27

For a positive integer $n$, let us define a set

$$A_n = \{ k\in\mathbb{N} \mid \sigma(k) = n \}$$

where $\sigma$ is the divisor-sum function (a well-known multiplicative number-theoretic function). Clearly $A_n \subseteq \{ 1,2,3,\ldots,n\}$ (since $\sigma(k)\ge k$ for all $k$).

For example

$$\begin{align} A_{119} & = \varnothing \\ A_{120} & = \{ 54, 56, 87, 95 \} \\ A_{121} & = \{ 81 \}. \end{align}$$

Now denote by $\Sigma_n$ the sum of the members of $A_n$, so $\Sigma_n = \sum_{k\in A_n}k$, so (continuing the example)

$$\begin{align} \Sigma_{119} & = 0 \\ \Sigma_{120} & = 292 \\ \Sigma_{121} & = 81. \end{align}$$

Note that $\Sigma_{119}<119$ and $\Sigma_{121}<121$, and on the other hand $\Sigma_{120}>120$.

This splits the natural numbers $n$ into three classes, according to whether $\Sigma_n<n$, $\Sigma_n=n$, or $\Sigma_n>n$. I find a lot of numbers in the first and the last of these classes. However, the only number with $\Sigma_n = n$ that I have found is the trivial case $n=1$.

Are there any numbers $n>1$ with $\Sigma_n = n$?

PS! I am planning on submitting new sequences to OEIS if people find this partition of $\mathbb{N}$ interesting.


Here are some statistics for all $n$ in $\left[ 1, 60000 \right]$:

$$\begin{array}{|r|r|r|r|} n \pmod{6} & \Sigma_n<n & \Sigma_n=n & \Sigma_n>n \\ \hline +1 \pmod{6} & 9993 & 1 & 6 \\ +2 \pmod{6} & 9020 & 0 & 980 \\ 3 \pmod{6} & 9992 & 0 & 8 \\ -2 \pmod{6} & 9415 & 0 & 585 \\ -1 \pmod{6} & 10000 & 0 & 0 \\ 0 \pmod{6} & 5958 & 0 & 4042 \\ \hline \mathrm{total} & 54378 & 1 & 5621 \\ \end{array}$$

Update: I searched a bit further, $\left[ 1,\quad 300\cdot 10^6 \right]$:

$$\begin{array}{|r|r|r|r|} n \pmod{6} & \Sigma_n<n & \Sigma_n=n & \Sigma_n>n \\ \hline +1 \pmod{6} & 49999688 & 1 & 311 \\ +2 \pmod{6} & 47797853 & 0 & 2202147 \\ 3 \pmod{6} & 49999279 & 0 & 721 \\ -2 \pmod{6} & 47343370 & 0 & 2656630 \\ -1 \pmod{6} & 49999985 & 0 & 15 \\ 0 \pmod{6} & 36529965 & 0 & 13470035 \\ \hline \mathrm{total} & 281670140 & 1 & 18329859 \\ \end{array}$$

The first $n$ with $n \equiv -1 \pmod{6}$ so that $\Sigma_n>n$ is $86831$. We have $A_{86831} = \{ 38416, 60025 \}$.

A value for which $\Sigma_n=n$ corresponds to an amicable tuple which comprises all numbers with that $\sigma$ value, i.e. $A_n$ is amicable. We could call that a total amicable tuple. This question then becomes if any total amicable tuples other than $\{ 1 \}$ exist.

I have now created A258913 in OEIS which gives what is called $\Sigma_n$ above. According to comment by Giovanni Resta there, any new $n$ with $\Sigma_n=n$ will exceed $2.5\cdot 10^{10}$.

