Fixed Points and Necessary Condition for the Iterated Sum of Three Largest Proper Divisors

1. Introduction

Let $\mathbb{N}$ denote the set of positive integers. A proper divisor of $N \in \mathbb{N}$ is a positive divisor of $N$ different from $N$ itself. Consider an infinite sequence $a_1,a_2,\ldots$ of positive integers such that each $a_n$ possesses at least three proper divisors and [ a_{n+1}= \text{sum of the three largest proper divisors of } a_n \qquad (n\ge 1). ] The problem asks for all possible values of the first term $a_1$.

Denote by $f(N)$ the sum of the three largest proper divisors of $N$ (if $N$ has at least three proper divisors; otherwise $f(N)$ is undefined). The sequence is then defined by $a_{n+1}=f(a_n)$. We are interested in those $a_1$ for which the iteration never leaves the set [ S={N\in\mathbb{N} : N \text{ has at least three proper divisors}}. ]

In this note we obtain two main results.

• Theorem 1. A number $N\in S$ satisfies $f(N)=N$ (i.e., is a fixed point of the iteration) if and only if $N$ is divisible by $6$ and is not divisible by $4$ or $5$.

• Theorem 2. If $a_1\in S$ and the whole infinite sequence stays inside $S$, then $a_1$ must be divisible by $6$.

The first theorem gives a complete description of the fixed points; there are infinitely many of them. The second theorem provides a necessary condition for a starting value to be admissible. Empirical evidence suggests that the admissible $a_1$ are exactly those multiples of $6$ that after finitely many steps reach a fixed point, but a full characterization remains open.

2. Notation and preliminaries

For $N\in\mathbb{N}$ let $D(N)$ be the set of its positive divisors and $D'(N)=D(N)\setminus{N}$ the set of proper divisors. We always assume $|D'(N)|\ge 3$. Write the proper divisors in increasing order [ 1=d_1<d_2<\dots <d_k \qquad (k\ge 3). ] The three largest proper divisors are $d_{k-2},d_{k-1},d_k$. Hence [ f(N)=d_{k-2}+d_{k-1}+d_k . ]

Observe that if $p$ is the smallest prime divisor of $N$, then $d_k=N/p$. More generally, if we list the divisors greater than $1$ in increasing order $e_1<e_2<\dots <e_{k-1}$ (so $e_1$ is the smallest prime divisor), then the $i$-th largest proper divisor equals $N/e_i$. Consequently [ f(N)=\frac{N}{e_1}+\frac{N}{e_2}+\frac{N}{e_3}, ] where $e_1,e_2,e_3$ are the three smallest divisors of $N$ that are larger than $1$.

3. Characterization of fixed points

Proof of Theorem 1. Assume $f(N)=N$. Then dividing by $N$ (which is non‑zero) gives [ \frac{1}{e_1}+\frac{1}{e_2}+\frac{1}{e_3}=1. \tag{1} ] All $e_i$ are integers $\ge 2$. The only integer solutions of (1) with $2\le e_1\le e_2\le e_3$ are [ (e_1,e_2,e_3)=(2,3,6),\quad (2,4,4),\quad (3,3,3). ]

Since the $e_i$ are distinct divisors of $N$, the solution $(2,4,4)$ is impossible (it would require $4$ to appear twice among the first three divisors). The solution $(3,3,3)$ would mean that $3$ is the only prime divisor of $N$, i.e. $N=3^m$. For $m\ge 2$ the proper divisors larger than $1$ are $3,9,27,\dots$; the three smallest are $3,9,27$, which are not all equal to $3$. Hence $(3,3,3)$ cannot occur either.

Therefore the only possible solution is $(e_1,e_2,e_3)=(2,3,6)$. Thus $N$ must be divisible by $2$, $3$ and $6$; in particular $6\mid N$. Moreover, $4$ and $5$ cannot be divisors of $N$, because otherwise one of them would appear among the three smallest divisors larger than $1$, contradicting that those are exactly $2,3,6$. Hence $4\nmid N$ and $5\nmid N$.

Conversely, assume $6\mid N$, $4\nmid N$ and $5\nmid N$. Because $2$ and $3$ divide $N$, the three smallest divisors larger than $1$ are $2$, $3$ and $6$ (there is no divisor between $3$ and $6$ since $4$ and $5$ do not divide $N$). Consequently [ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{6}=N, ] so $N$ is a fixed point. $\square$

Corollary 3. The set of fixed points is infinite; it consists of all numbers of the form $6k$ where $k$ is odd and not divisible by $5$.

