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This MSE question (from April 2020) asked whether the inequality $$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}$$ could be improved, where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$, $s(x)=\sigma(x)-x$ is the aliquot sum of $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.

In the accepted answer, MSE user Adam Ledger asserts that:

I am yet to find a counterexample for the following, I have not spent much time on your problem so will appreciate if this is not considered an improvement on your original inequality, but none the less I hope that it helps in some small way:

Denoting the Kronecker delta as follows: $$\delta \left( x,y \right) =\cases{1&$x=y$\cr 0&$x\neq y $\cr}\tag{ 0}$$

up to $n \leq 2 \cdot 10^7$ I have found the following to be satisfied: $${\frac {D \left( n \right) }{s \left( n \right) }}-{\frac { D \left( {n}^{2} \right) }{s \left( {n}^{2} \right) }}-\frac{1}{4} \delta \left( n-2\,\left\lfloor \frac{n}{2}\right\rfloor,1 \right) \lt \frac{3}{4} \tag{1}$$

Adam's result implies that we have the bounds $$0 < \frac{D(n)}{s(n)} - \frac{D(n^2)}{s(n^2)} < 1.$$

Here is my inquiry:

QUESTION: Does anybody here know how to prove Adam's Inequality $(1)$?

MOTIVATION FOR THE INQUIRY

Let $N = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

This answer shows that we have $$\frac{D(m^2)}{s(m^2)}=\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)},$$ which means that, conjecturally, we should have $$\frac{D(p^k m)}{D(p^k)s(m)}=\frac{D(m)}{s(m)}-\frac{D(m^2)}{s(m^2)} < 1$$ by Adam's result.

Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$.

Note that we have the numerical bounds $$1 < I(p^k) < \frac{5}{4} < \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < I(m) < 2,$$ from which we get $$0 < \frac{D(p^k m)}{D(p^k)s(m)}=\frac{2 - I(p^k)I(m)}{(2 - I(p^k))(I(m) - 1)} < \dfrac{2-\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}}{\dfrac{3}{4}\bigg(\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} - 1\bigg)} \approx 1.666929067.$$

Alas, this is where I get stuck!

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1 Answer 1

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This answer proves that for every $n\gt 1$, $$\dfrac{D(n)}{s(n)} - \dfrac{D(n^2)}{s(n^2)} < 1\tag2$$ holds.

To prove $(2)$, it is sufficient to prove the following claims.

Let $f(n):=(\sigma(n)-2n)\sigma(n^2)+n^3$.

Claim 1 : $(2)$ is equivalent to $f(n)\gt 0$.

Claim 2 : If $p$ is a prime number and $k$ is a positive integer, then $f(p^k)\gt 0$.

Claim 3 : If $f(n)\gt 0$ and $p$ is a prime number satisfying $\gcd(p,n)=1$ and $k$ is a positive integer, then $f(p^kn)\gt 0$.


Example : $f(2^4)\gt 0$ (claim 2) $\implies f(2^43^2)\gt 0$ (claim 3) $\implies f(2^43^25^3)\gt 0$ (claim 3).


Claim 1 : $(2)$ is equivalent to $f(n)\gt 0$.

Proof :

$$\frac{D(n)}{s(n)}-\frac{D(n^2)}{s(n^2)}=\bigg(\frac{n}{\sigma(n)-n}-1\bigg)-\bigg(\frac{n^2}{\sigma(n^2)-n^2}-1\bigg)$$ $$=\frac{n}{\sigma(n)-n}-\frac{n^2}{\sigma(n^2)-n^2}=\frac{n\sigma(n^2)-n^2\sigma(n)}{(\sigma(n)-n)(\sigma(n^2)-n^2)}$$

So, we have $$\begin{align}&\frac{D(n)}{s(n)}-\frac{D(n^2)}{s(n^2)}\lt 1\iff \frac{n\sigma(n^2)-n^2\sigma(n)}{(\sigma(n)-n)(\sigma(n^2)-n^2)}\lt 1 \\\\&\iff n\sigma(n^2)-n^2\sigma(n)\lt (\sigma(n)-n)(\sigma(n^2)-n^2) \\\\&\iff n\sigma(n^2)-n^2\sigma(n)\lt \sigma(n)\sigma(n^2)-n^2\sigma(n)-n\sigma(n^2)+n^3 \\\\&\iff (\sigma(n)-2n)\sigma(n^2)+n^3\gt 0\qquad\blacksquare\end{align}$$


Claim 2 : If $p$ is a prime number and $k$ is a positive integer, then $f(p^k)\gt 0$.

