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Let $$\sigma(x) = \sum_{e \mid x}{e}$$ denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, and the deficiency of $x$ by $D(x)=2x-\sigma(x)$. A positive integer $N$ is said to be deficient-perfect if $D(N) \mid N$.

Here is my question:

Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers $N > 1$? $$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$

(Note that the inequality $$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}$$ is true if and only if $N$ is deficient.)

References

A Criterion for Deficient Numbers Using the Abundancy Index and Deficiency Functions, Journal for Algebra and Number Theory Academia, Volume 8, Issue 1, February 2018, pages 1-9

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  • $\begingroup$ Which result does the paper prove ? Maybe, it can be used to prove your stronger statement. $\endgroup$ – Peter Feb 1 at 10:07
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    $\begingroup$ @Peter: The paper proves the inequality above, giving a criterion for deficient numbers in terms of the abundancy index and deficiency functions. I am currently trying to determine whether the inequality above could be improved to account for the case when $N > 1$ is deficient-perfect. $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 1 at 10:10
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    $\begingroup$ Perhaps one can use the fact that $$D(N) = 2N - \sigma(N) = \gcd(N, \sigma(N))$$ which holds when $N$ is deficient-perfect. $\endgroup$ – Jose Arnaldo Bebita-Dris Feb 10 at 22:31
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ILLUSTRATING VIA A TOY EXAMPLE

Let $M$ be an odd perfect number given in the so-called Eulerian form $$M = p^k m^2$$ (i.e. $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$).

It is known that the non-Euler part $m^2$ is deficient-perfect if and only if the Descartes-Frenicle-Sorli conjecture that $k=1$ holds. (See this paper for a proof of this fact.)

So, suppose that $k=1$. Then $m^2$ is deficient-perfect.

In particular, $m^2$ is deficient, so that the criterion in this paper applies.

We have $$\frac{2m^2}{m^2 + D(m^2)} < I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)}.$$

Under the hypothesis that $k=1$, $m^2$ is deficient-perfect, with deficiency $$D(m^2) = \frac{m^2}{(p+1)/2}.$$

We also have $$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p+1}.$$

Putting these all together, we have $$\frac{m^2}{D(m^2)} = \frac{p+1}{2}$$ $$\frac{2p}{p+1} = I(m^2) > \frac{2m^2}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg)}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg)}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+1}{\frac{p+3}{2}} = \frac{2(p+1)}{p+3}$$ which implies that $$p^2 + 3p = p(p+3) > (p+1)^2 = p^2 + 2p + 1$$ $$p > 1$$ (This last inequality is trivial as $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.) $$\frac{2p}{p+1} = I(m^2) < \frac{2m^2 + D(m^2)}{m^2 + D(m^2)} = \frac{2\bigg(\frac{m^2}{D(m^2)}\bigg) + 1}{\frac{m^2}{D(m^2)} + 1} = \frac{2\bigg(\frac{p+1}{2}\bigg) + 1}{\bigg(\frac{p+1}{2}\bigg) + 1} = \frac{p+2}{\frac{p+3}{2}} = \frac{2(p+2)}{p+3}$$ which implies that $$p^2 + 3p = p(p+3) < (p+1)(p+2) = p^2 + 3p + 2$$ $$0 < 2.$$

This example illustrates my interest in improvements to the bounds in terms of the abundancy index and deficiency functions of $N$, when $N > 1$ is deficient-perfect.

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Suppose that $N > 1$ is deficient-perfect. Since $N$ is deficient, then $$\frac{2N}{N + D(N)} < I(N) < \frac{2N + D(N)}{N + D(N)}.$$

I think that, since $D(N) \mid N$ when $N$ is deficient-perfect, then $N/D(N)$ is an integer, so that we have (since $\frac{N}{D(N)} \mid N$) $$I\bigg(\frac{N}{D(N)}\bigg) \leq I(N) < \frac{2N + D(N)}{N + D(N)} = \frac{2\bigg(\frac{N}{D(N)}\bigg) + 1}{\bigg(\frac{N}{D(N)}\bigg) + 1}.$$

CLAIM

$$\frac{2\bigg(\frac{N}{D(N)}\bigg)}{\bigg(\frac{N}{D(N)}\bigg) + 1} < I\bigg(\frac{N}{D(N)}\bigg)$$

This claim, if true, would prove that all deficient-perfect numbers $N$ correspond to almost perfect numbers $N/D(N)$.

Added April 18 2019 (6:13 PM - Manila time)

The claim is false. A counterexample is given by $$N = \bigg({3}\cdot{7}\cdot{11}\cdot{13}\bigg)^2.$$

Added April 18 2019 (6:17 PM - Manila time)

It appears that the claim is true when $D(N)=1$.

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