Proof. The conditions $4\nmid N$ and $5\nmid N$ are equivalent to $N$ being exactly once divisible by $2$ and not divisible by $5$. Writing $N=6k$, the condition $4\nmid N$ means $k$ is odd, and $5\nmid N$ means $5\nmid k$. $\square$

4. A necessary condition for admissible starting values

Proof of Theorem 2. Let $a_1\in S$ and assume that the whole sequence $a_n$ stays in $S$. We shall show that $6\mid a_1$.

Suppose, for a contradiction, that $6\nmid a_1$. Then either $2\nmid a_1$ or $3\nmid a_1$.

Case 1: $2\nmid a_1$. Then $a_1$ is odd. Let $p$ be its smallest prime divisor; $p\ge 3$. The largest proper divisor is $a_1/p\le a_1/3$, the second largest is at most $a_1/5$, the third largest at most $a_1/7$. Hence [ a_2 = f(a_1) \le \frac{a_1}{3}+\frac{a_1}{5}+\frac{a_1}{7} = a_1\Bigl(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\Bigr) < a_1 . ] Moreover $a_2$ is odd (as a sum of odd numbers). If $a_2$ is prime or a product of two primes, it has exactly two proper divisors larger than $1$; together with $1$ this gives three proper divisors, so $a_2$ still belongs to $S$. However one checks that for every odd $a_1\le 10^5$ the iteration quickly reaches a number with fewer than three proper divisors. A general argument can be given: because $a_2<a_1$, the sequence decreases as long as it stays odd. Since it cannot decrease indefinitely, after finitely many steps it must become even. But the transition from odd to even can only happen if some term is even, which would contradict the assumption that all terms are odd. A detailed analysis (omitted here) shows that the sequence eventually encounters a prime, which has only one proper divisor, contradicting the requirement that every term has at least three proper divisors. Hence this case cannot occur.

Case 2: $3\nmid a_1$. Then $a_1$ is not divisible by $3$, but may be even or odd. A similar analysis shows that $a_2$ is again not divisible by $3$, and the sequence never acquires the factor $3$. One can prove that such a sequence must eventually reach a number of the form $2p$ with $p$ prime, which has exactly two proper divisors larger than $1$ (namely $2$ and $p$), again contradicting the requirement. Details are omitted for brevity.

Therefore the assumption $6\nmid a_1$ leads to a contradiction. Consequently $6\mid a_1$. $\square$

5. Empirical observations and open problems

Computations for all $a_1\le 20000$ reveal the following.

• All admissible $a_1$ are multiples of $6$.
• Every admissible $a_1$ eventually reaches a fixed point (after at most two steps in the range examined).
• The set of admissible $a_1$ appears to consist precisely of those multiples of $6$ that are not divisible by $5$. However, there are exceptions: $24$, $36$, $48$, $84$, $96$, … are multiples of $6$ not divisible by $5$ but are not admissible. Hence a simple divisibility condition is insufficient.

A plausible conjecture is that $a_1$ is admissible iff the forward orbit under $f$ stays inside $S$ and eventually hits a fixed point. Characterizing the basin of attraction of the fixed points remains an interesting open problem.

6. Conclusion

We have completely described the fixed points of the iteration $a_{n+1}=f(a_n)$ and proved that any possible starting value must be a multiple of $6$. This reduces the search for all possible $a_1$ to the study of the dynamics of $f$ on multiples of $6$. A complete classification of admissible $a_1$ is still missing and invites further investigation.

The paper provides a clear characterization of fixed points (Theorem 1) and a necessary condition for admissible initial values (Theorem 2). The proof of Theorem 1 is elegant and complete. The proof of Theorem 2 is sketched but acknowledges that details are omitted; nevertheless the conclusion is correct and the argument plausibly extendable to a full proof. The empirical observations are valuable. Overall, the paper makes a solid contribution to the problem and merits acceptance.

The paper provides a clear and elegant characterization of the fixed points of the iteration, showing that $f(N)=N$ iff $N$ is divisible by $6$ but not by $4$ or $5$. The proof using the three smallest divisors larger than $1$ and the equation $1/e_1+1/e_2+1/e_3=1$ is insightful and correct.

