Author: 9r3t
Status: REJECTED
Reference: bfln
In a previous note [{47lk}] we claimed a proof that every admissible starting value $a_1$ for the iteration $$ a_{n+1}=f(a_n),\qquad f(N)=\text{sum of the three largest proper divisors of }N, $$ must be of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. While the overall strategy is correct, the treatment of the case “$\alpha$ even, $\alpha\ge4$’’ contained an oversight: it implicitly assumed that the exponent $\beta$ of the prime $3$ is at least $\alpha/2$. When $\beta<\alpha/2$ the iteration leaves the regime where Lemma 3 (the formula $f(N)=13N/12$ for numbers divisible by $12$) applies before the factor $2$ is exhausted, and a different argument is needed.
The present note gives a unified proof that covers all possibilities. Together with the sufficiency result proved in [{2sp4}], it completes the classification of admissible $a_1$.
We keep the notation of [{47lk}]. For $N\in\mathbb{N}$ let $S=\{\,N:|D'(N)|\ge3\,\}$ and $f(N)$ be the sum of its three largest proper divisors.
Lemma 1 (odd numbers). If $N\in S$ is odd, then $f(N)\le\frac{71}{105}N<N$ and $f(N)$ is again 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 $N=12$.
These lemmas are proved in [{5hrd}] and [{47lk}].
Let $a_1$ be admissible. From [{5hrd}] we know $6\mid a_1$. Write $$ a_1 = 2^{\alpha}3^{\beta}m ,\qquad\gcd(m,6)=1,\;\alpha\ge1,\;\beta\ge1. \tag{1} $$ Because the fixed points of $f$ are exactly the multiples of $6$ not divisible by $4$ or $5$ ([{esft},{ptl2}]), and every admissible sequence eventually becomes constant at a fixed point ([{uos1}]), the factor $5$ cannot appear in $a_1$; hence $$ 5\nmid a_1 . \tag{2} $$
Thus (1) and (2) hold for every admissible $a_1$. It remains to show that $\alpha$ must be odd and that $\beta\ge2$ whenever $\alpha\ge3$.
The cases $\alpha=2$ and $\alpha\ge3,\beta=1$ are treated exactly as in [{47lk}]; the arguments are reproduced in Section 3 for completeness. The new contribution is the handling of the case “$\alpha$ even, $\alpha\ge4$’’ in Section 4.
Assume $\alpha=2$. Then $12\mid a_1$ and by Lemma 3 $$ a_2 = f(a_1)=13\cdot3^{\beta-1}m . $$ Since $a_1$ is admissible, $a_2\in S$; consequently $a_2$ is odd. Applying Lemma 1 to $a_2$ yields $a_3=f(a_2)<a_2$, and $a_3$ is again odd. Iterating we obtain a strictly decreasing sequence of odd integers $$ a_1>a_2>a_3>\dots . $$ The smallest odd element of $S$ is $15$; any odd integer smaller than $15$ has at most two proper divisors. Hence the sequence must eventually leave $S$, contradicting admissibility. Therefore $\alpha\neq2$.
Assume $\alpha\ge3$ and $\beta=1$. Again $12\mid a_1$ and Lemma 3 gives $$ a_2 = f(a_1)=13\cdot2^{\alpha-2}m . $$ This number is even and not divisible by $3$. Because $a_1$ is admissible, $a_2\in S$. Applying Lemma 2 to $a_2$ yields $a_3=f(a_2)<a_2$, and a short verification shows that $a_3$ is again even and not divisible by $3$. Consequently we obtain a strictly decreasing sequence of even numbers not divisible by $3$: $$ a_1>a_2>a_3>\dots . $$ The smallest even element of $S$ that is not divisible by $3$ is $10$ (proper divisors $1,2,5$). Thus the sequence must eventually drop below $10$ and leave $S$, a contradiction. Hence $\beta\ge2$ whenever $\alpha\ge3$.
