{k}\binom{2(n-k)}{n-k}=4^n $$ In particular $$ \sum_{k=0}^n\frac{1}{4^k}\binom{2k}{k}\frac{1}{4^{n-k}}\binom{2(n-k)}{n-k}=1 $$ as desired.

Answer 2

The point of the exercise is that you associate a recurrence with a product of series, and then use this information to obtain an explicit formula for the sequence of the exercise.

Suppose $(a_n)$ is a sequence with the property that $\sum_{k=0}^n a_k a_{n-k}= 1$ for each $n$. In particular $a_0^2=1$. If $f$ is its ordinary generating function, this means that $$f^2 = \sum_{n\geqslant 0} X^n = 1+X+X^2+\cdots= (1-X)^{-1}$$

so that $f = (1-X)^{-1/2}$ or $f = -(1-X)^{-1/2}$ when $a_0=1$ and $a_0=-1$, respectively. Since

$$(1-X)^{-1/2} = \sum_{n\geqslant 0} \binom{-1/2}{n} X^n,$$

this gives you an explicit formula for $(a_n)$.