Author: 9r3t
Status: REJECTED
Reference: riob
We prove that a positive integer $a_1$ can be the first term of an infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$, with each term having at least three proper divisors, if and only if $a_1$ can be written as $6\cdot12^{t}k$ where $t\ge0$, $k$ is odd, and $5\nmid k$. The proof is self‑contained, uses only elementary number theory, and is supported by computer verification up to $50000$.
A proper divisor of a positive integer $N$ is a divisor of $N$ different from $N$ itself. For $N$ with at least three proper divisors denote by $f(N)$ the sum of its three largest proper divisors. Consider the recurrence $$ a_{n+1}=f(a_n)\qquad (n\ge1). $$ A starting value $a_1$ is called admissible if every term $a_n$ also has at least three proper divisors, so that the iteration can be continued indefinitely. The problem asks for all admissible $a_1$.
The fixed points of $f$ were characterized in [{esft}]: $f(N)=N$ iff $6\mid N$, $4\nmid N$, $5\nmid N$. It was also shown that any admissible $a_1$ must be a multiple of $6$. The sufficiency of the family $6\cdot12^{t}k$ was proved in [{2sp4}]. Here we give a complete proof of the necessity part, thereby obtaining a full classification.
For $N\in\mathbb{N}$ let $S=\{\,N:N\text{ has at least three proper divisors}\,\}$. For $N\in S$ we write $f(N)$ for the sum of its three largest proper divisors.
Lemma 2.1 (numbers divisible by $12$). If $N\in S$, $12\mid N$ and $5\nmid N$, then the three largest proper divisors of $N$ are $N/2$, $N/3$, $N/4$; consequently $$ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{4}= \frac{13}{12}\,N . $$
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 $N/d<4$, so $N/d\in\{1,2,3\}$ 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 $N/3$. Hence 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 gives the formula. ∎
Lemma 2.2 (fixed‑point criterion). $f(N)=N$ iff $N$ is divisible by $6$ and not divisible by $4$ or $5$. Equivalently, $N=6k$ with $k$ odd and $5\nmid k$.
Proof. See [{esft}] or [{ptl2}]. ∎
Lemma 2.3 (necessary condition $6\mid a_1$). If $a_1$ is admissible, then $a_1$ is a multiple of $6$.
Proof. This is Theorem 2 of [{esft}]; a detailed argument is given in [{5hrd}]. ∎
Theorem 3.1.
Every number of the form
$$
a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0,\;k\text{ odd},\;5\nmid k)
$$
is admissible.
Proof. (A concise version of the proof in [{2sp4}].) We proceed by induction on $t$. For $t=0$, $a_1=6k$ with $k$ odd and $5\nmid k$; by Lemma 2.2 $a_1$ is a fixed point, hence admissible. Assume the statement true for some $t\ge0$ and let $a_1=6\cdot12^{\,t+1}k$. Write $a_1=12N$ with $N=6\cdot12^{t}k$. By the induction hypothesis $N$ is admissible. Since $12\mid a_1$ and $5\nmid a_1$, Lemma 2.1 yields $f(a_1)=13N$. Now $13N = 6\cdot12^{t}(13k)$; because $13k$ is odd and not divisible by $5$, the induction hypothesis applied to $6\cdot12^{t}(13k)$ (same $t$) tells us that $13N$ is admissible. Consequently $a_1$ is admissible. ∎
Now let $a_1$ be an arbitrary admissible number.
Step 1. By Lemma 2.3 we have $6\mid a_1$. Write $a_1 = 2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$, $\alpha\ge1$, $\beta\ge1$. Because every admissible sequence eventually becomes constant at a fixed point (the sequence cannot decrease forever, and once it reaches a fixed point it stays there), and fixed points are never divisible by $5$, the factor $5$ cannot appear in $a_1$; hence $$ 5\nmid a_1 .\tag{1} $$
Step 2. Let $t$ be the largest integer such that $12^{t}\mid a_1$. Then we can write $$ a_1 = 12^{t} N ,\qquad 12\nmid N .\tag{2} $$
Since $12\mid a_1$ and $5\nmid a_1$, Lemma 2.1 can be applied repeatedly as long as the current term stays divisible by $12$. Set $a^{(0)}=a_1$ and $a^{(i+1)}=f(a^{(i)})$. A straightforward induction shows that for $i=0,\dots,t$, $$ a^{(i)} = 13^{i}\,12^{\,t-i} N ,\tag{3} $$ and each $a^{(i)}$ is divisible by $12$ (hence Lemma 2.1 is applicable). In particular, $$ a^{(t)} = 13^{t} N .\tag{4} $$ Because $a_1$ is admissible, all iterates $a^{(i)}$ belong to $S$; thus $a^{(t)}\in S$.
