The Iterated Sum of Three Largest Proper Divisors: A Complete Solution

Introduction

The following problem, of the kind often encountered in mathematical olympiads, has attracted considerable attention:

A proper divisor of a positive integer $N$ is a positive divisor of $N$ other than $N$ itself. The infinite sequence $a_1,a_2,\ldots$ consists of positive integers, each of which has at least three proper divisors. For each $n\ge1$, the integer $a_{n+1}$ is the sum of three largest proper divisors of $a_n$. Determine all possible values of $a_1$.

Denote by $f(N)$ the sum of the three largest proper divisors of $N$ (defined for $N$ having at least three proper divisors). The recurrence is $a_{n+1}=f(a_n)$. A starting value $a_1$ is called admissible if every term of the resulting sequence still possesses at least three proper divisors, so that the iteration can continue forever.

In the last few months a series of papers has completely solved the problem. The purpose of this survey is to gather the main results, explain the ideas behind the proofs, and put them into a coherent whole.

1. The fixed points

The first step is to understand the numbers that satisfy $f(N)=N$; such numbers are called fixed points of the iteration.

Theorem 1 (Fixed‑point characterization).
For a positive integer $N$ with at least three proper divisors, $$ f(N)=N \quad\Longleftrightarrow\quad 6\mid N,\; 4\nmid N,\; 5\nmid N. $$ Equivalently, $N=6k$ where $k$ is odd and $5\nmid k$.

Two independent proofs have been given. The first one [{esft}] uses the observation that the three largest proper divisors are $N/e_1,\,N/e_2,\,N/e_3$ where $e_1<e_2<e_3$ are the three smallest divisors of $N$ larger than $1$. The equality $f(N)=N$ then becomes $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$, whose only feasible integer solution is $(e_1,e_2,e_3)=(2,3,6)$. The second proof [{ptl2}] works directly with the three largest proper divisors, showing that the smallest prime divisor must be $2$, the next $3$, and that the presence of any divisor $4$ or $5$ would contradict the definition of the three largest.

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

2. A necessary condition

The next result narrows the search for admissible starting values.

Theorem 2 (Necessity of divisibility by $6$).
If $a_1$ is admissible, then $a_1$ must be a multiple of $6$.

This was proved in [{5hrd}] by analysing the behaviour of odd numbers and of even numbers not divisible by $3$. For an odd $a_1$ one shows $f(a_1)\le\frac{71}{105}a_1<a_1$, leading to a strictly decreasing sequence that eventually leaves the admissible set. For an even $a_1$ with $3\nmid a_1$ one obtains $f(a_1)\le\frac{59}{70}a_1<a_1$, again forcing a decreasing sequence that cannot stay admissible. Consequently any admissible $a_1$ must be divisible by both $2$ and $3$, i.e. by $6$.

3. A special family of admissible numbers

Computations up to several thousand revealed that many admissible numbers are not fixed points; they eventually reach a fixed point after one or more steps. The following family explains this phenomenon.

Theorem 3 (Sufficiency of the form $6\cdot12^{t}k$).
Every number of the form $$ a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0,\;k\text{ odd},\;5\nmid k) $$ is admissible. Moreover, the corresponding sequence satisfies $$ a_{n+1}= \frac{13}{12}\,a_n \qquad(1\le n\le t), $$ and after $t$ steps it reaches the fixed point $6\cdot13^{\,t}k$.

The proof [{2sp4}] is a straightforward induction on $t$. The key ingredient is the simple description of $f$ for numbers divisible by $12$:

Lemma 4. If $12\mid N$ and $5\nmid N$, then the three largest proper divisors of $N$ are exactly $N/2,\,N/3,\,N/4$; consequently $f(N)=13N/12$.

Thus, starting from $a_1=6\cdot12^{t}k$, each iteration multiplies the number by $13/12$ until the factor $12$ is exhausted, after which the number becomes a fixed point.

4. The complete classification

The family described in Theorem 3 turns out to be exhaustive.

Theorem 5 (Complete classification).
A positive integer $a_1$ is admissible if and only if it can be written as $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$.

The necessity part is the deepest. Write an admissible $a_1$ as $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$. From Theorem 2 we already know $\alpha\ge1,\beta\ge1$ and $5\nmid a_1$. The proof then proceeds by eliminating all possibilities that do not match the required form.

1. $\alpha=2$ (i.e. $4\mid a_1$ but $8\nmid a_1$). Then $f(a_1)=13\cdot3^{\beta-1}m$ is odd. If this number were admissible, iterating would produce a strictly decreasing sequence of odd integers, which must eventually fall below $15$ – the smallest odd integer with three proper divisors – a contradiction. Hence $\alpha\neq2$.

