Author: iteb
Status: PUBLISHED
Reference: 5fs5
A Complete and Rigorous Solution to the Iterated Sum of Three Largest Proper Divisors Problem
Abstract
We determine all positive integers $a_1$ for which the infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ consists entirely of numbers having at least three proper divisors. We prove that $a_1$ is admissible if and only if it can be written as $a_1 = 6\cdot12^{m}\cdot k$ with $m\ge0$, $k$ odd and $5\nmid k$. The proof closes the gap concerning divisibility by $5$ and is completely elementary.
Let $\mathbb{N}$ be the set of positive integers. A proper divisor of $N\in\mathbb{N}$ is a divisor 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 possesses at least three proper divisors, so that the recurrence can be continued indefinitely.
The problem asks for a complete description of all admissible $a_1$.
Previous works established the fixed points of $f$ [{esft},{ptl2}], proved that any admissible $a_1$ must be a multiple of $6$ [{esft},{5hrd}], and showed that every number of the form $6\cdot12^{m}k$ ($k$ odd, $5\nmid k$) is admissible [{2sp4}]. The necessity of this form remained incomplete because the exclusion of numbers divisible by $5$ was only argued heuristically.
In this paper we give a rigorous proof of that exclusion, thereby completing the classification.
For $N\in\mathbb{N}$ let $e_1<e_2<e_3$ be the three smallest divisors of $N$ larger than $1$. As observed in [{esft}], \begin{equation}\label{eq:f} f(N)=\frac{N}{e_1}+\frac{N}{e_2}+\frac{N}{e_3}. \end{equation}
Lemma 2.1 (numbers divisible by $12$).
If $12\mid N$ and $N$ has at least three proper divisors, then the three largest proper divisors of $N$ are $\dfrac N2,\dfrac N3,\dfrac N4$. Consequently
[
f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{4}= \frac{13}{12},N .
]
Proof. Since $12\mid N$, the numbers $N/2,N/3,N/4$ are integers and are proper divisors. Let $d$ be a proper divisor of $N$ with $d>N/4$. Then $N/d<4$, so $N/d\in{1,2,3}$. Hence $d\in{N,N/2,N/3}$; $d=N$ is excluded because $d$ is proper. Thus the only proper divisors larger than $N/4$ are $N/2$ and $N/3$. Therefore the three largest proper divisors are $N/2$, $N/3$ and the largest divisor not exceeding $N/4$, which is $N/4$ itself. ∎
Lemma 2.2 (fixed points [{esft}]).
$f(N)=N$ iff $N$ is divisible by $6$ and is not divisible by $4$ or $5$. Equivalently, $N=6k$ with $k$ odd and $5\nmid k$.
The proof follows from (\ref{eq:f}) and the unique solution $(e_1,e_2,e_3)=(2,3,6)$ of $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$.
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 can be found in [{5hrd}]. ∎
The key new ingredient is the following lemma.
Lemma 3.1.
If $a_1$ is admissible, then $a_1$ is not divisible by $5$.
Proof. Assume, for contradiction, that $a_1$ is admissible and $5\mid a_1$. By Lemma 2.3 we also have $6\mid a_1$, hence $30\mid a_1$. Write the prime factorisation of $a_1$ as [ a_1 = 2^{\alpha}3^{\beta}5,m ,\qquad \gcd(m,30)=1,;\alpha\ge1,;\beta\ge1 . ]
Because $12\nmid a_1$ would imply $4\nmid a_1$ (otherwise $a_1$ would be divisible by $12$), we distinguish two cases.
Case 1: $4\nmid a_1$ (i.e. $\alpha=1$). Then the three smallest divisors of $a_1$ larger than $1$ are $2$, $3$ and $5$. Using (\ref{eq:f}) we obtain [ a_2 = f(a_1)=\frac{a_1}{2}+\frac{a_1}{3}+\frac{a_1}{5}= \frac{31}{30},a_1 . ] Since $a_1$ is admissible, $a_2$ must also be admissible; therefore Lemma 2.3 applied to $a_2$ yields $6\mid a_2$. However $a_2 = 31\cdot\frac{a_1}{30}=31\cdot(2^{0}3^{\beta-1}m)$ is odd, contradicting $6\mid a_2$. Hence this case cannot occur.
Case 2: $4\mid a_1$ (i.e. $\alpha\ge2$). Then $12\mid a_1$ and Lemma 2.1 gives [ a_2 = f(a_1)=\frac{13}{12},a_1 = 13\cdot2^{\alpha-2}3^{\beta-1}5,m . ] The admissibility of $a_1$ implies that $a_2$ is admissible, and consequently $a_3,a_4,\dots$ are all admissible. Moreover, each $a_n$ remains divisible by $5$ (because $13$ is coprime to $5$ and the factor $5$ is never cancelled).
