Apéry congruence conjecture for the second-kind numbers modulo $9$

For $n\ge 0$, define the Apéry numbers of the second kind by

$$A^{(3)}_n=\sum_{k=0}^n \binom{n}{k}^2\binom{n+k}{k}^2.$$

Write the ternary expansion of $n$ as a finite word over $\{0,1,2\}$, and let $k$ be a nonnegative integer.

Second-kind Apéry congruence conjecture. The numbers $A^{(3)}_n$ satisfy the following congruences modulo $9$:

1. $A^{(3)}_n\equiv1\pmod 9$ if and only if the ternary expansion of $n$ contains $6k$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
2. $A^{(3)}_n\equiv2\pmod 9$ if and only if the ternary expansion of $n$ contains $6k+5$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
3. $A^{(3)}_n\equiv4\pmod 9$ if and only if the ternary expansion of $n$ contains $6k+4$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
4. $A^{(3)}_n\equiv5\pmod 9$ if and only if the ternary expansion of $n$ contains $6k+1$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
5. $A^{(3)}_n\equiv7\pmod 9$ if and only if the ternary expansion of $n$ contains $6k+2$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
6. $A^{(3)}_n\equiv8\pmod 9$ if and only if the ternary expansion of $n$ contains $6k+3$ digits equal to $1$ for some $k$, and otherwise contains only $0$'s and $2$'s.
7. In all other cases, $A^{(3)}_n$ is divisible by $9$; in particular, $A^{(3)}_n\not\equiv3,6\pmod 9$ for every $n$.

As with the first-kind numbers, the relevant differential equation is not suitable for the paper's method. The patterns were obtained conjecturally modulo $9$, and the supplied text gives no resolution.