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 $$n^3 A_n = (34n^351n^2+27n5)A_{n1} (n1)^3 A_{n2},$$ 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, $$A_{51},A_{52},A_{53},\ldots,A_{66},A_{67}\equiv 0 \pmod {17^2}.$$
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 base17 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 base17 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 $S$.
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 $????$.

1$\begingroup$ This is $3$, then $3\cdot 17,\ldots, 3\cdot 17+ 16$, then $3\cdot 17^2$,\ldots$, which might be suggestive. $\endgroup$ – Jeff Strom May 30 '14 at 11:26


$\begingroup$ That is it can be conjected that $k=3*17^m$ are all solutions? Or only for prime m? $\endgroup$ – Zurab Silagadze May 30 '14 at 11:45

$\begingroup$ @ZurabSilagadze maybe for all $m$, don't know. Don't have efficient computation of binomial coefficients modulo prime powers so far. $\endgroup$ – joro May 30 '14 at 12:03

1$\begingroup$ @joro  You're missing some. For example, $k = 224$, $k = 234$, $k = 285$, $k = 292$, $k = 343$, etc. $\endgroup$ – Jeremy Rouse May 30 '14 at 12:52