Proof :

$$\begin{align}f(p^k)&=(\sigma(p^k)-2p^k)\sigma(p^{2k})+p^{3k} \\\\&=\bigg(\frac{p^{k+1}-1}{p-1}-2p^k\bigg)\frac{p^{2k+1}-1}{p-1}+p^{3k} \\\\&=\frac{p^k-1}{(p-1)^2}\bigg(\underbrace{p^k(p^{k}-p +1) - 1}_{\text{positive}}\bigg) \\\\&\gt 0\qquad\blacksquare\end{align}$$


Claim 3 : If $f(n)\gt 0$ and $p$ is a prime number satisfying $\gcd(p,n)=1$ and $k$ is a positive integer, then $f(p^kn)\gt 0$.

Proof :

$$\begin{align}f(p^kn)&=(\sigma(p^kn)-2p^kn)\sigma(p^{2k}n^2)+p^{3k}n^3 \\\\&=(\sigma(p^k)\sigma(n)-2p^kn)\sigma(p^{2k})\sigma(n^2)+p^{3k}n^3 \\\\&=\bigg(\frac{p^{k+1}-1}{p-1}\sigma(n)-2p^kn\bigg)\frac{p^{2k+1}-1}{p-1}\sigma(n^2)+p^{3k}n^3\end{align}$$

So, we have $$\small\begin{align}&(p-1)^2f(p^kn) \\\\&=\bigg((p^{k+1}-1)\sigma(n)-2p^kn(p-1)\bigg)(p^{2k+1}-1)\sigma(n^2)+p^{3k}n^3(p-1)^2 \\\\&=\bigg((p^{k+1}-1)\sigma(n)-p^k(p-1)\sigma(n)+p^k(p-1)\sigma(n)-2p^k(p-1)n\bigg)(p^{2k+1}-1)\sigma(n^2) \\&\qquad +p^{3k}n^3(p-1)^2 \\\\&=\bigg((p^{k+1}-1)\sigma(n)-p^k(p-1)\sigma(n)\bigg)(p^{2k+1}-1)\sigma(n^2) \\&\qquad +\bigg(p^k(p-1)\sigma(n)-2p^k(p-1)n\bigg)(p^{2k+1}-1)\sigma(n^2)+p^{3k}n^3(p-1)^2 \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) \\&\qquad +p^k(p-1)\underbrace{(\sigma(n)-2n)\sigma(n^2)}_{\gt -n^3}(p^{2k+1}-1)+p^{3k}n^3(p-1)^2 \\\\&\gt(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) \\&\qquad +p^k(p-1)(-n^3)(p^{2k+1}-1)+p^{3k}n^3(p-1)^2 \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2) -n^3 (p - 1) p^k (p^{2 k} - 1) \\\\&=(p^k-1)(p^{2k+1}-1)\sigma(n)\sigma(n^2)-(p - 1) p^k (p^{2 k} - 1)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)\sigma(n)\sigma(n^2) - (p - 1) p^k (p^{2 k} - 1)n^3 \\\\&=\bigg((p^k-1)(p^{2k+1}-1)-(p - 1) p^k (p^{2 k} - 1)\bigg)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)(\sigma(n)\sigma(n^2) - n^3) \\\\&=(p^k - 1)\bigg( \underbrace{p^k(p^{k}-p +1) - 1}_{\text{positive}}\bigg)\sigma(n)\sigma(n^2) \\&\qquad +(p - 1) p^k (p^{2 k} - 1)\bigg(\underbrace{\sigma(n)\sigma(n^2) - n^3}_{\text{positive}}\bigg) \\\\&\gt 0\end{align}$$ Hence, $f(p^kn)\gt 0$ follows.$\quad\blacksquare$

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  • $\begingroup$ Allow me to review your answer in the next couple of days, @mathlove! An upvote for now. =) $\endgroup$ Commented Dec 29, 2022 at 22:06
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    $\begingroup$ Great answer. I was wondering about a justification of the $(σ(n)−2n)σ(n^2)>-n^3$ inequality and I think one is that it holds for all primes and is equivalent to $(I(n)-2)I(n^2)>-1$. Therefore the latter inequality holds for all primes, and as if $p$ divides $n$ then $I(p)\leq I(n)$, $(I(n)-2)I(n^2)\geq (I(p)-2)I(p^2)>-1$. $\endgroup$
    – SFA
    Commented Dec 31, 2022 at 22:29
  • $\begingroup$ I agree that it is a great answer. I am hereby accepting your answer, @mathlove! =) $\endgroup$ Commented Jan 1, 2023 at 4:30

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