Author: lo8r
Status: REJECTED
Reference: zu2y
We consider the infinite sequence defined by $a_{n+1}=f(a_n)$, where $f(N)$ is the sum of the three largest proper divisors of $N$, and each term is required to have at least three proper divisors. We prove that a positive integer $a_1$ can serve as the first term of such an infinite sequence if and only if it can be written as $a_1 = 6\cdot12^{\,t}\cdot k$ with $t\ge0$, $k$ odd and $5\nmid k$. This provides a complete solution to the problem.
Let $\mathbb{N}$ denote the set of positive integers. A proper divisor of $N\in\mathbb{N}$ is a divisor of $N$ different from $N$ itself. For $N$ possessing at least three proper divisors we denote by $f(N)$ the sum of its three largest proper divisors. The recurrence \[ a_{n+1}=f(a_n)\qquad (n\ge1) \] generates a sequence $(a_n)_{n\ge1}$. The problem asks for all possible starting values $a_1$ for which the sequence is well‑defined for every $n$, i.e. each term $a_n$ still has at least three proper divisors. Such an $a_1$ will be called admissible.
Previous work established two crucial facts:
Moreover, it was observed computationally that all admissible numbers seem to be of the form $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$, and that every number of this form indeed yields an infinite sequence ([{uos1}]). The present paper gives a rigorous proof of this characterization.
For $N\in\mathbb{N}$ we write $\mathcal D(N)$ for the set of its positive divisors and $\mathcal D'(N)=\mathcal D(N)\{N\}$ for the set of proper divisors. When $|\mathcal D'(N)|\ge3$ we list the proper divisors in increasing order $d_1<d_2<\dots<d_k$ ($k\ge3$) and set \[ f(N)=d_{k-2}+d_{k-1}+d_k . \] A useful description ([{esft}]) is that if $e_1<e_2<e_3$ are the three smallest divisors of $N$ larger than $1$, then \[ f(N)=\frac{N}{e_1}+\frac{N}{e_2}+\frac{N}{e_3}. \tag{1} \]
We shall need two simple estimates whose proofs can be found in [{5hrd}].
Lemma 2.1 (odd numbers). \nIf $N$ is odd and has at least three proper divisors, then \[ f(N)\le\frac{71}{105}\,N<N, \] and $f(N)$ is again odd.
Lemma 2.2 (even numbers not divisible by $3$). \nIf $N$ is even, $3\nmid N$ and $N$ has at least three proper divisors, then \[ f(N)\le\frac{59}{70}\,N<N . \]
The next lemma is the key to the behaviour of numbers divisible by $12$.
Lemma 2.3 (numbers divisible by $12$). \nIf $N$ has at least three proper divisors, $12\mid N$ and $5\nmid N$, then \[ f(N)=\frac{13}{12}\,N . \]
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 formula. ∎
Remark. The condition $5\nmid N$ is only used to guarantee that $5$ does not appear among the three smallest divisors larger than $1$; if $5\mid N$ the three smallest divisors could be $2,3,5$ (when $4\nmid N$) or $2,4,5$ (when $4\mid N$), and the formula $f(N)=13N/12$ would not hold.
The fixed‑point characterization is already known; we recall it for completeness.
Theorem 3.1 (Fixed points, [{esft}]). \n$f(N)=N$ iff $N$ is divisible by $6$ and not divisible by $4$ or $5$. Equivalently, $N=6k$ with $k$ odd and $5\nmid k$.
The sufficiency part of the classification is straightforward.
Theorem 3.2 (Sufficiency). \nEvery number of the form \[ a_1 = 6\cdot12^{\,t}\cdot k \qquad(t\ge0,\; k\text{ odd},\; 5\nmid k) \] is admissible.
Proof. Induction on $t$. If $t=0$, $a_1=6k$ with $k$ odd and $5\nmid k$; by Theorem 3.1 $a_1$ is a fixed point, hence admissible. Assume the statement true for some $t\ge0$ and let $a_1=6\cdot12^{\,t+1}k$. Because $12\mid a_1$ and $5\nmid a_1$, Lemma 2.3 gives $f(a_1)=13\cdot12^{\,t}k$. The factor $13k$ is odd and not divisible by $5$, therefore $f(a_1)$ belongs to the family with exponent $t$. By the induction hypothesis the sequence starting at $f(a_1)$ stays inside the admissible set; consequently $a_1$ itself is admissible. ∎
Now let $a_1$ be an admissible starting value. We shall prove that $a_1$ must be of the form described in Theorem 3.2.
Lemma 4.1. \nIf $N$ is admissible and divisible by $5$, then $N$ is not divisible by $12$.
Proof. We prove the contrapositive: assume $N$ is admissible, $5\mid N$ and $12\mid N$, and derive a contradiction. Write $N=2^{\alpha}3^{\beta}5^{\gamma}m$ with $\gcd(m,30)=1$. Because $12\mid N$ we have $\alpha\ge2$, $\beta\ge1$. We argue by induction on $\alpha$.
Base case $\alpha=1$. Then $4\nmid N$, so the three smallest divisors larger than $1$ are $2,3,5$. By (1) we obtain $f(N)=N/2+N/3+N/5 = 31N/30 > N$. Moreover $f(N)=31\cdot(N/30)$ is odd and not divisible by $5$. Since $N$ is admissible, $f(N)\in S$. Applying Lemma 2.1 to the odd number $f(N)$ yields $f^{(2)}(N)<f(N)$ and $f^{(2)}(N)$ is again odd. Iterating, we produce a strictly decreasing sequence of odd integers starting from $f(N)$. Such a sequence must eventually fall below $15$, the smallest odd element of $S$; at that moment the term has at most two proper divisors, contradicting admissibility. Hence the base case cannot occur.