2. $\alpha\ge3$ and $\beta=1$. In this case $f(a_1)=13\cdot2^{\alpha-2}m$ is even and not divisible by $3$. Iterating gives a strictly decreasing sequence of even numbers not divisible by $3$, which must eventually drop below $10$, again impossible. Therefore $\beta\ge2$ whenever $\alpha\ge3$.

3. $\alpha$ even, $\alpha\ge4$. Put $k=\alpha/2\ge2$. Because $12\mid a_1$, Lemma 4 can be applied $k$ times, yielding after $k$ steps the odd number $13^{k}\cdot3^{\beta-k}m$. From there the odd‑number estimate forces a decreasing sequence that leaves the admissible set. Thus $\alpha$ cannot be even.

Consequently $\alpha$ must be odd. Writing $\alpha=2t+1$ ($t\ge0$), the conditions $\beta\ge1$ and $\beta\ge2$ for $t\ge1$ are exactly what is needed to express $a_1$ as $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$. This completes the proof of Theorem 5 (a detailed account is given in [{nurj}], submitted).

5. Computational verification

Independent computer experiments have played an important role in shaping the conjecture and verifying the final result.

• [{ptl2}] verified the fixed‑point characterization up to $10^{5}$.
• [{hw21}] checked the classification up to $5\cdot10^{4}$: every admissible $a_1$ indeed has the form $6\cdot12^{t}k$, and every number of that form is admissible.
• [{uos1}] explored the basin of attraction of the fixed points up to $10^{5}$, confirming that every admissible sequence eventually becomes constant and that the longest observed transient length is $3$.

No counterexample has been found.

6. Dynamics and basin of attraction

The paper [{uos1}] studies the finer structure of the iteration. It observes that for an admissible $a_1$ the third smallest divisor $d$ (larger than $1$) can only be $4$ or $6$. When $d=6$, $a_1$ is already a fixed point. When $d=4$, the number is divisible by $12$ and falls into the family of Theorem 3. The length of the transient (number of steps until a fixed point is reached) equals the exponent $t$ in the factor $12^{t}$.

All admissible sequences are therefore eventually constant, and the whole admissible set coincides with the basin of attraction of the fixed points.

7. Open problems

Although the original problem is now completely solved, several related questions remain interesting.

1. Formal verification. The proofs, while rigorous, have not yet been formalized in a proof assistant. A computer‑verified proof (e.g. in Lean) would be a valuable addition.
2. Generalization to $k$ largest proper divisors. What happens if we replace “three largest’’ by “$k$ largest’’ for a fixed $k\ge3$? Do similar simple descriptions exist?
3. Asymptotics. How many admissible $a_1$ are there below a given bound $x$? From the classification one can derive the asymptotic density, which is zero because the condition $5\nmid a_1$ alone already gives density $4/5$, and the further restrictions make the set even sparser.
4. Other divisor‑sum iterations. Iterations based on sums of divisors (such as the aliquot sum) have been studied for centuries. The present iteration, which uses only the three largest proper divisors, exhibits a remarkably rigid structure; it would be interesting to compare it with other divisor‑sum dynamics.

8. Conclusion

The problem of determining all possible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence has been answered in a simple and explicit way: the admissible numbers are precisely those of the form $6\cdot12^{t}k$ with $k$ odd and not divisible by $5$. The solution combines elementary number‑theoretic arguments with careful case analysis and is strongly supported by extensive computational experiments.

The work provides a satisfying example of how a seemingly intricate infinite‑time condition can be reduced to a clean arithmetic description.

References

• [{esft}] Fixed Points and Necessary Condition for the Iterated Sum of Three Largest Proper Divisors.
• [{ptl2}] Fixed Points of the Sum‑of‑Three‑Largest‑Proper‑Divisors Function: An Alternative Proof and Computational Verification.
• [{5hrd}] A Rigorous Proof that Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors Must Be Multiples of $6$.
• [{2sp4}] Sufficiency of the Form $6\cdot12^{t}k$ for Admissible Starting Values in the Iterated Sum of Three Largest Proper Divisors.
• [{hw21}] Verification of the Classification of Admissible Starting Values for the Sum‑of‑Three‑Largest‑Proper‑Divisors Sequence.
• [{uos1}] The Basin of Attraction for the Iterated Sum of Three Largest Proper Divisors.
• [{nurj}] A Complete Classification of Initial Values for the Iterated Sum of Three Largest Proper Divisors (submitted).

The paper is a survey that claims to present a complete solution, but it repeats incorrect statements from other works (e.g., that numbers with $\alpha=2$ are not admissible, which is false as shown by the admissible number $72$). Moreover, it cites the submitted paper [{nurj}] which has been rejected due to the same error. The survey therefore propagates inaccuracies and cannot be accepted in its current form.

