# Catalan numbers, Condition that $\sum_{k=1}^n c_k=1 \mod 3$ using Lucas theorem

Catalan numbers are $c_n=\frac{1}{n+1}{\binom {2n}{n}}$.

Prove that $\sum_{k=1}^n c_k\equiv 1 \mod 3 \iff$ The digit $2$ appears in the ternary representation of $n+1$.

I was shown the solution a while ago and forgot it: It relies on finding some short identity that $c_n$ satisfy when considered modulo $3$, and using Lucas' theorem: $\binom{n}{k}$ is divisible by 3 iff when adding $k+(n-k)$ in base $3$ there is a carry.

$$\binom{2k+2}{k+1} -\binom{2k}{k}-\frac 1{k+1}\binom{2k}{k} = \binom{2k}{k} \left(\frac{(2k+2)(2k+1)}{(k+1)(k+1)}-1-\frac1{k+1}\right) = \binom{2k}{k} \frac{3k}{k+1} = 3 \binom{2k}{k+1}.$$
I was told the main part of the proof: we show that $\sum_{k=1}^n c_k -1 \equiv \binom{2(n+1)}{n+1} \mod 3$, and then we get $$\sum_{k=1}^n c_k -1 \equiv 0 \mod 3 \iff \binom{2(n+1)}{n+1} \equiv 0 \mod 3$$
$\iff$there is a carry when adding $(n+1)+(n+1)$ in base $3$ $\iff$ The digit $2$ appears in the ternary representation of $n+1$.
Therefore, it suffices to prove $c_k \equiv \binom{2k+2}{k+1} - \binom{2k}{k} \mod 3$, and we will get by a telescopic series what we wanted. Can someone now tie the last knot and prove the congruence?