Jeppe Stig Nielsen
  • 4,870
  • 18
  • 27
  • What made you think of this? – Asier Calbet Feb 01 '15 at 22:21
  • 1
    @Assaultous2 I was inspired by OEIS sequence [A085790](https://oeis.org/A085790) which I realized could be written in rows (corresponding to the sets $A_n$ above) just like [A152454](https://oeis.org/A152454). And I was inspired by [amicable tuples](http://en.wikipedia.org/wiki/Amicable_numbers#Amicable_tuples). If an example exists where $\Sigma_n=n$, the corresponding $A_n$ would constitute an amicable pair/tuple with the extra property that no numbers outside the tuple shared the same $\sigma$ value. – Jeppe Stig Nielsen Feb 01 '15 at 22:34
  • 2
    interesting... do you find more above or below n? – Asier Calbet Feb 01 '15 at 22:36
  • 1
    The sum $\sum_{n\leq X}\Sigma_n$ is asymptotically $C X^2$ for some $C>0$ (an euler product of some sorts). So you might expect the event $\Sigma_n = n$ to happen with probability roughly $1/n$, or even less given that the values $\Sigma_n$ are a bit spread out. How far did your numerical calculations go ? – Sary Feb 08 '15 at 16:15
  • @Sary I do not remember exactly what search limit I reached, but I had the computer search for many hours. The algorithm I used was not optimized at all, though. I also tried to check some known amicable triples and quadruples. I could not find an example. I am sure this question has been seen by people a lot smarter than me, and given the bounty I offered, I am sure someone would have posted an example if it was easy to find. Also nobody has offered a reason why this could be impossible. I guess this question is not going to have an answer soon. – Jeppe Stig Nielsen Feb 08 '15 at 21:24
  • 1
    @Assaultous2 I have edited the question to include some statistics. I find most with $\Sigma_n – Jeppe Stig Nielsen Feb 14 '15 at 22:47
  • 1
    @Sary You too might be interested in the statistics I now provide in the edited question. – Jeppe Stig Nielsen Feb 14 '15 at 22:59
  • @Assaultous2 That is $n=1$, as I said the only "equality" case I have been able to find is $\Sigma_1=1$. – Jeppe Stig Nielsen Feb 15 '15 at 14:31
  • oh yea, thats right, sorry – Asier Calbet Feb 15 '15 at 14:36
  • Instead of just writing "a well-known multiplicative number-theoretic function" is would be more useful to either shortly say what it is, or if that's not possible, link to an appropriate description (e.g. on Wikipedia or MathWorld). – celtschk Apr 09 '15 at 15:51
  • @celtschk This is the sum of all divisors of $k$, called $\sigma_1$ in [this Wikipedia article](http://en.wikipedia.org/wiki/Divisor_function). As an example, $\sigma(54)=1+2+3+6+9+18+27+54=120$. – Jeppe Stig Nielsen Apr 09 '15 at 15:58
  • It's amazing to me just how often $\Sigma_n = n-1$. For the first 100,001 numbers checked only 21398 have non-zero $\Sigma_n$ and of those 5838 have the values $\Sigma_n = n-1$. No other relationship like $\Sigma_n = n-k$ appears anywhere nearly as often. – Alexander Vlasev Apr 12 '15 at 10:51
  • 1
    @AleksVlasev In how many of those cases is $n-1$ a prime number? Because in that case the value could arise from just one numer, $A_n=\{ n-1 \}$. Of course, for a prime $p$, we have $\sigma(p)=p+1$. – Jeppe Stig Nielsen Apr 12 '15 at 21:12
  • Oh haha that's awesome. Every single one of them. – Alexander Vlasev Apr 13 '15 at 00:39
  • I assume the reason you chose $n $ (mod $ 6)$ numbers is because all multiples of $6$ are abundant, and so therefore if $\sigma(k) = n$ then $k$ must be less than $\frac{n}{2}$. This is desirable, of course. – boldbrandywine Apr 15 '15 at 23:43
  • @boldbrandywine I am not sure that was my exact reason. I guess one could have chosen modulo $10$ or something else also. I just wanted to illustrate that $\Sigma_n$ is more often large when $n$ has many (small) divisors (it seems). I chose $6$ because it is not too huge for a table, and because it is the product of the first two prime numbers. First I considered quoting odd and even $n$ only, but then I decided for a bit more detail. – Jeppe Stig Nielsen Apr 16 '15 at 06:39

2 Answers2

2

*Not an answer, but too long for a comment.

As pointed out in the comments, an obvious family of near misses (precisely, a family where $|\Sigma_n - n|\in O(1)$) comes from the primes: assuming $A_n = \{ p \}$, then $n=\sigma(p)=p+1$, and $\Sigma_n = p = n-1$. Not every prime $p$ works, because $A_{p+1}$ may contain other elements besides $p$, but infinitely many seem to.

A less-obvious family of near misses comes from pairs of primes $(p,q)$ for which $q=4p+3$. Assuming $A_n = \{ 12p, 4q \}$, then $$ n=\sigma(12p)=\sigma(2^2 \cdot 3 \cdot p)=(1+2+4)(1+3)(1+p)=28p+28,\\ n=\sigma(4q)=\sigma(2^2 q)=(1+2+4)(1+q)=7(1+q)=28p+28, $$ and $$ \Sigma_n=12p + 4q=12p+4(4p+3)=28p+12=n-16. $$ Again, this doesn't always work, because $A_{28p+28}$ may contain elements other than $12p$ and $4q$, but it appears to happen infinitely often.

Just inspecting the data doesn't turn up any other obvious families like this ($-1$ and $-16$ are by far the most recurrent values of $\Sigma_n - n$), but I wonder if they exist and are simply sparse. Can this be generalized? Is there another near-miss family for which $|A_n|=2$, or a near-miss family with $|A_n|=3$, for instance?

mjqxxxx
  • 37,360
  • 2
  • 50
  • 99
1

I have an idea that may be a partial answer. My hunch is that it is only true for $n=1$.

Let's rephrase your question: What you're really asking is "Can $k_1 + k_2 + \cdots + k_m = \sigma(k_1)$?" where $m = |A_{n}|$ and such that $\sigma(k_i) = n$ for all $i \leq m$. You may already know the following terminology, but for those who don't:

Consider the $\bf{abundance}$ of $k$, defined and denoted as $I(k) = \displaystyle \frac{\sigma(k)}{k}$.

Observe that $\displaystyle \frac{\sigma(k)}{k} > 1$ when $k > 1$.

If $I(k) > 2$, call $k$ $\bf{abundant}$.

