# If $L > 1$ is an odd almost perfect number with $\omega(L)=6$, then $L$ must be divisible by $3$.

Edited July 15 2016

Let $\mathbb{N}$ denote the set of positive integers.

Let $\sigma = \sigma_{1}$ denote the (classical) sum-of-divisors function. Let $I(x) = \dfrac{\sigma(x)}{x}$ denote the abundancy index of $x \in \mathbb{N}$. Denote the number of distinct prime factors of $x \in \mathbb{N}$ as $\omega(x)$.

A number $M \in \mathbb{N}$ is said to be almost perfect if $\sigma(M) = 2M - 1$.

By a criterion of Dris, we have that $$\sigma(M) = 2M - 1 \iff \dfrac{2M}{M + 1} \leq I(M) < \dfrac{2M + 1}{M + 1}.$$ Equality holds if and only if $M = 1$.

We now present a proof for the following theorem.

(Note: All succeeding results are only valid under the additional condition $\omega(L) = 6$.)

Theorem. If $L > 1$ is an odd almost perfect number with $\omega(L)=6$, then $L$ must be divisible by $3$.

PROOF

Suppose that $L > 1$ is almost perfect. It suffices to consider the case where we have $\omega(L) = N$, where $N$ is fixed.

Let $$L = \prod_{i=1}^{N}{{p_i}^{\alpha_i}}$$ be the canonical factorization of $L$. (That is, the $p_i$'s are primes with $p_1 < p_2 < \ldots < p_N$.) Assume to the contrary that $3 \nmid L$. Then $p_1 = \min(p_i) \geq 5$. (Note that the $\alpha_i$'s are all even, but it will not matter in the argument.) By the Dris criterion for almost perfect numbers, $L$ satisfies $$\dfrac{2L}{L+1} < I(L) < \dfrac{2L + 1}{L + 1}.$$ It will suffice to consider the inequality on the left, together with the following inequality for $I(L)$: $$I(L) = I\left(\prod_{i=1}^{N}{{p_i}^{\alpha_i}}\right) = \prod_{i=1}^{N}{I\left({p_i}^{\alpha_i}\right)} < \prod_{i=1}^{N}{\left(\dfrac{p_i}{p_i - 1}\right)}.$$ Consequently, we have the following: $$2\left(\prod_{i=1}^{N}{{p_i}^{\alpha_i}}\right) = 2L < \left(L+1\right)I(L) < \left(L+1\right)\cdot{\prod_{i=1}^{N}{\left(\dfrac{p_i}{p_i - 1}\right)}}$$ $$= \left(\left(\prod_{i=1}^{N}{{p_i}^{\alpha_i}}\right) + 1\right)\cdot{\prod_{i=1}^{N}{\left(\dfrac{p_i}{p_i - 1}\right)}}.$$ Now, we bound the quantity $$\prod_{i=1}^{N}{\left(\dfrac{p_i}{p_i - 1}\right)}$$ by using the result that "Odd almost perfect numbers have at least six distinct prime factors" (which is due to Kishore) together with our assumption that $\min(p_i) \geq 5$. We get $$\prod_{i=1}^{N}{\left(\dfrac{p_i}{p_i - 1}\right)} \leq \dfrac{5}{4}\cdot\dfrac{7}{6}\cdot\dfrac{11}{10}\cdot\dfrac{13}{12}\cdot\dfrac{17}{16}\cdot\dfrac{19}{18} = \dfrac{323323}{165888} = 1.949043933256\overline{172839506}.$$

This contradicts the Dris criterion for almost perfect numbers. Therefore, $p_1 = \min(p_i) = 3$.

It follows that odd almost perfect numbers greater than one must be divisible by 3. It follows that odd almost perfect numbers with exactly six distinct prime factors must be divisible by $3$.

QED

Questions

(1) By using a similar argument, it can be shown that, if $\omega(M)=6$ then:

$p_2 = 5 \implies 7 \leq p_3 \leq 23$

$p_2 = 7 \implies 11 \leq p_3 \leq 23$

$p_2 = 11 \implies 13 \leq p_3 \leq 17$

etc. etc. An obvious question at this point would be: To what extent can these bounds be improved, given that the lower bound $$\dfrac{2M}{M + 1} \leq I(M)$$ (for $M$ an almost perfect number) appears to be best-possible?

(2) Traditionally, sigma chaining has been used to produce additional (consequent) factors for a hypothetical odd perfect number $O = q^k n^2$, given that $$\sigma(q^k)\sigma(n^2) = \sigma(O) = 2O = 2{q^k}{n^2}.$$ Will it be possible to adapt the method of sigma/factor chains to the case of odd almost perfect numbers $M = X^2$ (where $\omega(X) \geq 6$)? Note that $$\sigma(X^2) = \sigma(M) = 2M - 1 = 2X^2 - 1.$$

Notice that these results improve on Kishore's bounds in On Odd Perfect, Quasiperfect, and Odd Almost Perfect Numbers from the year 1981.

• "the Dris criterion for almost perfect numbers" you are so arrogant to name it after yourself, but cool work dude hope you get published :) – terrace Aug 22 '17 at 14:40