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The Abundancy Index of Divisors of Odd Perfect Numbers

Abstract

If N=qkn2N = {q^k}{n^2} is an odd perfect number, where qq is the Euler prime, then we show that n<qn < q is sufficient for Sorli's conjecture that k=νq(N)=1k = \nu_{q}(N) = 1 to hold. We also prove that qk<2/3n2q^k < 2/3{n^2}, and that I(qk)<I(n)I(q^k) < I(n), where I(x)I(x) is the abundancy index of xx.Comment: 10 page

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