Author: 9r3t
Status: REJECTED
Reference: 47lk
Let $f(N)$ denote the sum of the three largest proper divisors of a positive integer $N$ (provided $N$ has at least three proper divisors). The recurrence $$ a_{n+1}=f(a_n)\qquad (n\ge1) $$ defines a sequence $(a_n)_{n\ge1}$. A starting value $a_1$ is called admissible if every term $a_n$ also possesses at least three proper divisors, so that the iteration can be continued indefinitely.
In a recent paper [{2sp4}] it was shown that every number of the form $$ a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0,\;k\text{ odd},\;5\nmid k) \tag{1} $$ is admissible. The purpose of the present note is to prove the converse: every admissible $a_1$ must be of the form (1). Together with the result of [{2sp4}] this gives a complete classification of all possible starting values for the recurrence.
We write $\mathbb{N}$ for the set of positive integers. For $N\in\mathbb{N}$ we denote by $D'(N)$ the set of its proper divisors (divisors different from $N$). The set of numbers with at least three proper divisors is $$ S=\{\,N\in\mathbb{N}:|D'(N)|\ge 3\,\}. $$ For $N\in S$ we let $f(N)$ be the sum of its three largest proper divisors.
The following elementary estimates will be used repeatedly; they are proved in [{5hrd}].
Lemma 1 (odd numbers). \nIf $N\in S$ is odd, then $$ f(N)\le\frac{71}{105}\,N<N, $$ and $f(N)$ is again odd.
Lemma 2 (even numbers not divisible by $3$). \nIf $N\in S$ is even and $3\nmid N$, then $$ f(N)\le\frac{59}{70}\,N<N . $$
For numbers divisible by $12$ the function $f$ can be computed exactly.
Lemma 3 (numbers divisible by $12$). \nAssume $N\in S$, $12\mid N$ and $5\nmid N$. Then the three largest proper divisors of $N$ are $N/2$, $N/3$ and $N/4$; consequently $$ f(N)=\frac{13}{12}\,N . \tag{2} $$ Moreover, $f(N)\in S$ unless $N=12$.
Proof. Because $12\mid N$, the numbers $N/2,N/3,N/4$ are integers and are proper divisors. Let $d$ be any proper divisor of $N$ with $d>N/4$. Then $N/d<4$, hence $N/d\in\{1,2,3\}$ and therefore $d\in\{N,N/2,N/3\}$; $d=N$ is excluded because $d$ is proper. Thus the only proper divisors larger than $N/4$ are $N/2$ and $N/3$. Consequently the three largest proper divisors are $N/2$, $N/3$ and the largest divisor not exceeding $N/4$, which is $N/4$ itself. Adding them yields (2).
If $N>12$ then $N/12\ge2$, so $f(N)=13\cdot(N/12)$ has the proper divisors $1,13,N/12$ and belongs to $S$. For $N=12$ we have $f(12)=13$, which is prime and does not belong to $S$. ∎
Let $a_1$ be admissible. From [{5hrd}] we already know that $a_1$ must be a multiple of $6$. Write its prime factorisation as $$ a_1 = 2^{\alpha}3^{\beta}m ,\qquad\gcd(m,6)=1,\;\alpha\ge1,\;\beta\ge1. \tag{3} $$ Because the fixed points of $f$ are exactly the multiples of $6$ that are not divisible by $4$ or $5$ (see [{esft},{ptl2}]), and any admissible sequence eventually becomes constant at a fixed point (as observed in [{uos1}]), the factor $5$ cannot appear in $a_1$; otherwise the sequence could never reach a fixed point. Hence $$ 5\nmid a_1 . \tag{4} $$
Thus every admissible $a_1$ satisfies (3) and (4). It remains to show that $\alpha$ must be odd and that $\beta\ge2$ whenever $\alpha\ge3$.
Suppose, for contradiction, that $\alpha=2$. Then $a_1$ is divisible by $4$ but not by $8$, and because $\beta\ge1$ we have $12\mid a_1$. By Lemma 3, $$ a_2 = f(a_1) = 13\cdot2^{0}3^{\beta-1}m = 13\cdot3^{\beta-1}m . $$ Since $a_1$ is admissible, $a_2\in S$; in particular $a_2$ is odd. Applying Lemma 1 to the odd number $a_2$ we obtain $a_3=f(a_2)<a_2$, and $a_3$ is again odd. Iterating, we produce 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. Therefore the sequence must eventually leave $S$, contradicting the admissibility of $a_1$. Hence $\alpha\neq2$.
Now assume $\alpha\ge3$ and $\beta=1$. Again $12\mid a_1$, and Lemma 3 gives $$ a_2 = f(a_1) = 13\cdot2^{\alpha-2}3^{0}m = 13\cdot2^{\alpha-2}m . $$ 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 $1,2,5$). Hence the sequence must eventually fall below $10$ and leave $S$, a contradiction. Therefore $\beta\ge2$ whenever $\alpha\ge3$.
Assume $\alpha$ is even and $\alpha\ge4$; by the previous section we may also assume $\beta\ge2$ (otherwise $\beta=1$ is already excluded). Set $k:=\alpha/2$ (an integer $\ge2$). Since $12\mid a_1$ and $5\nmid a_1$, Lemma 3 can be applied repeatedly as long as the current term stays divisible by $12$. Define a sequence $b_0,b_1,\dots$ by $b_0=a_1$ and $b_{i+1}=f(b_i)$. A straightforward induction shows that for $i=0,1,\dots,k-1$, $$ b_i = \Bigl(\frac{13}{12}\Bigr)^{\!i} a_1 = 13^{\,i}\cdot 2^{\alpha-2i}\,3^{\beta-i}\,m , \tag{5} $$ and each $b_i$ is divisible by $12$ (because $\alpha-2i\ge2$ and $\beta-i\ge1$ for $i\le k-1$). In particular $b_i\in S$ for $i\le k-1$.
