2

Motivation

Let $x, y$ and $z$ be positive integers.

Denote the sum of divisors of $x$ by $\sigma(x)$. Also, denote the deficiency of $y$ by $D(y)=2y-\sigma(y)$. Lastly, denote the abundancy index of $z$ by $I(z)=\sigma(z)/z$.

Let $N = p^k m^2$ be an odd perfect number with special/Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Since $p^k$ and $m^2$ are proper factors of the perfect number $N = p^k m^2$, then $p^k$ and $m^2$ are deficient.

From the referenced paper, and using the facts that $D(p^k) > 1$ and $D(m^2) > 1$, we have the bounds $$\frac{2p^k}{p^k + D(p^k)} < I(p^k) < \frac{2p^k + D(p^k)}{p^k + D(p^k)}$$ and $$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}$$ from which we obtain $$\frac{4p^k m^2}{(p^k + D(p^k))(m^2 + D(m^2))} < I(p^k)I(m^2) = 2 < \frac{(2p^k + D(p^k))(2m^2 + D(m^2))}{(p^k + D(p^k))(m^2 + D(m^2))}.$$

The LHS of the last inequality yields $$p^k m^2 < p^k D(m^2) + m^2 D(p^k) + D(p^k)D(m^2),$$ while the RHS results to $$D(p^k) D(m^2) < 2p^k m^2.$$

Dividing throughout the last two inequalities by $p^k m^2$, we get $$1 < \frac{D(p^k)}{p^k} + \frac{D(m^2)}{m^2} + \frac{D(p^k)}{p^k}\cdot\frac{D(m^2)}{m^2}$$ $$= \bigg(2 - I(p^k)\bigg) + \bigg(2 - I(m^2)\bigg) + \bigg(2 - I(p^k)\bigg)\bigg(2 - I(m^2)\bigg)$$ $$= 4 - \bigg(I(p^k) + I(m^2)\bigg) + 6 - 2\bigg(I(p^k) + I(m^2)\bigg)$$ $$= 10 - 3\bigg(I(p^k) + I(m^2)\bigg) \implies I(p^k) + I(m^2) < 3$$ and $$6 - 2\bigg(I(p^k) + I(m^2)\bigg) = \bigg(2 - I(p^k)\bigg)\bigg(2 - I(m^2)\bigg) = \frac{D(p^k)}{p^k}\cdot\frac{D(m^2)}{m^2} < 2$$ $$\implies 2 < I(p^k) + I(m^2)$$

Since it is known (Corollary 10, page 5) that $$\frac{57}{20} < I(p^k) + I(m^2) < 3$$ it is my impression that:

(1) The LHS of $$\frac{2n}{n + D(n)} < I(n) < \frac{2n + D(n)}{n + D(n)}$$ (for $D(n) > 1$) is best-possible.

<p><strong>(2)</strong> The RHS of
<span class="math-container">$$\frac{2n}{n + D(n)} &lt; I(n) &lt; \frac{2n + D(n)}{n + D(n)}$$</span>
(for <span class="math-container">$D(n) &gt; 1$</span>) is <em>not</em> best-possible.</p>

Question

Are my impressions correct?

Reference

Jose Arnaldo Bebita Dris, A Criterion for Deficient Numbers Using the Abundancy Index and Deficiency Functions, arXiv:1308.6767 [math.NT], 2013-2016; Journal for Algebra and Number Theory Academia, Volume 8, Issue 1 (February 2018), 1-9

Jose Arnaldo Bebita Dris
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1 Answers1

1

This answer proves that

$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n)\tag3$$

$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag4$$

hold for any $a>0,b\gt -1$.

Under the condition that $D(n)\gt 1$, we have

$$\begin{align}(3)&\iff \frac{2n}{n+D}<\frac{(2a+2)n−aD}{(a+1)n+D}\quad\text{and}\quad \frac{(2a+2)n−aD}{(a+1)n+D}<\frac{2n-D}{n} \\\\&\iff 2n(an+n+D)\lt (2an+2n-aD)(n+D)\quad\text{and} \\&\qquad\qquad n(2an+2n-aD)\lt (2n-D)(an+n+D) \\\\&\iff D(n)\lt n \\\\&\iff \sigma(n)\gt n\end{align}$$

and

$$\begin{align}(4)&\iff \frac{2n-D}{n}\lt\frac{(2b+4)n-bD}{(b+2)n+D}\quad\text{and}\quad \frac{(2b+4)n-bD}{(b+2)n+D}\lt \frac{2n+D}{n+D} \\\\&\iff (2n-D)(bn+2n+D)\lt n(2bn+4n-bD)\quad\text{and} \\&\qquad\qquad (2bn+4n-bD)(n+D)\lt (2n+D)(bn+2n+D) \\\\&\iff D(n)\gt 0\end{align}$$

It follows that both $(3)$ and $(4)$ hold under the condition that $D(n)\gt 1$.

Therefore, the answer is as follows :

  • The impression (1) is not correct.

  • The impression (2) is correct.

mathlove
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