Natural numbers $\ n\ $ as sums of the subsets of divisors of $\ n$

Question

QUESTION 1: What are the lower and upper bounds of numbers $\ |S|(n)\ $ of the number of representations of natural number $\ n\ $ as the sum of different divisors of $\ n$?

Full notation:

$$ D(n)\ :=\ \{d\in\mathbb N:\ n > d|n\}\,; $$ $$ S(n)\ :=\ \{A\subseteq D(n): \sum A = n\}; $$ $$ |S|(n)\ := |S(n)|. $$

Thus, $\ |S|(p^n) = 0\ $ for all natural $\ n\ $ and $\ p,\ $ where $\ p\ $ is prime. Hence $\ |S|(n)=0\ $ for $\ n<6,\ $ while $\ |S|(6)=1\ $ (see below).

Is function $\ |S|\ $ simply a mess? I feel that it has regular features:

QUESTION 2: Formulate and prove algebraic properties of function $\ |S|.$

Perhaps $\ |S|+1\ $ is -- so to speak -- more algebraic than $\ |S|$.

The said algebraic properties should show up especially for what I call baroque numbers (or multi-perfect is one of the classical names).

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Let $\ \sigma(n)\ :=\ \sum\{d\in\mathbb n:\ d|n\}\ $ be the sum of the divisors of natural $\ n.\ $ Let $\ \beta\in\mathbb N.\ $ Natural $\ n\ $ is said to be $\ \beta$-baroque $\ \Leftarrow:\Rightarrow$

$$ \frac{\sigma(n)}n\ =\ \beta $$ 2-Baroque numbers are simply the perfect numbers. Of course, $\ |S|(n)=1\ $ for every perfect $\ n$.

QUESTION 3 -- a conjecture: $\ |S|(n)\ge\beta-1\ $ for every $\beta$-baroque number $\ n$.

(I expect much sharper bounds).

EXAMPLE 1: $\ 120\ $ is 3-baroque, and:

$$ 120\ =\ 60+40+20\ =\ 60+30+20+10 $$ $$ =\ 60+30+20+8+2\ =\ 60+30+20+6+4 $$ $$ =\ 60+40+8+6+4+2 $$

EXAMPLE 2: $\ 672\ $ is 3-baroque, and:

$$ 672\ =\ 336+224+112\ =\ 336+224+84+ 28 $$ $$ \ 336+168+112+56\ =\ 336+168+112+32+24 $$

Answer 1

Since the term multiply perfect is as you acknowledge already in the literature, I'm not sure why you are introducing the baroque term. One noteworthy thing here is what are called pseudoperfect numbers, which are numbers which are your notation, numbers $n$ such that $|S|(n) \geq 1$, that is numbers which can be represented as a sum of some subset of their divisors.

In general for a lot of purposes with these functions, it is more useful to look at the set of all divisors which are at most $n$, and look at subsets of this. In standard notation then $\sigma(n)$ is the sum of the positive divisors of $n$ including $n$. A number is said to be pseudoperfect to be a number $n$ such that $2n$ is expressible as a sum of some subset of its set of positive divisors.

It is well known that if a number $n$ is pseudoperfect then so is $kn$ for any $k$. To see this note that if $d_1, d_2, \cdots d_k$ satisfy that $d_1 +d_2... +d_m =n$ then $kd_1 +kd_2... +kd_m =kn$, and if $d_1, d_2, \cdots d_m$ are divisors of $n$ then $kd_1, d_2 \cdots kd_m$ are all divisors of $kn$.

This motivates another standard notion, that of a primitive pseudoperfect number, which is a number is pseudoperfect and which has no primitive divisors. A number may be pseudoperfect in more than one way when the number is a multiple of different primitive pseudoperfect numbers. For example, $6$ is pseudoperfect, since $1+2+3=6$, and $20$ is pseudoperfect with $1+4+5+10=20$, We have $$1(20)+2(20)+3(20)=6(20)=6(1)+6(4)+6(5)+6(10).$$ Thus, $|S|(120) \geq 2$.

We also get from following this logic, that in general for any $a$ and $b$, that $|S|(ab) \geq \max \{|S|(a),|S|(b)\}$. I don't know if that is as algebraic a property as what you are looking for.

One can at least with this sort of idea then show the following, which is not as far as I'm aware in the literature, but certainly has to have been already known: Proposition: For any $k$ there exists an $n$ such that $|S1(n)$. Proof sketch: For any prime $p$, $(2^p-1)(2^{p-1})$ is pseudoperfect. (Just use the proof that this number would be perfect when $2^p-1$ is prime.) Now, note that for any two distinct primes $p$ and $q$, $2^p-1$ and $2^q-1$ are relatively prime. So for any set of distinct primes $\ p_1, p_2,\ldots,p_k,\ $ the number $$N = \prod_{i=1}^k $(2^{p_i}-1)(2^{p_i-1})$$ will satisfy $$|S|(N) \geq k.$$

Question 3 looks interesting, and I have not seen it discussed before. I suggest splitting it off and making a separate question for that. That said, my guess is that it is false, and that a computer search should turn up counterexamples.

Answer 2

Theorem:   Let $\ a\ $ and $\ b\ $ be relatively prime natural numbers. Then $$ |S|(a\cdot b)\ \ge\ |S|(a)\cdot |S|(b). $$

(Straightforward, just a little bit of checking to do).