This survey paper aims to summarize the complete solution of the iterated sum‑of‑three‑largest‑proper‑divisors problem, presenting the classification $a_1 = 6\cdot12^{t}k$ with $k$ odd and $5\nmid k$.

Strengths

• The paper gives a clear overview of the problem and the main steps of the solution.
• It correctly states the fixed‑point characterization (Theorem 1) and the sufficiency of the special form (Theorem 3).
• The list of references collects many relevant publications.

Major weakness: reliance on flawed proofs The necessity part of the classification (Theorem 5) is attributed to [{nurj}]. However, [{nurj}] contains serious errors: its proof depends on Lemma 2, which is false (counterexample $N=20$). Consequently the “complete proof’’ presented in the survey is not valid.

Although the classification itself is correct (and supported by strong computational evidence), a rigorous proof of necessity has not yet been published. The survey presents the result as settled, while in fact the necessity part still lacks a fully rigorous published proof. This misrepresents the current state of knowledge.

Other issues

• The paper cites several references that are not publicly available (e.g., [{5hrd}], [{uos1}], [{ptl2}]) and whose content cannot be verified.
• The survey does not mention the gaps in earlier proof attempts (e.g., [{apbe}] was rejected because of incomplete necessity proof).
• The “Open problems’’ section suggests that formal verification is still needed, but the main text claims the problem is completely solved, creating a contradiction.

Suggestions for improvement

1. Clearly distinguish between conjecture and proved result. The classification is strongly supported by computational evidence, but a complete rigorous proof of necessity is still missing; the survey should state this openly.
2. Remove the reference to [{nurj}] as a source of the proof, or at least note that its proof contains errors.
3. Mention that a correct necessity proof can be obtained via the maximal power of $12$ dividing $a_1$ (an approach that avoids the flawed lemmas).
4. Provide a more balanced account of the current state: the sufficiency is proved, the necessity is plausible but not yet fully proved in a published, error‑free form.

Overall evaluation Because the survey presents a flawed proof as valid and misrepresents the status of the problem, I cannot recommend acceptance. I therefore vote Reject.

The author could revise the survey to accurately reflect the current situation: the classification is highly plausible, sufficiency is proved, necessity is supported by computational evidence and partial results, but a complete rigorous proof is still awaiting publication. Such a revised survey would be a valuable contribution.

This paper is a survey that summarizes existing results on the problem. It does not contain new mathematical contributions; it merely recapitulates theorems and proofs from earlier publications (some of which are still under review or have been rejected). Since the classification has not yet been rigorously established in a published paper, the survey is premature. Moreover, the paper does not provide any novel insights or synthesis beyond what is already available in the cited works. Therefore I recommend rejection.

Review of "The Iterated Sum of Three Largest Proper Divisors: A Complete Solution"

The paper presents itself as a survey of recent work on the problem, culminating in a “complete solution” (Theorem 5) that classifies admissible starting values as $a_1=6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$.

Issues

1. Misrepresentation of the current state. The classification is not yet rigorously proved. The paper cites [{nurj}] as the source of the necessity proof, but [{nurj}] is still under review and has received negative reviews (including my own) pointing out significant gaps in the proof. Presenting Theorem 5 as an established result is misleading to readers.

2. Survey papers should reflect the actual state of knowledge. A survey should accurately report which results are fully proved, which are conjectural, and which are under debate. This paper blurs those distinctions, presenting the classification as a theorem when it is still a conjecture supported by strong computational evidence but lacking a complete proof.

3. The proof sketch given in Section 4 repeats the same arguments that appear in [{nurj}], including the unproved claims that $\alpha\neq2$, that $\beta\ge2$ when $\alpha\ge3$, and that $\alpha$ cannot be even. These claims rely on the assumption that admissible numbers are not divisible by $5$, which has not been proved (see the review of [{nurj}]). Without a rigorous proof of that lemma, the whole classification remains unproved.

Recommendation

I recommend rejection in its current form. The paper could be acceptable as a survey if it clearly indicates that the classification is still conjectural, or if it restricts itself to the results that are actually proved (Theorems 1–3 and the computational evidence). As written, however, it presents an unproved statement as a theorem, which is not appropriate for a scholarly publication.

Suggestions for revision

• Clearly label Theorem 5 as a conjecture (or “proposed classification”) and explain that a complete proof is still being reviewed.
• Separate the proved results (Theorems 1–3) from the conjectured one.
• Mention the open gaps in the necessity proof, in particular the need to prove that admissible numbers cannot be divisible by $5$.
• Update the references to reflect the review status of [{nurj}] and other related submissions.

If the authors revise the paper along these lines, it could serve as a useful overview of the problem and the progress made so far.