<s>[INST] What is so special about $a^3+b^3+c^3 = (13m)^3$? It seems rather surprising that, given the Diophantine equation,
$$a^3+b^3+c^3 = n^3\tag1$$
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.)
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 below. Oleg noticed a curious behavior:
\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 $218$ unsolvables,
$$a^3+b^3+c^3 = (7m)^3$$
and $300$,
$$a^3+b^3+c^3 = (13m)^3$$

Q: What could explain this "anomalous" behavior of $p=2^k,7^k,13^k$ for $k=1,2,3$ with respect to the cubic Diophantine equation $(1)$?   

 [/INST] A solution of (1) must contain $0, 2$ or $4$ terms divisible by $13$.  Essentially, this is because the only cubic residues mod $13$ are $0, 1, 5, 8, 12$, and there is no combination (with or without repetition) of three of the non-zero residues with a sum $s$ such that $s \equiv 0 \pmod{13}$.  A proof is in (A).
Although the same property applies to $2, 3$ and $7$, its effect on (1) for $13$, with $n^3 = (13m)^3$ is especially restrictive in the following sense.  If integers are randomly assigned to $a,b,c$ then the probability $P_0$ that none of them are divisible by $13$ is:
$$P_0 = (12/13)^3 = 1728/2197$$
The probability $P_2$ that exactly two are divisible by $13$ is:
$$P_2 = 3(1/13)^2(12/13) = 36/2197$$
So $1764/2197 ≈ 80\%$ of random assignments fail to meet the above condition.  The corresponding percentages for $2, 3, 7$ are respectively $50\%$, $52\%$ and $68\%$.
The much smaller number of cases for $3$ than for $2,7,13$ is partly explained by the fact that there are solutions of (1) with $n = 6,9$, which eliminates all cases other than those of the form $18k \pm 3$, and in particular all of the form $3^2k$.
Reference
A)  Bailey A (2009) Some Divisibility Properties of Cubic Quadruples, Mathematical Gazette 488 Nov 2009, Note 93.47; doi: 10.1017/S0025557200185262, jstor.
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