Congruence equation for Apery numbers Does the system of congruence equations
\begin{eqnarray} 
A_{17k}&\equiv& 0 \pmod {17^2}, \nonumber \\ 
A_{17k+1}&\equiv& 0 \pmod {17^2}, \tag{1}
\end{eqnarray}
has solutions other than $k=3$? Here $A_n$ are Apery numbers:
$$A_n=\sum\limits_{k=0}^n\binom{n}{k}^2\binom{n+k}{k}^2.$$
Thanks to the recurrence relation
$$n^3 A_n = (34n^3-51n^2+27n-5)A_{n-1}- (n-1)^3 A_{n-2},$$
if $k$ is a solution of (1), then all seventeen consecutive numbers beginning from $A_{17k}$ are congruent to zero modulo $17^2$. In particular,
$$A_{51},A_{52},A_{53},\ldots,A_{66},A_{67}\equiv 0 \pmod {17^2}.$$ 
 A: Results of Gessel can be used to solve this question. Gessel showed, in Theorem 1 here, that for any prime $p$, if $n=\sum d_i p^i$ is the base-$p$ expansion of $n$, then
$$(*) A_n \equiv \prod A_{d_i} \bmod p.$$
In Theorem 4 here he essentially refined this as follows. He proved that there is an (explicit) sequence $\{b_n\}_{n \ge 0}$ such that $$(**) A_{pn+k} \equiv A_{n} (A_{k} + pn b_{k}) \bmod {p^2},$$
where $p$ is a prime, $n \ge 0$ and $0 \le k < p$. The first values of $b_n$ are given by $b_0=0,b_1=12$.

Specializing $(*)$ to $p=17$, we see that $A_n$ is divisible by $17$ iff one of the digits $d_i$ in the base-17 expansion of $n$ satisfies $A_{d_i} \equiv 0 \bmod 17$. A finite check shows that the solutions to $A_d \equiv 0 \bmod 17$ in the range $0 \le d < 17$ are $d=3,13$. 
Specializing $(**)$ to $p=17$ and $k \in \{0,1\}$, we find that
$$A_{17n} \equiv A_{n} \bmod {17^2},$$
$$A_{17n+1} \equiv A_{n} (A_1 + 17\cdot 12\cdot n) \bmod {17^2}.$$
This immediately shows that the first equation in your system implies the second one, and that your system is equivalent to 
$$(***)A_n \equiv 0 \bmod {17^2}.$$
We now proceed to solve $(***)$. Let $S$ be the set of solutions to $(***)$. Writing $n$ as $17m+m'$ ($0 \le m' < 17$), we find that $(***)$ is equivalent to
$$A_{m} (A_{m'}+17mb_{m'}) \equiv 0 \bmod {17^2}.$$
There are three possible cases.


*

*$A_m \equiv 0 \bmod {17^2}$.

*$A_m \equiv 0 \bmod {17}$, $A_{m'}+17mb_{m'} \equiv 0 \bmod 17$.

*$A_{m'} + 17mb_{m'} \equiv 0 \bmod {17^2}.$


The first case says: if $n \in S$ then $17n+n' \in S$ ($0 \le n' < 17$).
The second case says: $m$ has $3$ or $13$ as one of its base-17 digits, and $m' \in \{3,13\}$.
The third case says: Either $m'=3$ and $m \equiv 0 \bmod 17$, or $m'=13$ and $m \equiv -1 \bmod 17$. 
(Here I've used the values $A_3=1445=17^2\cdot 5$, $A_{13} = 364713572395983725$, $b_{3}=4438$, $b_{13}=7475161875743183448469/6006$.)
This characterizes the set $S$.
A: According to a computation there are other solutions.
Solutions $k$ start:  $3 , 51 , 52 , 53 , 54 , 55 , 56 , 57 , 58 , 59 , 60 , 61 , 62 , 63 , 64 , 65 , 66 , 67,????, 51\cdot 17, 52 \cdot 17$

Jeremy Rouse suggests missing  $k=224, k=234, k=285, k=292, k=343$ and
I didn't search the range $????$.
