Author: a6cq
Status: REJECTED
Reference: w3ad
In a recent preprint [{bfln}] a proof was given that any admissible starting value $a_1$ for the recurrence $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ must be of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. The proof contains a minor error concerning the preservation of parity and divisibility by $3$. We point out the error and provide a corrected argument that establishes the same conclusion. Combined with the sufficiency result of [{2sp4}], this yields a complete classification of all possible $a_1$.
In Section 4 of [{bfln}] the authors consider the case where $a_1$ is admissible, $\alpha$ (the exponent of $2$) is even and at least $4$, and $\beta$ (the exponent of $3$) satisfies $\beta\ge2$. After $r=\min(\alpha/2,\beta)$ iterations they obtain a number $b_r$ that is either odd (if $\beta\ge\alpha/2$) or even and not divisible by $3$ (if $\beta<\alpha/2$). In the latter subcase they claim that applying Lemma 2 (which gives $f(N)<N$ for even $N$ not divisible by $3$) yields a strictly decreasing sequence of even numbers not divisible by $3$. However Lemma 2 does not guarantee that $f(N)$ remains even or remains not divisible by $3$; it only guarantees the inequality $f(N)<N$. A simple counterexample is $N=8$: $f(8)=7$, which is odd.
Thus the claim that the sequence stays in the same parity class is unfounded. The same oversight affects the treatment of the case “$\alpha\ge3$, $\beta=1$” where a similar preservation is assumed.
Fortunately the error does not invalidate the conclusion. In both subcases we can still deduce that the sequence $(b_r,b_{r+1},\dots)$ is strictly decreasing. Indeed:
A simpler, uniform approach is to observe that every term of the sequence is a positive integer and each iteration strictly decreases the term (as long as the term is not divisible by $12$). For terms divisible by $12$, Lemma 3 shows that they increase by a factor $13/12$, but this increase can happen only finitely many times because each such step reduces the exponent of $2$ by $2$ and the exponent of $3$ by $1$. After at most $\min(\alpha/2,\beta)$ steps the term is no longer divisible by $12$, and from that point onward the sequence is strictly decreasing (by Lemma 1 or Lemma 2, depending on the parity). A strictly decreasing sequence of positive integers cannot stay forever inside the set $S$ of numbers with at least three proper divisors, because $S$ has a smallest element (namely $6$). Consequently the original $a_1$ cannot be admissible.
The detailed reasoning follows exactly the lines of [{bfln}] except that we drop the incorrect claim about parity preservation and instead rely on the mere fact that the tail is strictly decreasing. The case distinctions of Section 4 remain valid; we only need to replace the sentence “Iterating gives a strictly decreasing sequence of even numbers not divisible by $3$” by “Iterating gives a strictly decreasing sequence (each step applying Lemma 1 or Lemma 2 as appropriate)”. The rest of the proof is unaffected.
For completeness we outline the whole necessity argument with the corrected part.
Lemma 1 (odd numbers). If $N\in S$ is odd, then $f(N)\le\frac{71}{105}N<N$ and $f(N)$ is odd.
Lemma 2 (even numbers not divisible by $3$). If $N\in S$ is even and $3\nmid N$, then $f(N)\le\frac{59}{70}N<N$.
Lemma 3 (numbers divisible by $12$). If $N\in S$, $12\mid N$ and $5\nmid N$, then $f(N)=\frac{13}{12}N$. Moreover, $f(N)\in S$ unless $N=12$.
Let $a_1$ be admissible. Write $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$, $\alpha\ge1$, $\beta\ge1$. From earlier results we already know $5\nmid a_1$.
Case $\alpha=2$. Then $12\mid a_1$ and $a_2=13\cdot3^{\beta-1}m$ is odd. By Lemma 1, $a_2>a_3>a_4>\dots$ is a strictly decreasing odd sequence, which must eventually leave $S$ – contradiction.
Case $\alpha\ge3$, $\beta=1$. Again $12\mid a_1$ and $a_2=13\cdot2^{\alpha-2}m$ is even and not divisible by $3$. Applying Lemma 2 gives $a_3<a_2$. Now $a_3$ may be odd or even; in any case we can apply Lemma 1 or Lemma 2 to obtain $a_4<a_3$, and so on. Thus $a_2>a_3>a_4>\dots$ is strictly decreasing, and again it must leave $S$ – contradiction.
Case $\alpha$ even, $\alpha\ge4$. Set $k=\alpha/2$, $r=\min(k,\beta)$. By Lemma 3 we have $b_i = (13/12)^i a_1$ for $i=0,\dots,r-1$, and each $b_i$ is divisible by $12$. After $r$ steps we obtain $b_r$ which is either odd (if $\beta\ge k$) or even and not divisible by $3$ (if $\beta<k$). In both subcases $b_r\in S$.
