Let $N = q^k n^2$ be an odd perfect number given in Eulerian form (i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$).
In this post, I would like to compute the general expression for the quantity
$$\frac{n^2}{D(n^2)}$$
where $D(x) = 2x – \sigma(x)$ is the deficiency of $x$. (Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.) I would also like to derive bounds for the said quantity, which are hopefully sharp.
It is known that, for an odd perfect number $N = q^k n^2$ given in Eulerian form, we have
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)},$$
where $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
We obtain, immediately from the middle equation, the equality
$$\frac{n^2}{D(n^2)}=\frac{\sigma(q^k)}{2s(q^k)}.$$
This quantity is an integer if and only if $k=1$ i.e. when the RHS is $(q+1)/2$. Here I am interested in deriving bounds for this quantity when $k>1$.
We rewrite the RHS as follows:
$$\dfrac{\sigma(q^k)/q^k}{2(s(q^k)/q^k)} = \frac{I(q^k)}{2(I(q^k) – 1)},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.
Now, use the following bounds (which hold since $k>1$ and $k \equiv 1 \pmod 4$ imply that $k \geq 5$):
$$I(q^5) \leq I(q^k) < \frac{q}{q-1}.$$
This gives
$$\dfrac{I(q^5)}{2\bigg(\dfrac{q}{q-1} – 1\bigg)} < \frac{n^2}{D(n^2)}=\frac{I(q^k)}{2(I(q^k) – 1)} < \dfrac{\dfrac{q}{q-1}}{2(I(q^5) – 1)}.$$
The LHS and RHS of the last inequality then simplifies to
$$\text{LHS} = \dfrac{I(q^5)}{2\bigg(\dfrac{q}{q-1} – 1\bigg)} = \frac{q}{2} – \frac{1}{2q^5}$$
while
$$\text{RHS} = \dfrac{\dfrac{q}{q-1}}{2(I(q^5) – 1)} = \frac{q^6}{2(q^5 – 1)}.$$
The function
$$f_1(q) = \frac{q}{2} – \frac{1}{2q^5}$$
is increasing, since its derivative is positive. Hence
$$f_1(q) \geq f_1(5) = \frac{7812}{3125}$$ since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$.
On the other hand, the function
$$f_2(q) = \frac{q^6}{2(q^5 – 1)}$$
is also increasing, since its derivative is also positive. Thus, we may rewrite $f_2(q)$ as follows:
$$f_2(q) = \frac{q\cdot{q^5}}{2(q^5 – 1)} = \frac{q(q^5 – 1)}{2(q^5 – 1)} + \frac{q}{2(q^5 – 1)} = \frac{q}{2} + \frac{q}{2(q^5 – 1)}.$$
Consequently, we have the following proposition:
PROPOSITION: If $N = q^k n^2$ is an odd perfect number with special/Euler prime $q$ and $k > 1$ holds, then we have
$$\frac{7812}{3125} \leq \frac{q}{2} – \frac{1}{2q^5} < \frac{n^2}{D(n^2)} < \frac{q}{2} + \frac{q}{2(q^5 – 1)}.$$
In particular:
$$k > 1 \implies \Bigg(\bigg(\frac{n^2}{D(n^2)} > \frac{7812}{3125}\bigg) \iff \bigg(\frac{D(n^2)}{n^2} < \frac{3125}{7812}\bigg) \iff \bigg(I(n^2) > 2 – \frac{3125}{7182} = \frac{12499}{7182} \approx 1.59997439836\bigg)\Bigg)$$
which does not improve on the currently known
$$I(n^2) > \frac{8}{5}.$$
Here are my:
QUESTIONS: Can you improve on the bounds for $n^2 / D(n^2)$ (in terms of $q$) when $k>1$? If it is no longer possible to refine the bounds given here, can you explain/show why?
Best Answer
Yes.
We have $$\frac{n^2}{D(n^2)}=\frac{\sigma(q^k)}{2s(q^k)}=\frac{q^{k+1}-1}{2(q-1)\bigg(\dfrac{q^{k+1}-1}{q-1}-q^k\bigg)}=\frac{q^{k+1}-1}{2q^k-2}$$
Let $f(k):=\dfrac{q^{k+1}-1}{2q^k-2}$. Then, we get $f'(k)=-\dfrac{(q-1)q^k\log q}{2(q^k-1)^2}$ which is negative.
Since $f(k)$ is decreasing, we obtain $$\lim_{k\to\infty}f(k)\lt f(k)\le f(5),$$ i.e. $$\frac q2\lt \frac{n^2}{D(n^2)}\le \frac q2+\dfrac{q-1}{2(q^5-1)}$$