2
$\begingroup$

(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:

PROBLEM STATEMENT

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:

MY ATTEMPT

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.