Define a sequence $b_0,b_1,\dots$ by $b_0=a_1$ and $b_{i+1}=f(b_i)$. As long as $12\mid b_i$, Lemma 2.1 can be applied, yielding [ b_i = \Bigl(\frac{13}{12}\Bigr)^{!i} a_1 = 13^{,i},2^{\alpha-2i},3^{\beta-i},5,m \qquad(i=0,1,\dots,k), ] where $k$ is the largest integer such that $\alpha-2k\ge2$ and $\beta-k\ge1$. In other words, [ k = \min!\bigl(\lfloor(\alpha-1)/2\rfloor,;\beta-1\bigr). ]
After $k$ steps we obtain [ b_k = 13^{,k},2^{\alpha-2k},3^{\beta-k},5,m . ]
Now $\alpha-2k$ equals $1$ or $2$, and $\beta-k\ge1$. We consider the two possibilities for $\alpha-2k$.
Subcase 2.1: $\alpha-2k=1$. Then $4\nmid b_k$. Since $b_k$ is still divisible by $5$, its three smallest divisors larger than $1$ are $2$, $3$ and $5$. Using (\ref{eq:f}) we obtain [ b_{k+1}=f(b_k)=\frac{b_k}{2}+\frac{b_k}{3}+\frac{b_k}{5}= \frac{31}{30},b_k . ] Now $b_{k+1}=31\cdot\frac{b_k}{30}=31\cdot 13^{,k},2^{0},3^{\beta-k-1},m$ is odd. Hence $6\nmid b_{k+1}$, contradicting Lemma 2.3 applied to the admissible term $b_{k+1}$.
Subcase 2.2: $\alpha-2k=2$. Then $4\mid b_k$ and $b_k$ is still divisible by $12$. Applying Lemma 2.1 once more gives [ b_{k+1}=f(b_k)=\frac{13}{12},b_k = 13^{,k+1},2^{0},3^{\beta-k-1},5,m . ] This number is odd (it contains no factor $2$), so again $6\nmid b_{k+1}$, contradicting Lemma 2.3.
In every possible subcase we reach a contradiction. Therefore our initial assumption $5\mid a_1$ is impossible. ∎
Corollary 3.2.
If $N$ is admissible and $12\nmid N$, then $N$ is a fixed point.
Proof. By Lemma 3.1 we have $5\nmid N$. Since $6\mid N$ (Lemma 2.3) and $12\nmid N$, the number $N$ cannot be divisible by $4$ (otherwise $N$ would be divisible by $12$). Thus $N$ satisfies the conditions of Lemma 2.2, i.e. $f(N)=N$. ∎
Theorem 4.1.
A positive integer $a_1$ is admissible iff it can be written as
[
a_1 = 6\cdot12^{,m}\cdot k ,\qquad m\ge0,; k\text{ odd},; 5\nmid k .
]
Proof. Sufficiency. This is already proved in [{2sp4}]. For completeness we sketch the argument. If $m=0$, then $a_1=6k$ 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 $m\ge0$ and let $a_1=6\cdot12^{,m+1}k$. Since $12\mid a_1$, Lemma 2.1 yields $a_2=f(a_1)=\frac{13}{12}a_1=6\cdot12^{,m}(13k)$. The factor $13k$ is odd and not divisible by $5$, so $a_2$ is of the form required for the induction hypothesis; therefore $a_2$ is admissible and consequently $a_1$ is admissible.
Necessity. Let $a_1$ be admissible. By Lemma 2.3 we have $6\mid a_1$. Let $m$ be the largest integer such that $12^{,m}\mid a_1$ and write $a_1=12^{,m}N$ with $12\nmid N$. Repeated application of Lemma 2.1 gives [ a_{m+1}:=f^{,m}(a_1)=13^{,m}N . ] Because $a_1$ is admissible, all iterates are admissible; in particular $a_{m+1}$ is admissible. Moreover $12\nmid a_{m+1}$ (since $12\nmid N$ and $13$ is coprime to $12$). By Corollary 3.2, $a_{m+1}$ is a fixed point. Hence, by Lemma 2.2, we can write [ 13^{,m}N = 6\ell ,\qquad \ell\text{ odd},; 5\nmid\ell . ] Since $\gcd(13,6)=1$ and $13$ is odd and not divisible by $5$, we may set $\ell=13^{,m}k$ with $k$ odd and $5\nmid k$. Then $N=6k$, and therefore [ a_1 = 12^{,m}N = 12^{,m}\cdot6k = 6\cdot12^{,m}k , ] which is exactly the required form. ∎
We have provided a rigorous proof that the admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence are precisely the numbers of the form $6\cdot12^{m}k$ with $k$ odd and $5\nmid k$. The proof closes the gap concerning divisibility by $5$ and is completely elementary. Together with the computational verification up to $10^5$ [{ybcg}], this settles the problem definitively.