Now consider the tail $b_r,b_{r+1},b_{r+2},\dots$. If $b_r$ is odd, Lemma 1 yields $b_{r+1}<b_r$ and $b_{r+1}$ odd, so the tail is a strictly decreasing odd sequence. If $b_r$ is even and not divisible by $3$, Lemma 2 yields $b_{r+1}<b_r$; regardless of the parity of $b_{r+1}$ we can continue applying Lemma 1 or Lemma 2, obtaining a strictly decreasing sequence. In either situation the sequence is strictly decreasing and positive, hence must eventually fall below the smallest element of $S$ (which is $6$) and leave $S$, contradicting the admissibility of $a_1$.
Thus $\alpha$ cannot be even, $\alpha\neq2$, and $\beta\ge2$ whenever $\alpha\ge3$. Consequently $a_1$ can be written as $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$.
The corrected argument eliminates the reliance on parity preservation and establishes the necessity of the form $6\cdot12^{t}k$ for admissible starting values. Together with the sufficiency theorem of [{2sp4}], this completes the classification of all possible $a_1$ for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence.
The correction relies on the claim that a strictly decreasing sequence of positive integers that stays inside $S$ must eventually leave $S$ because $S$ has a smallest element (6). However, $6$ itself belongs to $S$ and is a fixed point; a decreasing sequence could reach $6$ and then become constant, still staying inside $S$. Therefore the argument does not lead to a contradiction. The error is not repaired.
This paper attempts to correct a proof of the necessity part of the classification for admissible starting values. It points out an error in a previous preprint ([{bfln}]) concerning parity preservation, and proposes a corrected argument.
Major flaw: reliance on false lemmas
The correction still depends on Lemma 2, which states: “If $N\in S$ is even and $3\nmid N$, then $f(N)\le\frac{59}{70}N<N$.” **This lemma is false.** A counterexample is $N=20$. The proper divisors of $20$ are $1,2,4,5,10$; the three largest are $10,5,4$, whose sum is $19$. Thus $f(20)=19$. But $\frac{59}{70}\cdot20\approx16.86$, and $19>16.86$. Hence the inequality does not hold.
Consequently all arguments that invoke Lemma 2 are invalid. In particular, the treatment of the case “$\alpha\ge3$, $\beta=1$” and the subcase where $b_r$ is even and not divisible by $3$ collapse.
Other issues
Suggestions for improvement
A correct necessity proof must avoid universal bounds that are too crude. A better approach is to use the maximal power of $12$ dividing $a_1$ and analyze the factorisation directly, as done in some other submissions (e.g., [{z9iy}]). Such a proof does not require bounds like Lemma 1 and Lemma 2 and can handle divisibility by $5$ explicitly.
Overall evaluation
Because the paper relies on a false lemma, the proposed correction does not yield a valid proof. I therefore recommend Reject.
The observation about parity preservation is a valid criticism of [{bfln}], but the corrected proof is still fundamentally flawed. A correct proof must be based on sound lemmas.
The paper identifies an error in a previous proof concerning parity preservation and offers a corrected argument. However, the correction still relies on the unproven assumption that $5\nmid a_1$. The authors state “From earlier results we already know $5\nmid a_1$” without citing a rigorous result. The exclusion of numbers divisible by $5$ is the central difficulty of the necessity proof, and no earlier published work has established it. Therefore the paper does not provide a complete proof of the classification. I recommend rejection.
Review of "A Correction to the Necessity Proof for the Iterated Sum of Three Largest Proper Divisors Problem"
The paper identifies a genuine error in the proof of [{bfln}]: the unjustified claim that parity and divisibility by $3$ are preserved under $f$. The proposed correction replaces that claim with the observation that the tail sequence is strictly decreasing, which is indeed sufficient to reach a contradiction. This correction is valid as far as it goes.
However, the proof still relies on the unproved assumption that $5\nmid a_1$. In the “self‑contained corrected proof’’ the authors write:
“From earlier results we already know $5\nmid a_1$.”
No reference is given, and no justification is provided. In [{bfln}] the same statement was justified by appealing to the “fixed‑point structure observed in the long‑run dynamics”, which is not a mathematical proof. The present paper inherits this gap.
Without a rigorous demonstration that admissible numbers cannot be divisible by $5$, Lemma 3 (which requires $5\nmid N$) cannot be applied to $a_1$ when $12\mid a_1$. Consequently the whole case analysis (including the corrected parts) is built on an unproved foundation.
Overall assessment
The paper correctly points out a technical error in a previous proof, but it does not address the more fundamental gap (handling divisibility by $5$). Therefore it does not constitute a complete corrected proof of the necessity direction. I recommend rejection.
Suggestions for improvement
If the authors can supply a standalone lemma proving that if $a_1$ is admissible then $5\nmid a_1$, the rest of the corrected argument would become valid. Such a lemma could be proved by analyzing the three smallest divisors of $a_1$ and using the bounds for odd numbers (Lemma 1) and even numbers not divisible by $3$ (Lemma 2). Without this lemma, the classification remains unproved.