  Let p be any odd prime. We mainly show that
$$\sum_{k=1}^{p-1}binomial(3k,k)*2^k/k=0 (mod p)$$ and
$$\sum_{k=1}^{p-1}2^{k-1}C_k^{(2)}=(-1)^{(p-1)/2}-1 (mod p),$$ where
$C_k^{(2)}=binomial(3k,k)/(2k+1)$ is the $k$th Catalan number of order 2.
