# Tag Info

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form. Let $$I(x) = \dfrac{\sigma(x)}{x}$$ be the abundancy index of $x$. This is a partial answer to the original question, and proves the claim in the affirmative, subject to the validity of a recent proof claim by Patrick A. Brown that $q^k < n$ holds (in many cases). First, we show the ...