I am curious if there are some interesting sums of reciprocals of square integers that equal a reciprocal of a cubed integer besides simply modifying a geometric series.
I first checked to see if I could find a pattern in the infinite sums where the integers that are squared in the denominators increase as slowly as possible. For examples, $$\frac{1}{2^3}=\frac{1}{3^2}+\frac{1}{9^2}+\frac{1}{26^2}+\frac{1}{126^2}+\frac{1}{1036^2}+\cdots \\\frac{1}{3^3}=\frac{1}{6^2}+\frac{1}{11^2}+\frac{1}{32^2}+\frac{1}{235^2}+\frac{1}{2815^2}+\cdots \\\frac{1}{4^3}=\frac{1}{9^2}+\frac{1}{18^2}+\frac{1}{73^2}+\frac{1}{437^2}+\frac{1}{9012^2}+\cdots \\\frac{1}{5^3}=\frac{1}{12^2}+\frac{1}{31^2}+\frac{1}{259^2}+\frac{1}{3908^2}+\frac{1}{955662^2}+\cdots \\\frac{1}{6^3}=\frac{1}{15^2}+\frac{1}{74^2}+\frac{1}{624^2}+\frac{1}{22787^2}+\frac{1}{2691037^2}+\cdots $$ Is there some pattern in these sequences of numbers? I'm very curious to know.
But also, are there other non-geometric series that equal cube reciprocals and which follow a pattern?
And finally, what about finite series? For $1/4^3$ we can also say $$\frac{1}{4^3}=\frac{1}{8^2}=\frac{1}{9^2}+\frac{1}{18^2}+\frac{1}{72^2}$$ What similar finite sums exist for other reciprocals of cubes?