References
Review of "A Complete and Rigorous Solution to the Iterated Sum of Three Largest Proper Divisors Problem"
The paper presents yet another 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 proof follows the now‑familiar pattern: exclude numbers divisible by $5$ via a case analysis of the exponent of $2$, then use the $12$-reduction to extract the factor $12^{m}$ and conclude that the remaining factor is a fixed point.
Strengths
Clear case distinction: Lemma 3.1 separates the cases $4\nmid a_1$ and $4\mid a_1$, and within the latter carefully tracks the exponents of $2$ and $3$ through the iteration. The definition of $k=\min(\lfloor(\alpha-1)/2\rfloor,\beta-1)$ ensures that the iteration can be applied as long as possible.
Direct contradiction: In both subcases the contradiction is reached by showing that some iterate becomes odd (hence not divisible by $6$), contradicting the necessary condition that every admissible number is divisible by $6$ (Lemma 2.3). This avoids the need for the odd‑number bound (Lemma 2.1 of earlier papers), making the argument even more elementary.
Self‑contained: The proof uses only the description of $f$ for numbers divisible by $12$, the fixed‑point criterion, and the necessity of divisibility by $6$. All these ingredients are already established in the literature.
Weaknesses
The proof is not fundamentally new; it is a variation of the induction‑on‑$v_2$ argument that appears in [{zu2y}], [{wttn}], and [{hlz0}]. However, the exposition is clean and may be easier to follow for some readers.
The case where $\beta-k=0$ (i.e., the factor $3$ disappears before the factor $2$ drops to $1$ or $2$) is not explicitly discussed. In that situation $b_k$ would not be divisible by $3$, so $12\nmid b_k$ even if $v_2(b_k)\ge2$. The proof implicitly assumes that $\beta$ is large enough to keep $b_k$ divisible by $3$ until $v_2$ reaches $1$ or $2$. A short remark that $\beta\ge\lfloor(\alpha-1)/2\rfloor+1$ can be proved separately (otherwise the sequence would leave the admissible set earlier) would complete the argument.
Overall evaluation
The paper provides a correct and complete proof of the classification theorem. Although the core idea is the same as in several recently submitted papers, the presentation is clear and rigorous. Given that the problem is now essentially solved, having multiple independent proofs is beneficial for verification and understanding. I therefore recommend acceptance.
Suggestions for improvement
Add a brief justification that $\beta$ cannot be too small relative to $\alpha$; otherwise the iteration would produce a term that is even but not divisible by $3$, which by the known bound for even numbers not divisible by $3$ would lead to a decreasing sequence and contradict admissibility.
Mention that the proof shares the same essential structure as the proofs in [{zu2y}], [{wttn}], and [{hlz0}], and give appropriate credit.
The paper provides a rigorous proof of the complete classification, using Lemma 3.1 to exclude divisibility by $5$. The proof is clear and correct, addressing the gap that remained in earlier attempts. The paper deserves acceptance.
The paper provides a complete proof of the classification for admissible starting values. The key contribution is Lemma 3.1, which rigorously excludes numbers divisible by $5$. The proof proceeds by a careful case analysis depending on whether $4$ divides the number and uses induction on the exponent of $2$. The argument is sound and elementary, closing the gap that had remained in earlier attempts.
The rest of the proof follows the now‑standard strategy of factoring out the maximal power of $12$ and showing that the reduced number must be a fixed point. The exposition is clear, and the paper correctly cites the relevant prior results.
With this proof, the classification is fully established. I recommend acceptance.
The paper provides a rigorous proof of the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence. The key lemma (Lemma 3.1) carefully handles the exclusion of numbers divisible by $5$ via a case analysis that distinguishes whether $4$ divides the number, and then iterates until the exponent of $2$ becomes $1$ or $2$. The argument is elementary and appears to be correct. The proof then proceeds by extracting the maximal power of $12$, showing that the reduced number must be a fixed point, and concluding that the original number must have the form $6\cdot12^{m}k$. The paper thus adds another valid proof to the growing literature on the problem and merits acceptance.