<s>[INST] Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
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and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$
 [/INST] From $$\frac{\pi^2}8=\sum_{2\nmid n}\frac1{n^2}$$
we find that 
$$\frac{\pi^2}{72}=\frac1{3^2}\sum_{2\nmid n}\frac1{n^2}=\sum_{2\nmid n, 3\mid n}\frac1{n^2}$$
and hence 
$$\begin{align}\frac{\pi^2}9=\frac{\pi^2}8-\frac{\pi^2}{72}&=\sum_{2\nmid n, 3\nmid n}\frac1{n^2}\\&=\frac1{1^2}+\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+\frac1{13^2}+\frac1{17^2}+\frac1{19^2}+\frac1{23^2}+\frac1{25^2}+\ldots\end{align}$$
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