Can this inequality involving the deficiency and sum of aliquot divisors be improved?

Question

In what follows, we let $n > 1$ be a positive integer. The classical sum of divisors of $n$ is given by $\sigma_1(n)$. Denote the abundancy index of $n$ by $I(n)=\sigma_1(n)/n$.

Denote the deficiency of $n$ by $D(n)=2n-\sigma_1(n)$, and denote the sum of aliquot divisors of $n$ by $s(n)=\sigma_1(n)-n$.

CLAIM $$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}$$

PROOF $$I(n) < I(n^2) \implies 2 - I(n^2) < 2 - I(n) \implies D(n^2) < nD(n) \implies \frac{D(n^2)}{n^2}<\frac{D(n)}{n}$$

$$I(n) < I(n^2) \implies I(n) - 1 < I(n^2) - 1 \implies ns(n) < s(n^2) \implies \frac{s(n)}{n} < \frac{s(n^2)}{n^2}$$

From the last two inequalities, we get $$\bigg(\frac{D(n^2)}{n^2}<\frac{D(n)}{n}\bigg) \land \bigg(\frac{n^2}{s(n^2)}<\frac{n}{s(n)}\bigg).$$

Multiplying LHS and RHS of the two inequalities, we finally obtain $$\frac{D(n^2)}{s(n^2)} < \frac{D(n)}{s(n)}.$$

Here are my questions:

Can the inequality in the CLAIM be improved? If so, how?

Answer 1

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: $${\frac {D \left( n \right) }{s \left( n \right) }}-{\frac { D \left( {n}^{2} \right) }{s \left( {n}^{2} \right) }}-\frac{1}{4} \delta \left( n-2\,\left\lfloor \frac{n}{2}\right\rfloor,1 \right) \lt \frac{3}{4} \tag{1}$$

Answer 2

I am posting this answer in the context of odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Based from a formula in this recent question, we have $$D(x)D(y)-D(xy)=2s(x)s(y)$$ when $\gcd(x,y)=1$.

In particular, since $p^k m^2$ is perfect (and $\gcd(p,m)=\gcd(p^k,m^2)=1$), we obtain $D(p^k m^2) = 0$, so that $$D(p^k)D(m^2)=2s(p^k)s(m^2).$$ This last equation is equivalent to $$\frac{D(m^2)}{s(m^2)}=\frac{2s(p^k)}{D(p^k)}.$$ But using the same formula, since $\gcd(p,m)=\gcd(p^k,m)=1$ we obtain $$D(p^k)D(m)-D(p^k m)=2s(p^k)s(m).$$ Dividing throughout the last equation by $D(p^k)s(m)$, we get $$\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)}=\frac{2s(p^k)}{D(p^k)}.$$ Equating the two expressions for $$\frac{2s(p^k)}{D(p^k)}$$ we derive $$\frac{D(m^2)}{s(m^2)}=\frac{D(m)}{s(m)}-\frac{D(p^k m)}{D(p^k)s(m)}.$$

Answer 3

This answer adds further details to this earlier answer.

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

Note that we have the numerical bounds $$1 < I(p^k) < \frac{5}{4} < \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} < I(m) < 2.$$

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