Introduce the sequence (this is A047781 on OEIS) $$t_n=\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k}k$$ and denote the set $T(ij)=\{n\in\mathbb{N}: \text{the ternary digits of $n$ contain $i$ or $j$ only}\}$.
QUESTION. Is this true modulo $3$? $$t_n\equiv_3\begin{cases} 1 \qquad \text{if $\lfloor\frac{n}2\rfloor\in T(01)$} \\ 0 \qquad \text{otherwise}. \end{cases}$$
The answer is Yes.
The generating function for $t_n$ is $$\sum_{n\geq 0} t_n x^n = \frac14\big(\frac{1+x}{\sqrt{1-6x+x^2}}-1\big).$$ Correspondingly, $$\sum_{n\geq 0} t_n x^n \equiv \frac{1+x}{\sqrt{1+x^2}}-1 \pmod{3}.$$ It follows that for $n>0$, $$t_n \equiv \binom{-1/2}{\lfloor n/2\rfloor}\equiv (-1)^{\lfloor n/2\rfloor}\binom{2\lfloor n/2\rfloor}{\lfloor n/2\rfloor}\pmod{3},$$ from where Lucas' theorem gives the desired result.
Here is a comment following Max Alekseyev's resolution. It has to do with his generating function for $t_n$ and working out directly on the Taylor's expansion (Binomial Theorem). Namely, $$\frac14\left(\frac{1+x}{\sqrt{1-6x+x^2}}-1\right) =\sum_{n=1}^{\infty}\frac{h(n-1)+h(n)}4\,x^n$$ where $$h(n)=\sum_{k=\lfloor\frac{n}2\rfloor}^n(-1)^{n+k}\binom{2k}k\binom{k}{n-k}3^{2k-n}2^{-n}.$$ To get to the conclusion we seek, let's just take $n\rightarrow 2n$ for instance. This leads to \begin{align*}t_{2n}=\frac{h(2n-1)+h(2n)}4&\equiv_3h(2n-1)+h(2n) \\ &\equiv_3\sum_{k=n}^{2n-1}(-1)^{k-1}\binom{2k}k\binom{k}{2n-1-k}3^{2k-2n+1}2^{-2n+1} \\ &+\sum_{k=n}^{2n}(-1)^k\binom{2k}k\binom{k}{2n-k}3^{2k-2n}2^{-2n} \\ &\equiv_3\sum_{k=n}^{2n}(-1)^k\binom{2k}k\binom{k}{2n-k}3^{2k-2n}2^{-2n} \\ &\equiv_3(-1)^n\binom{2n}n2^{-2n} \\ &\equiv_3(-1)^n\binom{2n}n. \end{align*}
