The present question is tangentially related to this earlier one.
My question here is:
Does $2r - \sigma(r)$ divide $\sigma(r)$ if $r$ is deficient-perfect?
Recall that a positive integer $x$ is deficient-perfect if $x$ satisfies $D(x) \mid x$, where $D(x) = 2x - \sigma(x)$ is the deficiency of $x$ and $\sigma(x)$ is the sum of divisors of $x$.
My conjecture is that the answer is NO.
MY ATTEMPT
Suppose that $r$ is deficient-perfect.
This means that $2r - \sigma(r) = D(r) \mid r$, so that $r$ can be written in the form $$r = {R_1}D(r),$$ for some (positive) integer $R_1$.
This implies that $$r(2{R_1} - 1) = {R_1}\sigma(r).$$
Assume to the contrary that $D(r) \mid \sigma(r)$. Then it follows that $$\sigma(r) = {R_2}D(r) = {R_2}(2r - \sigma(r))$$ for some (positive) integer $R_2$.
This implies that $$r(2{R_1} - 1) = {R_1}\sigma(r) = {R_1}{R_2}D(r) = {R_1}{R_2}(2r - \sigma(r))$$ from which we obtain $${R_1}{R_2}\sigma(r) = 2r{R_1}{R_2} - r(2{R_1} - 1).$$ Dividing through by ${R_1}{R_2}$, we get $$\sigma(r) = r\Bigg(2 - \bigg(\frac{2{R_1} - 1}{R_1 R_2}\bigg)\Bigg).$$
Alas, this is where I get stuck.
