\frac{1}{16}\sum\limits_{i=0}^3\sum\limits_{j=0}^3 (81-(2i/4)^2-3(2j/4)^2) \frac{1}{16}\sum\limits_{i=0}^3\sum\limits_{j=0}^3 (81-(2i/4)^2-3(2j/4)^2)

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