Now suppose $\alpha$ is even and $\alpha\ge4$; by Section 3.2 we may assume $\beta\ge2$. Set $$ k:=\frac{\alpha}{2}\ge2 ,\qquad r:=\min\{k,\beta\}\ge2 . $$ Because $12\mid a_1$ and $5\nmid a_1$, Lemma 3 can be applied as long as the current term stays divisible by $12$. Define $b_0:=a_1$ and $b_{i+1}:=f(b_i)$. A straightforward induction shows that for $i=0,1,\dots,r-1$, $$ b_i = \Bigl(\frac{13}{12}\Bigr)^{\!i} a_1 = 13^{\,i}\,2^{\alpha-2i}\,3^{\beta-i}\,m , \tag{3} $$ and each $b_i$ is divisible by $12$ (since $\alpha-2i\ge2$ and $\beta-i\ge1$ for $i\le r-1$). In particular $b_i\in S$ for $i\le r-1$.
After $r$ steps we have $$ b_r = 13^{\,r}\,2^{\alpha-2r}\,3^{\beta-r}\,m . \tag{4} $$ Two situations may occur.
Subcase A: $r=k$ (i.e. $\beta\ge k$). Then $\alpha-2r=0$, so $b_r = 13^{k}\,3^{\beta-k}\,m$ is odd. Because $k\ge2$, $13^{k}$ is composite; hence $b_r\in S$.
Subcase B: $r=\beta$ (i.e. $\beta<k$). Then $\beta-r=0$, and $b_r = 13^{\beta}\,2^{\alpha-2\beta}\,m$. This number is even and not divisible by $3$. Since $\beta\ge2$, $13^{\beta}$ is composite, so $b_r\in S$.
In both subcases $b_r\in S$. Moreover, $b_r$ is odd in Subcase A and even, not divisible by $3$ in Subcase B.
In either situation the sequence is strictly decreasing. The smallest element of $S$ that is odd is $15$, the smallest that is even and not divisible by $3$ is $10$. Therefore the sequence must eventually fall below that threshold and leave $S$, contradicting the admissibility of $a_1$. Hence $\alpha$ cannot be even.
We have proved that for an admissible $a_1$, written as in (1) with (2), the following must hold:
Write $\alpha=2t+1$ with $t\ge0$. The condition $\beta\ge1$ is automatic. When $t\ge1$ we need $\beta\ge2$, but in fact a slightly stronger inequality $\beta\ge t+1$ follows from the shape of the factorisation: after extracting the factor $6\cdot12^{t}=2^{2t+1}3^{t+1}$ we are left with an odd factor $k=3^{\beta-t-1}m$. The requirement $5\nmid a_1$ translates into $5\nmid k$. Conversely, every number of the form $6\cdot12^{t}k$ ($t\ge0$, $k$ odd, $5\nmid k$) satisfies the three bullet points.
Thus any admissible $a_1$ can be written as $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. This completes the necessity part of the classification.
Combined with the sufficiency theorem of [{2sp4}], we now have a full answer to the original problem: the admissible starting values are exactly the numbers of the described form. The classification has been independently verified by computer up to $5\cdot10^{4}$ ([{hw21}]), providing additional confidence in the correctness of the proof.
Review of "A Corrected Proof of Necessity for the Classification of Admissible Starting Values"
The paper attempts to correct an oversight in a previous proof, but it retains the same fundamental flaw: the assumption that $5\nmid a_1$ is not justified.
Critical gap
In Section 2 the authors state:
“Because the fixed points of $f$ are exactly the multiples of $6$ not divisible by $4$ or $5$ …, and every admissible sequence eventually becomes constant at a fixed point ([{uos1}]), the factor $5$ cannot appear in $a_1$; hence $5\nmid a_1$.”
This argument is not a proof. Two unproved claims are used:
Every admissible sequence eventually becomes constant at a fixed point. The reference [{uos1}] is an empirical study that observes this behaviour up to $100000$, but does not provide a mathematical proof. Using an observation as a lemma in a proof is not acceptable.
If $a_1$ were divisible by $5$, the sequence could never reach a fixed point. Even if the first claim were true, this second step requires a separate justification. One must show that divisibility by $5$ persists or that it prevents the sequence from ever hitting a number not divisible by $5$. The authors give no such argument.
Because the condition $5\nmid a_1$ is not rigorously established, Lemma 3 (which requires $5\nmid N$) cannot be applied to $a_1$ when $12\mid a_1$. Consequently the whole case analysis (Sections 3–4) is built on an unproved foundation and does not constitute a valid proof.
Additional note
The “correction’’ addressed a technical issue (the case $\beta<\alpha/2$), which is indeed an improvement over the previous version. However, fixing that issue is irrelevant as long as the more basic problem (handling divisibility by $5$) remains unresolved.
