It seems rather surprising that, given the Diophantine equation,

$$a^3+b^3+c^3 = n^3$$

then a good $\color{red}{99.8\%}$ of $n<1000000$ are solvable in positive integers $a,b,c$. (See the discussion in this [MSE post][1].)

Oleg567 gave a list of the $1867$ unsolvable $n$ in that range. More than half were primes, while the divisibility by prime $p$ of the unsolvable composites are given by the table:

\begin{array}{|l|c|c|c|c|}
\hline
p & \mbox{# divisible  by } p & ...\; p^2 & ...\;p^3 \\
\hline
2 & \color{blue}{177} &  41 & 22\\
3 & 31 &-&-\\
5 & 53 &-&-\\
7 & \color{blue}{218} & 52 & 11 \\
11 & 43 &-&-\\
13 & \color{blue}{300} & 46 & 4 \\
17 & 47 &-&-\\
23 & 31 &-&-\\
31 & 34 &-&-\\
37 & \color{blue}{82} &-&-\\
43 & 39 &-&-\\
47 & 29 &-&-\\
59 & 24 &1&-\\
61 & 13 &-&-\\
73 & 16 &-&-\\
79 & 48 &4&-\\
\hline
\end{array}

Thus, in that range, there are just $31$ unsolvables of form,

$$a^3+b^3+c^3 = (3m)^3$$

but a good $300$ unsolvables,

$$a^3+b^3+c^3 = (13m)^3$$

and $82$,

$$a^3+b^3+c^3 = (37m)^3$$

>Q: What could explain this behavior of $p=7,13,37$ with respect to cubes?   

  [1]: https://math.stackexchange.com/questions/2514643/statistics-for-n-in-sum-of-cubes-a3b3c3-n3