Given the Diophantine equation,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$

there is the rather curious observation that the **smallest** positive solutions for $k=5$ or $6$ is *multi-grade*.

$$24^k+28^k+67^k=3^k+54^k+62^k,\quad k = 1,5$$

$$15^k + 10^k + 23^k = 3^k + 19^k + 22^k,\quad k = 2,6$$

Duncan Moore has exhaustively searched $(1)$ for all positive and **primitive** solutions below a bound $Z$. [Table 1][1] is for $k=5$, while [Table 2][2] is for $k=6$. We summarize the data below.

**I. Table 1:**

$$\begin{array}{|c|c|c||}
\text{# of solns}&\color{blue}{A:=\text{(% of}\; k = 1,5)}&\text{diff}\\
2^0\cdot168&63.7\text{%}& \\
2^1\cdot168&65.8\text{%}&+2.7\\
2^2\cdot168&65.6\text{%}&-0.3 \\
2^3\cdot168&63.6\text{%}&-2.0\\
2^4\cdot168&61.0\text{%}&-2.6\\
2^5\cdot168&59.1\text{%}&-1.9\\
\end{array}$$

*Note:* To address one comment below, $A$ is the percentage of solns given in the first column that is valid for both $k=1,5$. For example, out of the first $2^5\cdot168 = 5376$ solns, then $59.1\text{%}$ are for $k=1,5$.

Each row doubles the $\text{#}$. Since Moore's database has $5393$ solns, and $5393/2^5\approx168.53$, then I used that as the base value. 

**II. Table 2:**

$$\begin{array}{|c|c|c|}
\text{# of solns}&\color{blue}{B:=\text{(% of}\; k = 2,6)}&\text{diff}\\
50&80\text{%}& \\
100&85\text{%}&+5.0\\
200&89\text{%}&+4.0\\
400&91.7\text{%}&+2.7\\
\end{array}$$

*Note:* Thus, out of the first $400$ solns, then a whopping $91.7\text{%}$ of them are actually multi-grade for $k=2,6$. (I'm not sure if excluding non-primitive solutions below the bound $Z$ is relevant. Program-wise, it seems easier to just include them.)

**Questions:**

1. Why is the percentage of $A$ *decreasing*, while that of $B$ is *apparently increasing*? Or will $B$ eventually have a negative diff like $A$? (The data is too small to be conclusive.)
2. If both are decreasing, will $A,B \to 0$? Or will it taper off to some constant? 


**P.S.** [This answer][3] to a related post might be informative. Incidentally, the smallest solutions to,

$$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+x_4^k\tag2$$

are **also** multigrades as $k=1,5$, and $k=2,6$, though there are no exhaustive tables for these.


  [1]: http://homepage.ntlworld.com/duncan-moore/taxicab/3_3_5.txt
  [2]: http://homepage.ntlworld.com/duncan-moore/taxicab/3_3_6.txt
  [3]: http://mathoverflow.net/a/150430/12905