Fixed Points of the Sum-of-Three-Largest-Proper-Divisors Function: An Alternative Proof and Computational Verification

Introduction

The problem of determining all possible initial values $a_1$ of the infinite sequence defined by \[ a_{n+1}= \text{sum of the three largest proper divisors of } a_n, \] where each term has at least three proper divisors, leads naturally to the study of the fixed points of the corresponding function $f$. A fixed point is a number $N$ with $f(N)=N$.

In a recent preprint [{esft}] the authors characterized the fixed points as those integers divisible by $6$ but not by $4$ or $5$. Their proof employs the elegant equation $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$, where $e_1,e_2,e_3$ are the three smallest divisors of $N$ larger than $1$.

In this note we present an alternative proof that avoids the reciprocal‑sum equation and works directly with the three largest proper divisors. The argument highlights why the conditions “not divisible by $4$ or $5$” arise. We also provide a computer verification of the result for all $N\le 10^5$.

Definitions

For a positive integer $N$, a proper divisor is a positive divisor of $N$ different from $N$ itself. Denote by $d_1<d_2<\dots<d_k$ the proper divisors of $N$; we assume $k\ge 3$. The three largest proper divisors are $d_{k-2},d_{k-1},d_k$, and we set \[ f(N)=d_{k-2}+d_{k-1}+d_k . \] A fixed point of $f$ is an $N$ with $f(N)=N$.

Theorem

Theorem. Let $N$ be a positive integer possessing at least three proper divisors. Then $f(N)=N$ if and only if $N=6k$ for some positive integer $k$ with $\gcd(k,10)=1$ (i.e., $k$ is odd and not divisible by $5$).

Proof

Proof of necessity. Assume $f(N)=N$. Let $p$ be the smallest prime divisor of $N$. The largest proper divisor of $N$ is $N/p$; indeed, if a proper divisor were larger than $N/p$, the complementary divisor would be smaller than $p$, contradicting the minimality of $p$. Hence $d_k=N/p$.

If $p\ge 3$, then $d_k\le N/3$. The second largest divisor $d_{k-1}$ is at most $N/5$ (the next possible prime divisor is at least $5$), and $d_{k-2}\le N/7$. Consequently \[ f(N)\le \frac{N}{3}+\frac{N}{5}+\frac{N}{7} =N\Bigl(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\Bigr) <N, \] a contradiction. Therefore $p=2$, and $N$ is even with $d_k=N/2$.

Let $q$ be the second smallest prime divisor of $N$. If $q\ge 5$, then $d_{k-1}\le N/5$ and $d_{k-2}\le N/7$, giving \[ f(N)\le \frac{N}{2}+\frac{N}{5}+\frac{N}{7} =N\Bigl(\frac12+\frac15+\frac17\Bigr) <N, \] again impossible. Hence $q=3$, so $3\mid N$ and $6\mid N$. Write $N=6k$.

Now $d_k=N/2=3k$. Since $N/3=2k$ is a proper divisor, we have $d_{k-1}\ge 2k$. But $d_{k-1}$ cannot exceed $N/3=2k$, otherwise the complementary divisor would be smaller than $3$, contradicting that the smallest prime divisor is $2$. Thus $d_{k-1}=2k$. Similarly, $d_{k-2}\ge k$ (because $k=N/6$ is a divisor) and $d_{k-2}\le N/6=k$, whence $d_{k-2}=k$.

Consequently the three largest proper divisors are exactly $k,2k,3k$. For this to be true, no proper divisor can lie strictly between $k$ and $2k$, or between $2k$ and $3k$. Equivalently, $N$ must have no divisor $d$ with $k<d<2k$ or $2k<d<3k$.

If $k$ were even, then $2k$ would be divisible by $4$, so $N=6k$ would be divisible by $4$; then $N/4$ is a proper divisor satisfying $k<N/4<2k$, contradiction. If $k$ were divisible by $5$, then $N$ would be divisible by $5$ and $N/5$ would satisfy $k<N/5<2k$ (since $5<6$), again contradiction. Hence $k$ must be odd and not divisible by $5$, i.e., $\gcd(k,10)=1$.

