Finding a generating function of a series

Question

So say if you have a sequence defined as, for $a\in\mathbb{Z}$, $$ c_n = \binom{a}{0} \binom{a}{n} - \binom{a}{1} \binom{a}{n-1} + \cdots+ (-1)^n \binom{a}{n} \binom{a}{0} = \sum_{i=0}^n (-1)^i \binom{a}{i} \binom{a}{n-i}$$ How would you find the generating function? It's easy to see that for $n = 2k+1$, $c_n= 0$, and for $n = 2k$ we can make something like this: $$ 2 \sum_{i=0}^k \binom{a}{i} \binom{a}{2k-i} + (-1)^{k+1} \binom{a}{k+1} $$ If $k > a$ we can use Voldemort's identity and turn the sum into $2 \binom{2a}{2k}$ That makes it more specific though so possibly I'm looking at it too narrowly

Answer 1

Here we show how to find the exponential generating function of the sum as requested by the OP.

Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

Now re-write your sum as follows: $$Q_n = \sum_{k=0}^n (-1)^k {a\choose k}{a\choose n-k} = \frac{1}{n!} \sum_{k=0}^n{n\choose k} \frac{ (-1)^k \times a!}{(a-k)!} \frac{a!}{(a-(n-k))!}$$ Now we clearly have $$A(z) = \sum_{q=0}^a \frac{(-1)^q \times a!}{(a-q)!}\frac{z^q}{q!} = \sum_{q=0}^a {a\choose q} (-z)^q = (1-z)^a.$$ Similarly we get $$B(z) = \sum_{q=0}^a \frac{a!}{(a-q)!}\frac{z^q}{q!} = \sum_{q=0}^a {a\choose q} z^q = (1+z)^a.$$ It follows that the value of the original sum is given by $$ \frac{1}{n!} \times n! [z^n] A(z) B(z) = [z^n] (1-z)^a (1+z)^a = [z^n] (1-z^2)^a.$$ This means that $$Q_n = \begin{cases} 0 \quad && \text{if} \quad n \quad \text{is odd} \\ (-1)^{n/2} {a\choose n/2} \quad && \text{if} \quad n \quad \text{is even.} \\ \end{cases}.$$