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

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.