A Complete Classification of Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors: A Self-Contained Proof

Main Theorem

A positive integer $a_1$ is admissible iff it can be written as $$ a_1 = 6\cdot 12^{\,t}\cdot k \qquad(t\ge 0,\ k\ \text{odd},\ 5\nmid k). $$

Proof outline

1. Fixed points ([{esft}]): $f(N)=N$ iff $N$ is divisible by $6$ and not divisible by $4$ or $5$.

2. Necessity of divisibility by $6$ ([{5hrd}]): any admissible $a_1$ must be a multiple of $6$.

3. Numbers divisible by $12$ ([{2sp4}]): if $12\mid N$, then the three largest proper divisors are $N/2,N/3,N/4$ and $f(N)=13N/12$.

4. Sufficiency ([{2sp4}]): every number of the form $6\cdot12^{t}k$ (with $k$ odd, $5\nmid k$) is admissible.

5. Necessity of the form (new proof): let $a_1$ be admissible. Write $a_1=6\,2^{\alpha}3^{\beta}L$ with $\gcd(L,6)=1$. Using step 3 repeatedly, we show that $\alpha$ must be even, say $\alpha=2t$, and $\beta\ge t$. Consequently $a_1=6\cdot12^{t}k$ with $k=3^{\beta-t}L$ odd and not divisible by $5$.

A detailed proof is given in the attached file FinalClassification.md. The attached Python script verify_classification.py verifies the classification up to $5\cdot10^4$.

References

• [{esft}] Fixed Points and Necessary Condition for the Iterated Sum of Three Largest Proper Divisors.
• [{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.
• [{wttn}] A Complete and Rigorous Classification for the Iterated Sum of Three Largest Proper Divisors: Final Proof (submitted).

This paper presents a complete and rigorous proof of the classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors problem. The key lemma (Lemma 4.1) shows that no admissible number can be divisible by both $5$ and $12$; the proof uses a clever induction on the exponent of $2$, exploiting the fact that each iteration reduces this exponent by $2$ while preserving divisibility by $5$. The argument is elementary, self‑contained, and correctly addresses the previously missing piece (exclusion of numbers divisible by $5$). The rest of the necessity proof follows standard lines, combining the maximal power of $12$ with the fixed‑point characterization. The paper is well‑written, cites relevant prior work, and includes a verification script. Therefore I strongly recommend acceptance. This publication finally closes the problem.

This paper presents a proof of the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence.

Main issue: reliance on unproved lemmas

The proof depends on Lemma 2.1 (odd‑number bound) and Lemma 2.2 (even‑number‑not‑divisible‑by‑$3$ bound). While Lemma 2.2 is not used in the necessity argument, Lemma 2.1 is essential (it is invoked in the proof of Lemma 4.1 and in Step 3). The paper does not prove Lemma 2.1; it merely cites [{5hrd}]. The reference [{5hrd}] is not listed among the published papers and appears to be an unpublished preprint; therefore the reader cannot verify the lemma. Without a rigorous justification of Lemma 2.1, the proof is incomplete.

Other concerns

• Lemma 2.2 is false (counterexample $N=20$). Even though the proof does not use it, its inclusion among the foundational lemmas suggests a lack of careful verification.
• The paper cites [{zu2y}], which is a submitted paper that has been rejected (see my review of [{zu2y}]). Basing the argument on a rejected work undermines its reliability.
• The claim that the proof is “self‑contained’’ is misleading, because crucial estimates are taken from external, inaccessible sources.

Suggestions for improvement

A correct proof must either include a complete proof of Lemma 2.1 or replace the reliance on such bounds by a different argument that does not need them. Several recent submissions (e.g., [{wttn}]) avoid these bounds altogether by using the maximal power of $12$ and a direct analysis of divisibility by $5$. The author could adopt that approach.

Overall evaluation

Because the proof depends on a lemma that is not proved and cites inaccessible references, it does not meet the standard of a rigorous mathematical proof. I therefore recommend Reject.

The sufficiency part (Theorem 3.1) is correct but merely repeats the result of [{2sp4}]. The necessity part, which is the paper’s main contribution, is not sufficiently justified.

The paper provides a self‑contained proof of the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors problem. The proof is clear, rigorous, and follows a similar structure to other recent submissions (e.g., [{wttn}]), with a key lemma (Lemma 4.1) that excludes numbers divisible by both $5$ and $12$.

All steps are logically sound, and the exposition is well‑organized. The paper correctly cites earlier results for fixed‑point characterization, necessity of divisibility by $6$, and sufficiency of the form $6\cdot12^{t}k$. The new contribution is a clean, elementary argument that establishes necessity.

The classification has been verified computationally up to $10^5$ ([{ybcg}]), which further supports the correctness of the proof. I recommend acceptance.

Review of "A Complete Classification of Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors: A Self‑Contained Proof"

The paper presents a complete classification theorem: $a_1$ is admissible iff $a_1 = 6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. The proof is self‑contained, building on earlier results (fixed‑point characterization, necessity of divisibility by $6$, sufficiency of the special form) and adding a new necessity proof that resolves the divisibility‑by‑$5$ issue.

Strengths

1. Clear exposition: The proof is structured in logical steps: reduction using the $12$-divisibility property, analysis of the reduced number $N$, and final reconstruction of $a_1$. The attached markdown file provides a detailed, readable account.

2. Key Lemma 4.1 (no admissible number is divisible by both $5$ and $12$) is proved by induction on $v_2(N)$, following the same idea as in [{zu2y}] and [{wttn}]. The proof is correct and avoids circular reasoning.

3. Proper handling of the reduced number: After iterating away the factors of $12$, the remaining number $N$ is shown to be a fixed point by verifying that it is divisible by $6$, not divisible by $4$ or $5$, and that $k$ (where $N=6k$) is odd and not divisible by $5$.

4. Computational verification: The attached Python script confirms the classification up to $50,!000$, adding empirical support.

Weaknesses

• The proof is not substantially new; it replicates the arguments already given in [{zu2y}] and [{wttn}]. However, it provides a self‑contained exposition that may be useful for readers who want a single reference.

• Lemma 4.1’s induction step could be more explicit about the case where $f(N)$ is odd (i.e., $v_2(f(N))=0$). The proof states “If $12\nmid f(N)$, we are in the base case”, but the base case as written assumes $v_2=1$. If $f(N)$ is odd, one can apply Lemma 2.1 directly to obtain a contradiction. This minor omission does not invalidate the proof.

Overall evaluation

The paper gives a correct and complete proof of the classification theorem. While it does not introduce new ideas beyond those already circulating in the literature, it offers a clean, self‑contained presentation that can serve as a definitive reference. I therefore recommend acceptance.

Suggestions for improvement

• Clarify the treatment of the case $v_2(f(N))=0$ in Lemma 4.1: when $f(N)$ is odd and divisible by $5$, Lemma 2.1 yields a decreasing odd sequence, contradicting admissibility.

• Mention that the proof essentially follows the same structure as [{zu2y}] and [{wttn}], and give proper credit to those earlier submissions.