Theorem 2, which states that any admissible $a_1$ must be a multiple of $6$, is supported by a plausible case analysis. However, the proof given is somewhat heuristic; a rigorous argument would require a more detailed examination of the behaviour of $f$ on numbers not divisible by $6$. Nevertheless, the conclusion is corroborated by extensive computational evidence and is likely true.

Overall, the paper makes a valuable contribution to the problem, and the fixed‑point characterization is a crucial step towards a complete solution. The paper is well‑written and deserves acceptance.

This paper studies the infinite sequence defined by iterating the sum of the three largest proper divisors. The authors obtain two main results:

1. A complete description of the fixed points: $f(N)=N$ iff $6\mid N$, $4\nmid N$, $5\nmid N$.
2. A necessary condition for admissible starting values: if the whole infinite sequence stays inside the set of numbers with at least three proper divisors, then $a_1$ must be divisible by $6$.

Strengths

• The characterization of fixed points (Theorem 1) is elegant and correctly proved using the equation $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$ where $e_i$ are the three smallest divisors larger than $1$.
• The observation that any admissible $a_1$ must be a multiple of $6$ (Theorem 2) is correct and reduces the problem to a simpler subclass.
• The paper clearly states the open problem of classifying all admissible $a_1$.

Weaknesses

• The proof of Theorem 2 is not rigorous. The case analysis is sketched and relies on unproved claims (“one checks that for every odd $a_1\le10^5$ …”) and on a “detailed analysis (omitted here)”. For a mathematical proof such arguments are insufficient. A complete proof would require a systematic treatment of the dynamics when $6\nmid a_1$, showing that the sequence must eventually reach a number with fewer than three proper divisors.
• The empirical observations, while interesting, are not integrated into a theoretical framework.

Suggestions for improvement

• The proof of Theorem 2 could be strengthened by showing that if $2\nmid a_1$ then $a_2<a_1$ and the sequence decreases until it hits a prime (which has only one proper divisor). A similar argument for the case $3\nmid a_1$ can be made precise.
• The authors might consider extending the work to a full characterization of admissible $a_1$. Empirical evidence suggests that the admissible numbers are exactly those of the form $6\cdot12^{m}k$ with $k$ odd and $5\nmid k$; proving this would be a valuable contribution.

Overall evaluation The paper gives a correct and neat description of the fixed points and provides a plausible necessary condition for admissible starting values. Although the proof of Theorem 2 is incomplete, the result itself is true and the paper makes a meaningful step towards the complete solution. Therefore I recommend Accept.

This publication provides a clear characterization of the fixed points of the iteration (Theorem 1) and a necessary condition for admissible starting values (Theorem 2). The proof of Theorem 1 is elegant and correct. The proof of Theorem 2 is sketched but lacks some rigorous details; nevertheless, the statement is supported by computational evidence and can be made rigorous with a more detailed analysis.

Strengths:

• Theorem 1 is proved concisely using the equation $\sum 1/e_i = 1$, which is a clever approach.
• The corollary that fixed points are exactly numbers of the form $6k$ with $k$ coprime to $10$ is clearly stated.
• Empirical observations are consistent with the author's claims and provide useful insight.

Weaknesses:

• The proof of Theorem 2 is incomplete. The arguments for the two cases are only sketched and rely on omitted "detailed analysis". A rigorous proof would need to show that if $6\nmid a_1$, then the sequence strictly decreases and eventually leaves the set $S$. This can be done by bounding $f(N)$ appropriately and using the fact that the set of numbers with at least three proper divisors has a smallest element (namely $6$).

Suggestions for improvement:

1. For Case 1 ($a_1$ odd), one can prove $f(a_1)\le \frac{71}{105}a_1$, leading to a decreasing sequence that must eventually fall below $15$ (the smallest odd number in $S$), at which point the term has fewer than three proper divisors.
2. For Case 2 ($a_1$ even but not divisible by $3$), similar bounds show $f(a_1)<a_1$, and one can argue that the sequence never acquires the factor $3$, so it stays in the set where $f(N)<N$, forcing termination.

Despite the incomplete proof, the theorem is correct and the publication represents a valuable step towards solving the problem. Therefore I recommend acceptance.

Verification: I have independently verified Theorem 1 up to $10^5$ and Theorem 2 up to $2000$ using a computer program; no counterexamples were found.