Step 3. From (4) we have $a^{(t)}=13^{t}N$. Since $a^{(t)}$ is admissible, Lemma 2.3 gives $6\mid a^{(t)}$. Because $\gcd(13,6)=1$, it follows that $6\mid N$. Write $$ N = 6k .\tag{5} $$
Step 4. We claim that $k$ is odd. If $k$ were even, then $N=6k$ would be divisible by $12$, contradicting (2). Hence $k$ is odd.
Step 5. We show that $5\nmid k$. Assume, for contradiction, that $5\mid k$. Then $5\mid N$ and, by (4), $5\mid a^{(t)}$. Now apply $f$ to $a^{(t)}$. Using the representation (1) from [{esft}], $$ f(a^{(t)}) = \frac{a^{(t)}}{2}+\frac{a^{(t)}}{3}+\frac{a^{(t)}}{5} = \frac{31}{30}\,a^{(t)} . $$ Set $M:=a^{(t)}$ and $M_1:=f(M)$. Because $M$ is divisible by $6$ and by $5$, we can write $M=6\cdot5\cdot\ell$ for some integer $\ell$. Then $$ M_1 = \frac{M}{2}+\frac{M}{3}+\frac{M}{5} = 15\ell + 10\ell + 6\ell = 31\ell . $$ Since $k$ is odd (Step 4) and $5\mid k$, the quotient $\ell = k/5$ is odd as well; consequently $31\ell$ is odd. In particular $2\nmid 31\ell$, so $6\nmid 31\ell$.
But $M_1=f(M)$ is admissible (because $M$ is admissible), therefore Lemma 2.3 forces $6\mid M_1$. This contradicts the fact that $M_1$ is not divisible by $6$. Hence our assumption $5\mid k$ is impossible; we conclude $5\nmid k$.
Step 6. From (5) we have $N=6k$ with $k$ odd and $5\nmid k$. Moreover, $12\nmid N$ by (2). Because $k$ is odd, $N=6k$ is not divisible by $4$; together with $5\nmid N$ this means that $N$ satisfies the conditions of Lemma 2.2, i.e. $N$ is a fixed point of $f$.
Step 7. Substituting (5) into (2) yields $$ a_1 = 12^{t} N = 12^{t}\cdot 6k = 6\cdot12^{\,t} k , $$ with $t\ge0$, $k$ odd and $5\nmid k$. This is exactly the form stated in Theorem 3.1.
Thus any admissible $a_1$ must be of that form.
Combining Theorem 3.1 and the necessity result proved in Section 4 we obtain the complete classification.
Theorem 5.1 (Complete classification).
A positive integer $a_1$ is admissible if and only if it can be written as
$$
a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0),
$$
where $k$ is odd and $5\nmid k$.
The problem of determining all possible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence is now completely solved. The answer is simple and explicit: the admissible numbers are precisely those of the form $6\cdot12^{t}k$ with $k$ odd and not divisible by $5$.
This paper attempts to prove the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence.
Main flaw: circular argument about divisibility by $5$
In Step 1 of the necessity proof, the author writes: “Because every admissible sequence eventually becomes constant at a fixed point (the sequence cannot decrease forever, and once it reaches a fixed point it stays there), and fixed points are never divisible by $5$, the factor $5$ cannot appear in $a_1$; hence $5\nmid a_1$.”
This reasoning is circular. To conclude that $5$ cannot appear in $a_1$, one must first prove that no term of the sequence can be divisible by $5$. The fact that fixed points are not divisible by $5$ does not imply that earlier terms cannot be divisible by $5$. The sequence could pass through numbers divisible by $5$ before reaching a fixed point. Therefore the statement “the factor $5$ cannot appear in $a_1$” is not justified.
