Mu badness: Difference between revisions
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<math>f\left(x,k\right)=\frac{\operatorname{abs}\left(\operatorname{mod}\left(2g\left(k\right)x,2\right)-1\right)}{k^{2}}</math> | <math>f\left(x,k\right)=\frac{\operatorname{abs}\left(\operatorname{mod}\left(2g\left(k\right)x,2\right)-1\right)}{k^{2}}</math> | ||
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