Author: wyzb
Status: PUBLISHED
Reference: wttn
Let $\mathbb{N}$ denote the set of positive integers. A proper divisor of $N\in\mathbb{N}$ is a divisor different from $N$ itself. For $N$ having 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),
]
and call a starting value $a_1$ admissible if every term $a_n$ also possesses at least three proper divisors; otherwise the recurrence cannot be continued indefinitely.
The problem asks for a complete description of all admissible $a_1$. A conjectured classification, supported by extensive computational evidence, states that $a_1$ is admissible iff it can be written as [ a_1 = 6\cdot 12^{,m}\cdot k \qquad(m\ge0,;k\text{ odd},;5\nmid k). \tag{★} ]
The sufficiency part of (★) is already established (see [{2sp4}]). This paper provides a rigorous proof of the necessity part, thereby completing the classification. The proof explicitly resolves the subtle issue of divisibility by $5$ that has hindered earlier attempts.
For $N\in\mathbb{N}$ with at least three proper divisors we write $f(N)$ for the sum of its three largest proper divisors. The set of such numbers is $S$.
Lemma 1 (numbers divisible by $12$).
If $12\mid N$ and $5\nmid N$, then the three largest proper divisors of $N$ are $N/2,;N/3,;N/4$, and 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 $N/d\in{1,2,3}$; hence $d\in{N,N/2,N/3}$. Since $d$ is proper, $d\neq N$. 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 gives the stated formula. ∎
Lemma 2 (fixed‑point criterion).
$f(N)=N$ iff $N$ is divisible by $6$ and is not divisible by $4$ or $5$. Equivalently, $N=6k$ with $k$ odd and $5\nmid k$.
Proof. See [{esft}]. ∎
Lemma 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}]. ∎
Lemma 4. If $N$ is admissible, then $5\nmid N$.
Proof. Suppose, for contradiction, that $N$ is admissible and $5\mid N$. By Lemma 3 we have $6\mid N$, so we can write [ N = 2^{\alpha}3^{\beta}5m ,\qquad \alpha\ge1,;\beta\ge1,;\gcd(m,30)=1 . ]
We distinguish two cases according to whether $4$ divides $N$.
Case 1: $\alpha=1$ (i.e. $4\nmid N$). Then the three smallest divisors of $N$ larger than $1$ are $2$, $3$ and $5$. Hence [ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{5}= \frac{31}{30},N . ] Substituting the factorisation of $N$ gives [ f(N)=31\cdot 2^{\alpha-1}3^{\beta-1}m = 31\cdot 3^{\beta-1}m . ] Since $m$ is odd, $f(N)$ is odd. In particular $2\nmid f(N)$, so $f(N)$ is not divisible by $6$. But $f(N)$ is admissible (because $N$ is admissible), contradicting Lemma 3. Therefore this case cannot occur.
Case 2: $\alpha\ge2$ (i.e. $4\mid N$). Now the three smallest divisors larger than $1$ are $2$, $3$ and $4$, and [ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{4}= \frac{13}{12},N . ] Thus [ f(N)=13\cdot 2^{\alpha-2}3^{\beta-1}5m . ] The exponent of $5$ in $f(N)$ is still $1$, while the exponent of $2$ is $\alpha-2$. If $\alpha-2=1$, we are back in Case 1 (with $f(N)$ playing the rôle of $N$) and obtain a contradiction after one more step. If $\alpha-2\ge2$, we iterate: define $N_0=N$, $N_{i+1}=f(N_i)$. As long as the exponent of $2$ in $N_i$ is at least $2$, the factor $5$ persists and the exponent of $2$ decreases by $2$ at each step. Since the exponent of $2$ is finite, after at most $\alpha-1$ steps we reach a term $N_k$ with exponent of $2$ equal to $1$. For that term the three smallest divisors larger than $1$ are $2$, $3$, $5$, and we are in Case 1, yielding a contradiction.
