I understand. Given integers $1 \leq a \leq b,$ we can travel out of the $(1,1,1)$ solution of $$ x^2 + a y^2 + b z^2 = (1+a+b) \; xyz $$ with "jumps" $$ x' = (1+a+b) \; yz - x \; , \; $$ $$ y' = \left(\frac{1+a+b}{a} \right) \; zx - y \; , \; $$ $$ z' = \left(\frac{1+a+b}{b} \right) \; xy - z \; . \; $$ In order for this to always work, we need integer coefficients, or both $$ a | (b+1) \; , \; $$ $$ b | (a+1) \; . \; $$ If $b > 1,$ we need $b = a+1,$ hence $a | (a+2)$ and $a | 2.$ That is, $$ a \leq 2 \; , \; \; b = a+1 \; \; . $$

References include KAP and me and HURWITZ. I imagine Waldschmidt knows some more references related to the $(1,a,b)$ examples. I think I will do a few layers of the 1,2,3 tree, this is new to me. $$ x^2 + 2 y^2 + 3 z^2 = 6 xyz $$ The three jumps are $$ x' = 6yz - x \; , \; \; y' = 3 zx - y \; , \; \; z' = 2xy - z \; \; . $$

I understand. Given integers $1 \leq a \leq b,$ we can travel out of the $(1,1,1)$ solution of $$ x^2 + a y^2 + b z^2 = (1+a+b) \; xyz $$ with "jumps" $$ x' = (1+a+b) \; yz - x \; , \; $$ $$ y' = \left(\frac{1+a+b}{a} \right) \; zx - y \; , \; $$ $$ z' = \left(\frac{1+a+b}{b} \right) \; xy - z \; . \; $$ In order for this to always work, we need integer coefficients, or both $$ a | (b+1) \; , \; $$ $$ b | (a+1) \; . \; $$ If $b > 1,$ we need $b = a+1,$ hence $a | (a+2)$ and $a | 2.$ That is, $$ a \leq 2 \; , \; \; b = a+1 \; \; . $$

References include KAP and me and HURWITZ. I imagine Waldschmidt knows some more references related to the $(1,a,b)$ examples. I think I will do a few layers of the 1,2,3 tree, this is new to me. $$ x^2 + 2 y^2 + 3 z^2 = 6 xyz $$ The three jumps are $$ x' = 6yz - x \; , \; \; y' = 3 zx - y \; , \; \; z' = 2xy - z \; \; . $$