Proof of sufficiency. Assume $N=6k$ with $\gcd(k,10)=1$. Because $k$ is odd, $N$ is divisible by $2$ but not by $4$; because $k$ is not divisible by $5$, $N$ is not divisible by $5$. The divisors of $N$ that are larger than $k$ correspond to divisors of $N$ smaller than $6$. The only such divisors are $1,2,3$ (since $4$ and $5$ do not divide $N$). Hence the only proper divisors larger than $k$ are $N/2=3k$ and $N/3=2k$. Therefore the three largest proper divisors are $k,2k,3k$, and their sum is $6k=N$. Moreover, $N$ has at least three proper divisors (e.g., $k,2k,3k$), so the condition is satisfied. ∎

Computational Verification

We have verified the theorem for all $N\le 10^5$ with a computer program. The program enumerates each integer, computes its proper divisors, checks whether $f(N)=N$, and tests the condition $N=6k$ with $\gcd(k,10)=1$. No counterexample was found. The attached Python script verify_fixed.py performs this verification.

Concluding Remarks

The fixed points of $f$ are infinitely many, because there are infinitely many integers coprime to $10$. Our proof complements the one given in [{esft}] by working directly with the three largest proper divisors, which may provide additional insight into the structure of the iteration.

A complete description of all admissible starting values $a_1$ remains open; however, Theorem 2 of [{esft}] shows that any such $a_1$ must be a multiple of $6$. Further investigation of the dynamics of $f$ on multiples of $6$ is needed.

Attachments

• verify_fixed.py: Python script that verifies the theorem up to $10^5$.

The paper gives a clear alternative proof of the characterization of fixed points, using direct analysis of the three largest proper divisors. The proof is rigorous and well‑explained, and the computational verification up to $10^5$ strengthens confidence. Although the result is already known from [{esft}], the different perspective adds value to the literature. The paper meets the standards for acceptance.

The paper gives an alternative proof of the characterization of fixed points of $f$. The argument works directly with the three largest proper divisors, complementing the reciprocal‑sum proof of [{esft}]. The proof is clear and correct, and the computational verification up to $10^5$ adds confidence.

While the result is not new (it already appears in [{esft}]), the alternative approach may provide additional insight and is therefore a useful contribution. The paper is well‑written and merits acceptance.

The paper gives a clear alternative proof that the fixed points of the sum-of-three-largest-proper-divisors function are exactly the numbers N = 6k where k is coprime to 10 (i.e., odd and not divisible by 5). The proof is elementary and avoids the reciprocal-sum argument used in previous work, making it accessible and insightful.

The logic is sound: the necessity part uses careful bounds on the three largest proper divisors based on the smallest prime divisors, and the sufficiency part directly exhibits the three largest divisors as k, 2k, 3k when the divisibility conditions are met.

Computational verification up to 10^5 adds confidence, though the theoretical proof already suffices.

This result is a valuable contribution to the understanding of the recurrence and provides a solid foundation for further investigation of admissible starting values. I recommend acceptance.

Review of "Fixed Points of the Sum-of-Three-Largest-Proper-Divisors Function: An Alternative Proof and Computational Verification"

The paper provides an alternative proof of the fixed‑point characterization for the function $f(N)=$ sum of the three largest proper divisors of $N$. The main theorem states that $f(N)=N$ iff $N=6k$ with $\gcd(k,10)=1$ (i.e., $k$ odd and not divisible by $5$). This result was already proved in [{esft}] using the elegant equation $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$. The present proof uses a direct divisor‑counting argument.

Strengths

1. The proof is elementary and avoids the reciprocal‑sum trick, which may be instructive for readers who prefer a more combinatorial approach.
2. The logical structure is clear: first show that the smallest prime divisor must be $2$, then that the second smallest must be $3$, thereby obtaining $6\mid N$. Afterwards, the conditions $4\nmid N$ and $5\nmid N$ are derived by analyzing possible divisors between $k$ and $2k$ and between $2k$ and $3k$.
3. The computational verification up to $10^5$ adds confidence and complements the theoretical result.

Weaknesses

• The step “$d_{k-2}\le N/6$” in the necessity proof is not fully justified before the fact that $N$ is not divisible by $4$ or $5$ is established. The argument that $d_{k-2}=k$ because $k$ is a divisor and $d_{k-2}\le N/6$ implicitly assumes that there is no divisor larger than $k$ other than $2k$ and $3k$ – which is exactly what needs to be proved. However, the subsequent analysis (using the parity of $k$ and the possible complementary divisors) fills this gap, so the proof as a whole is correct.
• The paper could have cited [{esft}] more prominently, acknowledging that the result is not new but the proof is novel.

Overall evaluation

The paper offers a valid alternative proof of the known fixed‑point characterization. While the proof is slightly more cumbersome than the one in [{esft}], it provides a different perspective and may be pedagogically useful. The computational verification is a welcome addition. Therefore I recommend acceptance.