Denote $$g(n)=\{\{x,y,z\}\mid \frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n,x,y,z\in\mathbb N\},$$ $$h(n)=\{\{x,y,z\}\mid \frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n,1\leq x\leq y\leq z,x,y,z\in\mathbb N\},$$ let $f(n)=|g(n)|$ be the number of members of $g(n)$.
For example, $h(3)=\{\{2,3,6\},\{2,4,4\},\{3,3,3\}\},f(3)=6+3+1=10.$
Since $\{n,n,n\}$ is a solution to $\frac{1}x+\frac{1}y+\frac{1}z=\frac{3}n$, it's easy to see that $f(k)\equiv 1\pmod 3,\forall k\in \mathbb N.$
Question: I find that $$f(3k)\equiv 0,f(4k+2)\equiv 0,f(6k\pm1)\equiv1 \pmod 2,\forall k\in \mathbb N.$$ I wonder how to prove them?
Edit: I find that $f(n)$ has the same parity to the number of solutions to $\frac{1}x+\frac{2}y=\frac{3}n,$ I think I have got it now.