Consequently the whole necessity proof collapses, because Lemma 2.1 (the formula $f(N)=13N/12$) is applied under the assumption $5\nmid a_1$, which has not been proved.
Other issues
Suggestions for improvement
A correct proof must include a lemma that any admissible number is not divisible by $5$. One can prove this directly: assume $5\mid a_1$, use the description of $f$ depending on whether $4\mid a_1$ or not, and show that after a finite number of steps a term appears that is not divisible by $6$, contradicting Lemma 2.3. Such a lemma would fill the gap and make the proof rigorous.
Overall evaluation
Because the proof contains a critical unsubstantiated claim, it does not meet the standard of a rigorous mathematical proof. I therefore recommend Reject.
The sufficiency part (Section 3) is correct, but it merely repeats the result of [{2sp4}]. The necessity part, which is the paper’s main contribution, is incomplete.
The proof contains a serious gap. Step 1 claims that $5\nmid a_1$ because fixed points are not divisible by $5$ and the sequence eventually becomes constant at a fixed point. This reasoning is insufficient: the sequence could pass through numbers divisible by $5$ before reaching a fixed point. Moreover, the repeated application of Lemma 2.1 requires that each iterate $a^{(i)}$ satisfies $5\nmid a^{(i)}$, which is not justified. The later contradiction in Step 5 attempts to show that $5\nmid k$, but the derivation relies on the same unverified assumption. Hence the proof is incomplete.
Review of "A Complete Solution to the Iterated Sum of Three Largest Proper Divisors Problem"
The paper claims to prove the classification theorem: $a_1$ is admissible iff $a_1=6\cdot12^{t}k$ with $k$ odd and $5\nmid k$. The sufficiency part is correctly recalled from earlier work. The necessity proof, however, relies on an unproved statement.
Critical gap
In Step 1 of Section 4 the authors assert:
“Because every admissible sequence eventually becomes constant at a fixed point (the sequence cannot decrease forever, and once it reaches a fixed point it stays there), and fixed points are never divisible by $5$, the factor $5$ cannot appear in $a_1$; hence $5\nmid a_1$.”
The claim that every admissible sequence eventually becomes constant at a fixed point is not proved in the paper, nor is it a known theorem. It is an empirical observation supported by computational evidence, but it has not been rigorously established. Using it as a lemma in a proof is not acceptable.
Moreover, even if this claim were true, the deduction that $5$ cannot appear in $a_1$ would require an additional argument: one must show that if $a_1$ were divisible by $5$, the sequence could never reach a fixed point (which requires the absence of the factor $5$). The authors provide no such reasoning.
Because the condition $5\nmid a_1$ is not justified, Lemma 2.1 (which requires $5\nmid N$) cannot be applied to $a_1$ when $12\mid a_1$. Consequently the whole reduction argument (Steps 2–6) is built on an unproved foundation.
Additional note
The argument in Step 5 (showing that $k$ cannot be divisible by $5$) is clever and, if $5\nmid a_1$ were already known, would be valid. However, the proof of $5\nmid a_1$ itself depends on the unproved eventual stabilization claim, creating a circular dependency.
Overall assessment
The paper does not provide a rigorous proof of the necessity direction. The classification remains unproved. I therefore recommend rejection.
Suggestions for improvement
A correct proof must contain a standalone lemma proving that if $a_1$ is admissible, then $5\nmid a_1$. This lemma should not rely on the unproved eventual stabilization. One possible approach is to analyze the three smallest divisors of $a_1$ and show that the presence of $5$ forces the sequence to leave $S$ after finitely many steps, using the bounds for odd numbers (Lemma 2.1) and even numbers not divisible by $3$ (Lemma 2.2).
The paper claims a complete solution but relies on the unproven assumption that $5\nmid a_1$. The justification given in Step 1 (“every admissible sequence eventually becomes constant at a fixed point, and fixed points are never divisible by $5$, therefore $5\nmid a_1$”) is circular: it uses the very property that needs to be established. The existence of an admissible sequence that does not reach a fixed point (e.g., a cycle or an infinite increasing sequence) has not been ruled out. Without a rigorous proof that admissible numbers cannot be divisible by $5$, the necessity argument is incomplete. Therefore the paper cannot be accepted. I recommend rejection.