If $I(k) = 2$, call $k$ $\bf{perfect}$.

If $I(k) < 2$, call $k$ $\bf{deficient}$.

Here's a start with cases:

1) WLOG, assume $k_1$ is deficient. Thus $\displaystyle \frac{k_2}{k_1} + \cdots + \frac{k_m}{k_1} < 1$. Thus, all $k_i$ are more abundant than $k_1$. WLOG, assume $k_2 < k_1$. Further, assume $k_2$ is still deficient. Then $\displaystyle \frac{k_1}{k_2} + \frac{k_3}{k_2} + \cdots + \frac{k_m}{k_2} < 1$, but this is a contradiction since $\displaystyle \frac{k_1}{k_2} > 1$. Thus, if such $A_n$ is to exist, it cannot have more than one deficient element. In fact, it's similar to show that such $A_n$ cannot have together a deficient element and a perfect element. So the other elements must be abundant.

2) WLOG, assume $k_1$ is perfect. Then $\displaystyle \frac{k_2}{k_1} + \cdots + \frac{k_m}{k_1} = 1$, and, for similar reasoning as before, each $k_i$ must be abundant. So, if such $A_n$ is to exist, it can have at most one perfect element, and the rest must be abundant.

3) WLOG, assume all $k_i$ are abundant. Thus $\sigma(k_i) > 2k_i$ for all $i$. This implies each $k_i < \displaystyle\frac{n}{2}$. Thus we know from this fact that $A_n$ needs at least $3$ elements.

I'm currently still working on this case to (hopefully) show a contradiction. I'm working with many different methods, so it might be a few days.

Overall, we've shown that if such an $A_{n}$ exists, it needs to have at least three primarily abundant elements; "primarily" meaning "all but one."

boldbrandywine
  • 614
  • 4
  • 13
  • What you denote by $I(k)$ I call the **abundancy (index)** of $k$. If $(k_1,k_2)$ is an amicable pair, and we agree that it is sorted ascendingly by $k$, so $k_1I(k_2)$. Since $k_1+k_2=n$, we have $\frac{k_1}{n}+\frac{k_2}{n}=1$ or $\frac12 ( \frac{k_1}{n}+\frac{k_2}{n} ) =\frac12$, so on (harmonic) *average* the abundancy $\frac{n}{k}$ is $2$. So $k_1$ is abundant and $k_2$ is deficient. Example: $(k_1,k_2)=(220,284)$. However this pair is not "total" by my definition from the question. – Jeppe Stig Nielsen Apr 16 '15 at 21:38
  • Similarly, for an amicable triple like $(k_1,k_2,k_3)=(1980, 2016, 2556)$ with $k_1I(k_2)>I(k_3)$. So here on (harmonic) average the abundancy of the three $k_i$ is $3$. So we could have that all three are abundant. (Or maybe the first two, $k_1$ and $k_2$, are very abundant, and the last one, $k_3$, is perfect/deficient, in a way so that the average abundancy I describe is still $3$.) Again this example triple is not "total". For an example of an amicable (not total) quadruple ("average" abundancy is $4$), check $(k_1,k_2,k_3,k_4)=(3270960,3361680,3461040,3834000)$. – Jeppe Stig Nielsen Apr 16 '15 at 21:52
  • I understand. When I say $A_n$, what I mean is the "total" set as you have defined. Upon analyzing the situation further, I have found that if $k \in A_n$ and $k$ is part of a total tuple, then $2k \notin A_n$; moreover, I've proven that if $k \in A_n$ and $k$ is to be part of total tuple, then $\displaystyle\frac{u}{v} k \notin A_n$, where $\displaystyle \frac{u}{v} \in \mathbb{Q}$ and completely reduced. Maybe this should help? – boldbrandywine Apr 17 '15 at 03:03
  • Additionally, $v$ should divide $k$, and $1 < \displaystyle\frac{u}{v} <2$. I've also shown that $2\sigma(k) < \sigma(2k)$ for all $k$, and $\sigma(2k) = 3\sigma(k)$ when $k$ is odd. – boldbrandywine Apr 17 '15 at 03:05
  • When $k$ is even, we have $\sigma(2k) = \displaystyle \frac{2^{t_1 +2}-1}{2^{t_1 +1}-1} \sigma(k)$, where $t_1$ is the first power showing up on the prime $2$ in the prime decomposition of $k$. – boldbrandywine Apr 17 '15 at 03:08
  • Sorry for the string of comments... But, if my proof about the less-than-2 fractional part is true, then I suspect that implies all elements in a total tuple need to be relatively prime. I need to think on it more, but that would seem to lead to a contradiction. – boldbrandywine Apr 17 '15 at 03:15
  • Not sure I understand all you say. But if you want to prove a **total** amicable tuple cannot exist for $m>1$, be sure to use the "totality" in a crucial way. Because plenty of amicable tuples are known. If for example you could prove that for every non-trivial amicable tuple there exist a "new" integer which is not in the tuple, and whose $\sigma$ value is also $n$, it would suffice. Is there a way to construct such a witness for non-totality? – Jeppe Stig Nielsen Apr 17 '15 at 07:20