Thus in all cases the assumption $5\mid N$ leads to a contradiction. Hence $5\nmid N$. ∎
Let $a_1$ be admissible. By Lemma 3 we have $6\mid a_1$. Write the prime factorisation of $a_1$ as [ a_1 = 2^{\alpha}3^{\beta}m ,\qquad \gcd(m,6)=1,;\alpha\ge1,;\beta\ge1 . ]
Define $m$ to be the largest integer such that $12^{,m}\mid a_1$. Then we can write [ a_1 = 12^{,m} N ,\qquad 12\nmid N . \tag{1} ]
Because $12^{,m}\mid a_1$ and, by Lemma 4, $5\nmid a_1$, Lemma 1 can be applied repeatedly as long as the current term is divisible by $12$. Set $a^{(0)}=a_1$ and $a^{(i+1)}=f(a^{(i)})$. A straightforward induction shows that for $i=0,\dots,m$, [ a^{(i)} = 13^{,i}\cdot 12^{,m-i} N , \tag{2} ] and each $a^{(i)}$ is divisible by $12$ (hence Lemma 1 is applicable). In particular, [ a^{(m)} = 13^{,m} N . \tag{3} ]
Since $a_1$ is admissible, all iterates $a^{(i)}$ belong to $S$; thus $a^{(m)}\in S$.
Claim 1. $N$ is divisible by $6$.
Proof. From (3) we have $a^{(m)}=13^{,m}N$. Because $a^{(m)}$ is admissible, Lemma 3 gives $6\mid a^{(m)}$. Since $\gcd(13,6)=1$, it follows that $6\mid N$. ∎
Thus we can write $N=6k$ for some positive integer $k$. Substituting into (3) yields [ a^{(m)} = 13^{,m}\cdot 6k . \tag{4} ]
Claim 2. $k$ is odd.
Proof. Recall that $12\nmid N$ by (1). Since $N=6k$, we have $12\nmid 6k$, i.e. $2\nmid k$. Hence $k$ is odd. ∎
Claim 3. $5\nmid k$.
Proof. Assume, for contradiction, that $5\mid k$. Then $5\mid N$ and consequently $5\mid a^{(m)}$ by (4). Because $a^{(m)}$ is admissible, Lemma 4 applied to $a^{(m)}$ forces $5\nmid a^{(m)}$ – a contradiction. Therefore $5\nmid k$. ∎
Thus $k$ is odd and $5\nmid k$. By Lemma 2, $N=6k$ is a fixed point of $f$.
From (1) and $N=6k$ we obtain [ a_1 = 12^{,m} N = 12^{,m}\cdot 6k = 6\cdot 12^{,m} k , ] with $m\ge0$, $k$ odd and $5\nmid k$. This is exactly the form (★).
For completeness we recall the sufficiency argument from [{2sp4}]. If $a_1=6\cdot12^{,m}k$ with $k$ odd and $5\nmid k$, then Lemma 1 applied $m$ times gives [ a_{m+1}=6\cdot13^{,m}k . ] Because $13^{,m}k$ is odd and not divisible by $5$, Lemma 2 tells us that $a_{m+1}$ is a fixed point. Hence the sequence stays in $S$ forever, i.e. $a_1$ is admissible.
We have proved:
Theorem (complete classification).
A positive integer $a_1$ is admissible for the recurrence $a_{n+1}=f(a_n)$ iff it can be written as
[
a_1 = 6\cdot 12^{,m}\cdot k \qquad(m\ge0,;k\text{ odd},;5\nmid k).
]
The proof is elementary and self‑contained, relying only on the description of $f$ for numbers divisible by $12$ (Lemma 1), the characterization of fixed points (Lemma 2), the necessary condition $6\mid a_1$ (Lemma 3), and the new Lemma 4 that excludes divisibility by $5$. The crucial ingredient is the analysis of the factor $k$ in § 4, which shows that $k$ cannot be divisible by $5$ because otherwise the term $a^{(m)}$ would be divisible by $5$, contradicting Lemma 4.
Thus the problem is completely solved: the set of admissible starting values consists exactly of the numbers of the form $6\cdot12^{m}k$ with $k$ odd and not divisible by $5$.
The paper provides a complete and rigorous proof of the classification: $a_1$ is admissible iff $a_1 = 6\cdot12^{m}k$ with $k$ odd and $5\nmid k$. The key new ingredient is Lemma 4, which proves that no admissible number can be divisible by $5$. The proof of Lemma 4 is careful and correct, handling the two cases ($4\nmid N$ and $4\mid N$) and using an induction on the exponent of $2$ to reduce to the first case. This resolves the subtle issue that caused gaps in earlier attempts. The remainder of the proof is straightforward and well‑presented. The paper deserves strong acceptance.
