# Can these bounds in terms of the abundancy index and deficiency functions be improved for deficient-perfect numbers?

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

• Which result does the paper prove ? Maybe, it can be used to prove your stronger statement. Feb 1 '19 at 10:07
• @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. Feb 1 '19 at 10:10
• Perhaps one can use the fact that $$D(N) = 2N - \sigma(N) = \gcd(N, \sigma(N))$$ which holds when $N$ is deficient-perfect. Feb 10 '19 at 22:31

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.

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 $$D(N)=1$$.