I need to find the summation of the series given below: $$ \sum_{i=0}^{\lfloor n/3 \rfloor}\binom {n}{3i}\binom {m}{3i} , \textrm{where} \lfloor n/3 \rfloor =\lfloor m/3 \rfloor $$
I know how to find summation of every third binomial and the summation of squares of binomials in a series, but I can't seem to figure out the exact answer of this problem, any help is appreciated.
Let $\omega=\exp(2\pi i/3)$ be the primitive cube root of unity. Then $$\frac{1+\omega^k + \omega^{2k}}{3} = \begin{cases}1 &\text{if $3 \mid k$}\\0 &\text{otherwise.}\end{cases}$$ So $$\sum_{k \ge 0} a_{3k} = \sum_{k \ge 0} \frac{1+\omega^k + \omega^{2k}}{3} a_k.$$ Taking $a_k = \binom{n}{k}\binom{m}{k}$ yields \begin{align} \sum_{k \ge 0} \binom{n}{3k}\binom{m}{3k} &= \sum_{k \ge 0} \frac{1+\omega^k + \omega^{2k}}{3} \binom{n}{k}\binom{m}{k} \\ &= \frac{1}{3}\sum_{k \ge 0} \binom{n}{k}\binom{m}{k} + \frac{1}{3}\sum_{k \ge 0} \omega^k \binom{n}{k}\binom{m}{k} + \frac{1}{3}\sum_{k \ge 0} \omega^{2k}\binom{n}{k}\binom{m}{k} \\ &= \frac{1}{3}\binom{n+m}{n} + \frac{1}{3}{}_2 F_1(-m, -n; 1; \omega) + \frac{1}{3}{}_2 F_1(-m, -n; 1; \omega^2). \end{align} The first term explains the asymptotic behavior you observed.