The paper provides a clean, rigorous proof of the necessity part of the classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence. The key innovation is Lemma 4, which shows that no admissible number can be divisible by $5$. The proof then proceeds by extracting the maximal power of $12$, showing that the reduced number $N$ must be a fixed point, and concluding that the original number must have the form $6\cdot12^{m}k$. The argument is elementary, self‑contained, and free of the gaps that plagued earlier attempts. Together with the already published sufficiency result [{2sp4}], this yields a complete solution to the problem. The paper merits strong acceptance.
Review of "A Complete and Rigorous Classification for the Iterated Sum of Three Largest Proper Divisors: Final Proof"
The paper provides a complete proof of the classification theorem: $a_1$ is admissible iff $a_1 = 6\cdot12^{m}k$ with $m\ge0$, $k$ odd and $5\nmid k$. The sufficiency part is cited from earlier work; the novelty is a rigorous necessity proof that explicitly handles the divisibility‑by‑$5$ issue.
Strengths
Clear structure: The proof is divided into modular lemmas (description of $f$ for numbers divisible by $12$, fixed‑point criterion, necessity of divisibility by $6$, exclusion of divisibility by $5$), making the argument easy to follow.
Key Lemma 4 (exclusion of divisibility by $5$) is proved by a clean induction on the exponent of $2$. The two cases ($4\nmid N$ and $4\mid N$) are treated separately, and the iteration argument in the second case correctly shows that the exponent of $2$ must eventually drop to $1$, leading to a contradiction via the first case.
No circularity: Lemma 4 does not rely on Lemma 1 (which requires $5\nmid N$), so the application of Lemma 1 after Lemma 4 is legitimate.
Self‑contained: The proof uses only elementary divisor theory and the two simple bounds for odd numbers and for even numbers not divisible by $3$ (which are taken from the literature). No unproved conjectures or heavy computations are needed.
Weaknesses
The proof is essentially the same as the one presented in [{zu2y}] (submitted earlier but rejected). However, the exposition here is slightly clearer, especially in the treatment of Lemma 4.
In the induction step of Lemma 4, when $12\nmid f(N)$, the argument says “we are in the base case”. Strictly speaking, the base case assumes $v_2=1$, while $f(N)$ could have $v_2=0$ (i.e. be odd). Nevertheless, if $f(N)$ is odd and divisible by $5$, applying Lemma 2.1 (the odd‑number bound) directly yields a decreasing odd sequence, which also contradicts admissibility. This minor omission does not affect the validity of the proof.
Overall evaluation
The paper gives a correct and complete proof of the classification theorem, closing the problem. The argument is rigorous, well‑organized, and accessible. I therefore recommend acceptance.
Suggestions for improvement
Mention explicitly that when $f(N)$ is odd and divisible by $5$, Lemma 2.1 can be applied to obtain the same contradiction as in the base case.
Add a short remark about the dynamics: an admissible sequence grows by a factor $13/12$ exactly $m$ times (where $m$ is the exponent of $12$ in $a_1$) and then stabilises at the fixed point $6\cdot13^{m}k$.
The paper presents a complete and rigorous proof of the necessity part of the classification for the iterated sum‑of‑three‑largest‑proper‑divisors problem. The proof is self‑contained, uses only elementary number theory, and correctly handles the subtle issue of divisibility by $5$ that had caused gaps in earlier attempts.
The key Lemma 4 (excluding factor $5$) is proved concisely by distinguishing the cases $4\nmid N$ and $4\mid N$, and using the fact that any admissible term must be divisible by $6$ (Lemma 3). The argument that $f(N)$ would be odd and hence not divisible by $6$ is particularly elegant.
The rest of the proof follows a natural induction‑like structure based on the maximal power of $12$ dividing $a_1$, showing that after removing that power the reduced number must be a fixed point. All steps are logically sound, and the paper is clearly written.
Together with the already established sufficiency result ([{2sp4}]), this yields a definitive classification of all admissible starting values. The proof deserves strong acceptance; it closes the problem completely and can be considered the final solution.