(In what follows, we let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. $I(x)=\sigma(x)/x$ is the abundancy index of $x$, while $D(x)=2x-\sigma(x)$ is the deficiency of $x$.)

From this MSE question, I have the equation $$D(xy) - D(x)D(y) = 2xy - \sigma(xy) - (2x - \sigma(x))(2y - \sigma(y))$$ $$=2xy - \sigma(xy) - 4xy + 2y\sigma(x) + 2x\sigma(y) - \sigma(x)\sigma(y)$$ $$=-2xy - 2\sigma(x)\sigma(y) + 2y\sigma(x) + 2x\sigma(y) + (\sigma(x)\sigma(y) - \sigma(xy))$$ $$=2(x - \sigma(x))(\sigma(y) - y) + (\sigma(x)\sigma(y) - \sigma(xy)).$$

Note that, in general we have $$2(x - \sigma(x))(\sigma(y) - y) \leq 0$$ and $$\sigma(x)\sigma(y) - \sigma(xy) \geq 0.$$

Thus, we obtain:

Proposition: If $x$ and $y$ are relatively prime (i.e., $\gcd(x, y) = 1$), then it follows that $D(xy) \leq D(x)D(y)$.

Note that we cannot apply the preceding result to our present problem:


If $D(x)=2x-\sigma(x)$, then if $n>1$ is deficient which is bigger, $D(n)$ or $D(n^2)$?

Nonetheless, note that we have the following partial result:


Since $n>1$, we obtain $$\sqrt{I(n^2)}<I(n)<I(n^2),$$ and $$I(n)=2-\frac{D(n)}{n}$$ $$I(n^2)=2-\frac{D(n^2)}{n^2}.$$ It follows that $D(n^2)<nD(n)$, and $$\sqrt{2-\frac{D(n^2)}{n^2}}<2-\frac{D(n)}{n}<2-\frac{D(n^2)}{n^2}.$$ Consequently, $$2-\frac{D(n)}{n}<2-\frac{D(n^2)}{n^2}<\bigg(2-\frac{D(n)}{n}\bigg)^2.$$ Since $n>1$ is deficient, then $$1<2-\frac{D(n)}{n}<\bigg(2-\frac{D(n)}{n}\bigg)^2$$ so that $$0<\bigg(2-\frac{D(n)}{n}\bigg)\bigg(1-\frac{D(n)}{n}\bigg)$$ whence it follows that $$0<1-\frac{D(n)}{n}$$ which implies that $$D(n)<n.$$ Consequently, $$D(n^2)<nD(n)<n^2.$$ Alas, this is where I get stuck.

  • $\begingroup$ it look like the past few question are related and never answered, why not try MW instead of MSE ? (+1) $\endgroup$ – Ahmad Aug 27 '17 at 11:54

The result depends on $n$.

For example $$D(3^k)=\frac{3^k+1}{2}<\frac{3^{2k}+1}{2}=D(3^{2k}),$$ $$D(2^k)=1=D(2^{2k}).$$ And for any prime $p\in(2^{k+1},2^{2k+1})$, we have $$D(2^kp)=p+1-2^{k+1}>0>1+(p-2^{2k+1})(p+1)=D(2^{2k}p^2).$$ All three families give infinitely many instances.

Also, I'd like to note a fact that $D(494)=148>11=D(494^2)$, which is an instance of $D(n)>D(n^2)>0$. However, I have no idea whether there are other instances.

EDIT: More $n$ with the property $D(n)>D(n^2)>0$: \begin{align} &2\cdot13\cdot19\cdot44371\\ &2\cdot13\cdot19\cdot44381\\ &2\cdot13\cdot19\cdot44383\\ &2\cdot13\cdot19\cdot44371\cdot2406128687\\ &2\cdot13\cdot19\cdot44371\cdot2406128689\\ &2\cdot13\cdot19\cdot44371\cdot2406128687\cdot2466259112360839463\\ \end{align} To get this list, I just make use the known instances of $D(n)>D(n^2)>0$ and multiply it with an appropriate prime $p$ such that $D(np)>D(n^2p^2)>0$. I believe we may obtain more instances by relax the constrains.

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