I need to know if the expression $$\sum_{i=0}^{a-1}\binom{a}{i}\binom{2a-i}{a} $$ is zero mod $2$.
I know that I have to look at binary expansion of numbers $a$, $2a$ and $2a-i$, then use the theorem that
If $m=a_0+a_p+\cdots+a_rp^r$ and $n=b_0+b_1p+\cdots+b_rp^r$, then $$\binom{m}{n}\equiv\prod_{i=0}{r}\binom{a_i}{b_i}$$ in mod $p$
In the case $i=0$, we have $\binom{2a}{a}$ which is even. This is because, if $a=2^{j_1}+\cdots+2^{j_k}$, then $2a=2^{j_1+1}+\cdots+2^{j_k+1}$, thus $$\binom{2a}{a}\equiv \binom{0}{1}(\text{other binomial coefficients})=0 $$
For the rest, I have no idea.
Any help will be appreciated.
Yes, it is even. Actually, ALL terms in your sum are even! For $i=0,\dots,a-1$, $$\begin{align} \binom{a}{i}\binom{2a-i}{a}&=\frac{a!}{i!(a-i)!}\cdot\frac{(2a-i)!}{a!(a-i)!}=\frac{(2(a-i))!}{(a-i)!(a-i)!}\cdot\frac{(2a-i)!}{i!(2(a-i))!} \\ &=\binom{2(a-i)}{a-i}\binom{2a-i}{i} =2\binom{2(a-i)-1}{a-i-1}\binom{2a-i}{i}\equiv 0\pmod 2. \end{align}$$ Note that $\binom{a}{i}\binom{2a-i}{a}$ is $T(a,i)$ in OEIS's sequence A104684.