What are the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$?

Here, $\sigma_{1}$ is the classical sum-of-divisors function. For example, $\sigma_{1}(3^2) = 1 + 3 + {3^2} = 13$.

(The function $D(x) = 2x - \sigma_{1}(x)$ is called the deficiency of $x$ [see OEIS Sequence A033879].)

Now, for the motivation behind the question:

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e., $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$).

Since $N$ is perfect, we have the equation

$$\sigma_{1}(N) = 2N \Longleftrightarrow \sigma_{1}({q^k}{n^2}) = \sigma_{1}(q^k)\sigma_{1}(n^2) = 2{q^k}{n^2}.$$

We can express $\sigma_{1}(q^k)$ as

$$\sigma_{1}(q^k) = 1 + q + \ldots + q^{k-1} + q^k = \left(1 + q + \ldots + q^{k-1}\right) + q^k = \sigma_{1}(q^{k-1}) + q^k$$

whereupon we have

$$\sigma_{1}(q^k)\sigma_{1}(n^2) = \left(\sigma_{1}(q^{k-1}) + {q^k} \right)\sigma_{1}(n^2) = 2{q^k}{n^2}.$$

Dividing through by $q^k$ in

$$\left(\sigma_{1}(q^{k-1}) + {q^k} \right)\sigma_{1}(n^2) = 2{q^k}{n^2},$$

we get

$$2n^2 - \sigma_{1}(n^2) = I(q^{k-1})\cdot{\frac{\sigma(n^2)}{q}},$$


$$I(x) = \frac{\sigma_{1}(x)}{x}$$

is the abundancy index of $x$.

From the last equation, since $q \geq 5$ implies $1 \leq I(q^{k-1}) < \frac{5}{4} = 1.25$, we have the biconditional

$$k = 1 \Longleftrightarrow 2n^2 - \sigma_{1}(n^2) = \frac{\sigma(n^2)}{q}.$$

In other words, we have the biconditional

$$k = 1 \Longleftrightarrow 2n^2 - \sigma_{1}(n^2) = \frac{\sigma(n^2)}{q^k}.$$

[In the paper The Abundancy Index of Divisors of Odd Perfect Numbers, the assertion $k = 1$ for odd perfect numbers $N = {q^k}{n^2}$ was called as Sorli's conjecture -- although, in a more recent (proceedings) paper by B. Beasley of Presbyterian College, titled Euler and the Ongoing Search for Odd Perfect Numbers, he points out that it was in fact Descartes who initially made this conjecture (in a letter to Marin Mersenne in 1638), "with Frenicle's subsequent observation occurring in 1657".]

Now, following Broughan, et. al., it is known that the index $\frac{\sigma_{1}(n^2)}{q^k}$ of the Euler factor $q^k$ satisfies

$$\frac{\sigma_{1}(n^2)}{q^k} \geq 315,$$



cannot take any of the $11$ forms

$$\{p, p^2, p^3, p^4, p^5, p^6, {p_1}{p_2}, {p_1}^2{p_2}, {p_1}^3{p_2}, {p_1}^2{p_2}^2, {p_1}{p_2}{p_3}\}$$

where $p$ is any odd prime and $p_1, p_2, p_3$ are any distinct odd primes.

Now back to the original question: We want to know the divisors of $2n^2 - \sigma_{1}(n^2)$ for composite $n$.

Note that, if $n$ is prime, say $r$, then we have

$$2n^2 - \sigma_{1}(n^2) = 2r^2 - \sigma_{1}(r^2) = 2r^2 - (r^2 + r + 1) = r^2 - r - 1 = r(r - 1) - 1,$$

which is, in general, odd (so that $2r^2 - \sigma_{1}(r^2) \equiv 1 \pmod 2$ for $r$ prime.) [In fact, since $\sigma_{1}(n^2) \equiv 1 \pmod 2$ for all positive integers $n$, then we have $2n^2 - \sigma_{1}(n^2) \equiv 1 \pmod 2$ for all positive integers $n$.]

Of course, the situation is more complicated when $\omega(n) \geq 2$ (where $\omega(x)$ is the number of distinct prime factors of $x$).

  • 1
    $\begingroup$ I suppose as $n$ varies the divisors vary also, what is exactly your question? I doubt closed form for the divisors of the OP. $\endgroup$ – joro Oct 1 '13 at 16:55
  • $\begingroup$ Thank you for your clarification, @joro. I am likewise highly skeptical that a closed form exists for the divisors of $2n^2 - \sigma_{1}(n^2)$, but obtaining results similar to Broughan, et. al. (and Chen and Chen) would be more to the spirit of my question. $\endgroup$ – Arnie Bebita-Dris Oct 1 '13 at 17:00
  • $\begingroup$ Chen and Chen have a new paper titled On the index of an odd perfect number, where they extend Broughan, et. al.'s results. $\endgroup$ – Arnie Bebita-Dris Aug 12 '14 at 13:48
  • $\begingroup$ These MSE links are related to this MO question: On Descartes numbers and On odd perfects and spoofs. $\endgroup$ – Arnie Bebita-Dris Apr 5 '15 at 19:23

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