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Let $N \in \mathbb{N}$. (That is, let $N$ be a positive integer.)

This is in reference to two of my earlier questions here at MSE:

  1. Does the following inequality hold true, in general?

  2. Does this inequality have any solutions in $\mathbb{N}$?

In particular, my question in this post is as follows:

Does the following inequality hold if and only if $N$ is an odd deficient number?

$$I(N) = \frac{\sigma(N)}{N} < \frac{2N}{N + 1}$$

(Note that $I(N)$ is the abundancy index of $N$ and $\sigma(N)$ is the sum of the divisors of $N$.)

It is a relatively easy exercise to prove that if $I(N) < \frac{2N}{N + 1}$, then $N$ is deficient. Showing the other direction appears to be difficult.

(I think we can generalize Tharsis's observation, that powers of $2$ are deficient and do not satisfy the inequality above, to something like:

"If $M = {2^k}{m}$ is an even deficient number with $\gcd(2,m)=1$ and $k \geq 1$, then $$I(M) \geq \frac{2M}{M+1}.")$$

Attempt at proof:

Let $M = {2^k}{m}$ be an even deficient number with $\gcd(2,m)=1$ and $k \geq 1$.

Then since $I(x)$ is a weakly multiplicative function of $x$, we have

$$I(M) = I(2^k)I(m) = \frac{2^{k + 1} - 1}{{2^k}(2 - 1)}I(m) = \left(2 - \left(\frac{1}{2}\right)^k\right)I(m) \geq \left(2 - \frac{1}{2}\right)I(m) = \frac{3}{2}I(m).$$

We want to show that $I(M) \geq \frac{2M}{M+1}$.

Now suppose to the contrary that

$$\frac{2M}{M + 1} > I(M) \geq \frac{3}{2}I(m).$$

But $M = {2^k}{m}$ is a positive integer with $\gcd(2,m)=1$ and $k \geq 1$. Therefore, we have

$$2 > \frac{2M}{M + 1} = \frac{2({2^k}{m})}{{2^k}{m} + 1} = \frac{2}{1 + \frac{1}{{2^k}{m}}} \geq \frac{12}{7} = 1.\overline{714285},$$

since we can assume without loss of generality that the inequality $m \geq 3$ is true.

However, what we actually need is a decent upper bound for $\frac{2M}{M + 1}$ (i.e., an upper bound that is sharper than $2$).

Consequently, assume that we do have a sharper upper bound

$$2 - \epsilon > \frac{2M}{M + 1} \geq \frac{12}{7},$$

for some positive real number $\epsilon < \frac{2}{7}$.

(Notice that this assumption means that there is an upper bound

$$M < \frac{2 - \epsilon}{\epsilon} = \frac{2}{\epsilon} - 1$$

for $M$ in terms of $\epsilon$. However, while we can easily obtain the lower bound

$$6 \leq {2^k}{m} = M < \frac{2}{\epsilon} - 1,$$

it seems very difficult to try to obtain an upper bound. Please refer to this link for the WolframAlpha verification.)

If we content ourselves with the upper bound

$$2 > \frac{2M}{M + 1} > I(M) \geq \frac{3}{2}I(m),$$

we obtain

$$I(m) < \frac{4}{3} = 1.\bar{3},$$

whence we do not get any further (i.e., there are no contradictions, so far).

Alas! This is as far as I could go with my elementary methods. Perhaps somebody with fresh insights (and better ideas!) could suggest something new, or point out papers containing a similar approach (or the criteria in my answer below) in the existing literature.


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Oh my, I feel so naive! =( I want to start a bounty on this question, but I currently do not know how, as I have not tried doing that before. Anybody willing to walk me through the process? Thanks! – user11235813 Aug 7 '13 at 22:08
Geez! I just read the FAQ, and it says I need to wait for at least 48 hours, before I could start a bounty on this question. =) – user11235813 Aug 7 '13 at 22:17
for what it's worth, if $n$ is a power of 2, then $\sigma(n) = 2n-1 > 2 n^2 / (n+1). $ Indeed, the conditions $2 n^2 / (n+1) \leq w < 2n$ for $w$ an integer force $w=2n-1.$ – Will Jagy Aug 7 '13 at 22:34
The correct way to produce italics is *italics*. See here for a guide to formatting with Markdown. – Zev Chonoles Aug 7 '13 at 22:40
That's interesting @WillJagy! Can you write out that comment into an actual answer and include more details? Thank you! – user11235813 Aug 7 '13 at 22:44
up vote 2 down vote accepted

You are asking about what can be said when $$ \frac{2n^2}{n+1} \leq \sigma(n) < 2 n. $$ However, $$ \frac{2n^2}{n+1} = 2n - 2 + \frac{2}{n+1}. $$ The only integer at least this large but smaller than $2n$ is $2n-1.$ So, you are asking about all solutions of $\sigma(n) = 2n-1.$

Meanwhile, if $n$ is a power of 2, then $\sigma(n) = 2n-1.$

I think those are all. So does my computer. Should be provable.

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Wow! I didn't realize that numbers $n$ with $\sigma(n) = 2n - 1$ are called least-deficient or almost perfect numbers! And that this is an open problem... =( A quick Google search yields the following post by PPollack in the MathForum@Drexel. I wonder if there have been more recent (and maybe, stronger) results? I'll take it from here and keep digging. Thanks again @WillJagy! =) – user11235813 Aug 7 '13 at 23:15
See also Wikipedia: almost perfect number. – Jeppe Stig Nielsen Jan 31 '15 at 18:48

In comparison and contrast to WillJagy's answer, here is the situation when

$$I(N) < \frac{2N}{N + 1}.$$

Following WillJagy's reasoning, we have

$$\sigma(N) < \frac{2{N^2}}{N + 1} = 2N - 2 + \frac{2}{N + 1},$$

from which it follows that

$$\sigma(N) < 2N - 2 + \frac{2}{N + 1} < 2N - 2 + 1 = 2N - 1,$$

since $N = 1$ does not satisfy

$$I(N) < \frac{2N}{N + 1}.$$

In other words, we have the biconditional "$N$ is an Odd Almost Perfect Number [OAPN] if and only if $I(N) \geq \frac{2N}{N + 1}$." [Edit: The numbers $2^k, k \geq 0$ also satisfy this inequality.]

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On further thought, the numbers $O={2^k}, k \geq 0$ also satisfy $I(O) \geq \frac{2O}{O+1}$. – user11235813 Aug 10 '13 at 23:08

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