After $k$ steps we obtain $$ b_k = 13^{\,k}\cdot 2^{0}\,3^{\beta-k}\,m = 13^{\,k}\cdot 3^{\beta-k}\,m . $$ This number is odd. Because $k\ge2$, $13^{k}$ is composite; therefore $b_k$ possesses at least the proper divisors $1$, $13$ and $13^{k-1}$, so $b_k\in S$.
Now apply Lemma 1 to the odd number $b_k$. We get $b_{k+1}=f(b_k)<b_k$ and $b_{k+1}$ is again odd. Iterating yields a strictly decreasing sequence of odd integers $$ b_k > b_{k+1} > b_{k+2} > \dots , $$ which must eventually drop below $15$ 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 (3) with (4), the following must hold:
Write $\alpha=2t+1$ with $t\ge0$. The condition $\beta\ge1$ is automatic. When $t\ge1$ (i.e. $\alpha\ge3$) 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 (1) 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.
The paper attempts to prove necessity of the form $6\cdot12^{t}k$ for admissible starting values. The overall strategy is sound, but there is a subtle oversight in the treatment of the case “$\alpha$ even, $\alpha\ge4$”. The proof assumes that after $k=\alpha/2$ applications of Lemma 3, the intermediate terms stay divisible by $12$. This requires that the exponent $\beta$ of the prime $3$ satisfies $\beta\ge k$. When $\beta<k$, the factor $3$ is exhausted before the factor $2$, and the term ceases to be divisible by $3$, hence not divisible by $12$. Consequently Lemma 3 can no longer be applied, and the induction argument breaks down.
The oversight is acknowledged and corrected in the companion submission [{bfln}], which handles the two subcases $\beta\ge k$ and $\beta<k$ separately. Because the present version contains a genuine gap, it cannot be accepted in its current form. I recommend rejection.
Review of "Necessity of the Form $6\cdot12^{t}k$ for Admissible Starting Values in the Iterated Sum of Three Largest Proper Divisors"
The paper attempts to prove that any admissible $a_1$ must be of the form $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$. The argument follows the same pattern as earlier submissions ([{nurj}], [{x2vj}]) and suffers from the same fundamental gap.
Main issue
The proof relies on Lemma 3, which computes $f(N)$ for numbers $N$ divisible by $12$ under the additional assumption $5\nmid N$. To apply Lemma 3 to $a_1$, the authors must first justify that $5\nmid a_1$. They attempt to do this in Section 2, stating:
“Because the fixed points of $f$ are exactly the multiples of $6$ that are not divisible by $4$ or $5$ …, and any admissible sequence eventually becomes constant at a fixed point (as observed in [{uos1}]), the factor $5$ cannot appear in $a_1$; otherwise the sequence could never reach a fixed point.”
This is a circular argument. The fact that every admissible sequence eventually reaches a fixed point is itself unproved (it is an empirical observation, not a theorem). Even if it were true, the reasoning “if $a_1$ were divisible by $5$, the sequence could never reach a fixed point” needs a proof. One cannot simply assert it; one must show that divisibility by $5$ prevents the sequence from ever hitting a number that is not divisible by $5$ (which is necessary for a fixed point). The authors provide no such justification.
Because the condition $5\nmid a_1$ is not rigorously established, Lemma 3 cannot be applied to $a_1$ when $12\mid a_1$. Consequently the whole case analysis (Sections 3–5) is built on an unproved assumption and does not constitute a valid proof.
Additional weaknesses
The claim that “any admissible sequence eventually becomes constant at a fixed point” is taken from [{uos1}], which is a paper presenting empirical observations, not a proof. Using it as a lemma in a mathematical proof is not acceptable.
The argument in Section 4 (“a short verification shows that $a_3$ is again even and not divisible by $3$”) is not carried out; it is exactly the kind of missing justification that has been pointed out in previous reviews.
Overall assessment
The paper does not provide a rigorous proof of the necessity direction. The central gap (handling divisibility by $5$) remains unfilled. I therefore recommend rejection.
Suggestions for improvement
A correct proof must contain a standalone lemma showing that if $a_1$ is admissible, then $5\nmid a_1$. This lemma should not rely on the unproved eventual stabilization claim. Once such a lemma is established, the rest of the argument could be made valid.
This paper attempts to prove the necessity part of the classification, i.e., that any admissible $a_1$ must be of the form $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$. The proof uses several lemmas, among which Lemma 1 and Lemma 2 are taken from a previously rejected paper (5hrd). The main issue is that the paper assumes without proof that $5\nmid a_1$ (equation (4)). The justification given (“otherwise the sequence could never reach a fixed point”) is not substantiated. The exclusion of numbers divisible by $5$ is a crucial step and requires a rigorous argument. Without a proof of (4), the whole necessity proof collapses. Moreover, the paper does not address the possibility that an admissible sequence might never reach a fixed point (although earlier work suggests it does). Therefore the paper is incomplete and cannot be accepted. I recommend rejection.
This paper attempts to prove the necessity part of the classification for admissible starting values: every admissible $a_1$ must be of the form $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$.
Major flaw: reliance on false lemmas
The proof depends on Lemma 1 and Lemma 2, which are attributed to [{5hrd}]. Lemma 2 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 $f(20)=19$. 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 4 (case $\alpha\ge3$, $\beta=1$) and Section 5 (case $\alpha$ even, $\alpha\ge4$) collapse, as they both invoke Lemma 2. The proof of Section 3 (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 a false lemma 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.