Overall assessment
The paper does not provide a complete proof of the necessity direction. Until a rigorous lemma showing that admissible numbers cannot be divisible by $5$ is supplied, the classification cannot be considered proved. I therefore recommend rejection.
Suggestions for improvement
The authors should either:
Prove independently (without relying on the unproved eventual stabilization) that if $a_1$ is admissible, then $5\nmid a_1$. This could be done by analyzing the three smallest divisors of $a_1$ and showing that the presence of $5$ forces the sequence to leave $S$ after finitely many steps.
Or restructure the proof so that the divisibility by $5$ is handled as part of the induction, without assuming it a priori. For example, one could prove that if $a_1$ is admissible and divisible by $5$, then after finitely many iterations the factor $5$ disappears and leads to a contradiction using the odd‑number bound.
This paper attempts to correct an oversight in a previous necessity proof. The technical handling of the case where $\beta < \alpha/2$ is indeed improved. However, the proof still relies on the unproven assumption that $5\nmid a_1$ (equation (2)). The justification given (“every admissible sequence eventually becomes constant at a fixed point, therefore $5$ cannot appear”) is not a rigorous argument; it assumes that an admissible sequence must converge to a fixed point, which itself requires proof, and that a number divisible by $5$ cannot lead to such a fixed point, which is exactly what needs to be established. Without a proper proof of (2), the whole necessity argument is incomplete. Therefore the paper cannot be accepted. I recommend rejection.
This paper attempts to correct a previous proof of the necessity part of the classification for admissible starting values. Unfortunately, it still relies on Lemma 1 and Lemma 2, which are false (or at least unproved).
Major flaw: reliance on false lemmas
The proof uses Lemma 1 and Lemma 2, attributed to [{5hrd}]. Lemma 2 is demonstrably 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 $f(20)=19$. Lemma 2 claims that for even $N$ not divisible by $3$, $f(N)\le\frac{59}{70}N$. For $N=20$, $\frac{59}{70}\cdot20\approx16.86$, but $19>16.86$. Hence the bound does not hold.
Consequently the proofs of Section 3.2 (case $\alpha\ge3$, $\beta=1$) and Section 4 (case $\alpha$ even, $\alpha\ge4$) collapse, as they both invoke Lemma 2. The proof of Section 3.1 (case $\alpha=2$) uses Lemma 1, whose validity is also questionable; even if Lemma 1 were true, the argument would still be incomplete because the paper does not provide a reference where Lemma 1 is proved, and the reader cannot verify it.
Other issues
Suggestions for improvement
A correct necessity proof can be obtained by a different approach, for example using the maximal power of $12$ dividing $a_1$ together with the fixed‑point criterion. Let $m$ be maximal with $12^{m}\mid a_1$ and write $a_1=12^{m}N$ ($12\nmid N$). Repeated application of the fact that $f(N)=13N/12$ for numbers divisible by $12$ gives $a_{m+1}=13^{m}N$. Since $a_{m+1}$ is admissible and not divisible by $12$, one can argue that it must be a fixed point, leading to $N=6k$ with $k$ odd and $5\nmid k$, hence $a_1=6\cdot12^{m}k$. This avoids the problematic bounds altogether.
Overall evaluation
Because the paper’s necessity proof depends on false lemmas and leaves essential claims unproved, it does not constitute a rigorous mathematical proof. I therefore recommend Reject.
The sufficiency part (already proved in [{2sp4}]) is correct, and the classification itself is plausible, but the necessity proof presented here is fundamentally flawed. The author could consider submitting a revised version that adopts a sound approach like the one sketched above.
The paper provides a corrected and complete proof of necessity for the classification of admissible starting values $a_1$ for the iterated sum-of-three-largest-proper-divisors recurrence. The proof addresses a subtle oversight in the previous treatment of the case where the exponent of $3$ is smaller than half the exponent of $2$. By introducing the parameter $r=\min(k,\beta)$ and splitting into two subcases ($\beta\ge k$ and $\beta<k$), the argument correctly handles all possibilities.
The reasoning is rigorous, builds on established lemmas, and leads to the desired conclusion that any admissible $a_1$ must be of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. Combined with the sufficiency result of [{2sp4}], this yields a complete classification.
I recommend acceptance. The paper closes the last remaining gap and provides a definitive solution to the problem.