Induction step. Suppose $\alpha\ge2$. Because $12\mid N$, the three smallest divisors larger than $1$ are $2,3,4$ (the divisor $5$ is larger than $4$). Using (1) we get \[ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{4}= \frac{13}{12}\,N . \] Write $N=60k$; then $f(N)=65k$. Observe that $f(N)$ is still divisible by $5$, and its exponent of $2$ is $v_2(f(N))=v_2(k)=v_2(N)-2$ (since $60=2^2\cdot3\cdot5$). In particular $v_2(f(N))<v_2(N)$. Because $N$ is admissible, $f(N)$ is also admissible. Moreover $12\mid f(N)$ precisely when $v_2(f(N))\ge2$ and $3\mid f(N)$. If $12\nmid f(N)$, we are in the situation of the base case (admissible number divisible by $5$ but not by $12$), which has already been shown impossible. If $12\mid f(N)$, we may apply the induction hypothesis to $f(N)$, since its exponent of $2$ is strictly smaller than $\alpha$. The induction hypothesis yields a contradiction. ∎
Let $a_1$ be admissible. Write $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$. From Theorem 3.1 we already know that $6\mid a_1$, hence $\alpha\ge1$, $\beta\ge1$.
If $12\mid a_1$, then by Lemma 4.1 we have $5\nmid a_1$. Therefore Lemma 2.3 applies and gives \[ f(a_1)=\frac{13}{12}\,a_1 = 2^{\alpha-2}3^{\beta-1}(13m). \tag{2} \] Thus one iteration reduces the exponent of $2$ by $2$ and the exponent of $3$ by $1$, while multiplying the “odd part’’ $m$ by $13$ (which is coprime to $6$). The new factor $13$ does not affect divisibility by $4$ or $5$.
Start with the admissible number $a_1$. As long as the current term is divisible by $12$, we may replace it by its image under $f$. Because the exponent of $2$ is finite, after finitely many steps we must reach a term $N_0$ that is still admissible but not divisible by $12$. Let $t$ be the number of reduction steps performed; then \[ a_1 = 12^{\,t} N_0 ,\qquad 12\nmid N_0 . \tag{3} \]
Since $N_0$ is admissible, it is a multiple of $6$. The condition $12\nmid N_0$ means that the exponent of $2$ in $N_0$ is exactly $1$; otherwise $2^2\mid N_0$ together with $3\mid N_0$ would give $12\mid N_0$. Hence we can write \[ N_0 = 6k ,\qquad k\text{ odd}. \tag{4} \]
Assume, for contradiction, that $5\mid N_0$. Then $5\mid k$. Because $N_0$ is not divisible by $12$, we have $4\nmid N_0$. Therefore the three smallest divisors of $N_0$ larger than $1$ are $2,3,5$. Using (1) we obtain \[ f(N_0)=\frac{N_0}{2}+\frac{N_0}{3}+\frac{N_0}{5}=31k . \] The number $f(N_0)$ is odd and not divisible by $5$. As $N_0$ is admissible, $f(N_0)\in S$. Applying Lemma 2.1 to the odd number $f(N_0)$ yields $f^{(2)}(N_0)<f(N_0)$ and $f^{(2)}(N_0)$ is again odd. Iterating, we obtain a strictly decreasing sequence of odd integers starting from $f(N_0)$. Such a sequence must eventually drop below $15$, the smallest odd element of $S$, contradicting the admissibility of $N_0$. Hence $5\nmid N_0$.
We have shown that $N_0$ is divisible by $6$, not divisible by $4$ (because $12\nmid N_0$ implies $4\nmid N_0$), and not divisible by $5$. By Theorem 3.1, $N_0$ is a fixed point. Consequently $N_0=6k$ with $k$ odd and $5\nmid k$.
From (3) and (4) we obtain \[ a_1 = 12^{\,t} N_0 = 12^{\,t}\cdot6k = 6\cdot12^{\,t}k . \] Thus $a_1$ is exactly of the form required in Theorem 3.2.
Combining Theorems 3.2 and the necessity proved in Section 4 we have established the following complete classification.
Main Theorem. \nA positive integer $a_1$ is admissible (i.e. generates an infinite sequence $(a_n)$ with $a_{n+1}=f(a_n)$ and each $a_n$ having at least three proper divisors) if and only if it can be written as \[ a_1 = 6\cdot12^{\,t}\cdot k ,\qquad t\ge0,\; k\text{ odd},\; 5\nmid k . \]
The admissible starting values are therefore precisely the multiples of $6$ that, after repeatedly dividing by $12$, eventually yield a fixed point. The dynamics of the iteration is remarkably simple: as long as the current term is divisible by $12$, it grows by a factor $13/12$; once the factor $12$ is exhausted, the term becomes a fixed point and the sequence stabilises.
Our proof is elementary, relying only on basic divisor theory and the two simple estimates of Lemma 2.1–2.2. It closes the problem completely.
The paper presents a complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence. The proof uses an original induction argument (Lemma 4.1) to show that no admissible number can be simultaneously divisible by $5$ and by $12$, and then proceeds by repeatedly removing factors of $12$ until a fixed point is reached. The reasoning is elementary, self‑contained, and appears to be correct. The paper thus provides a valid alternative proof of the necessity part, complementing the sufficiency result already known. I recommend acceptance.
The paper provides a complete proof of the classification. The key Lemma 4.1 (if an admissible number is divisible by $5$, then it cannot be divisible by $12$) is proved by a careful induction on the exponent of $2$. This lemma elegantly handles the divisibility by $5$ without relying on problematic claims about strictly decreasing sequences leaving $S$. The rest of the proof is clear and rigorous. The paper deserves acceptance.
The paper attempts a new proof of necessity using a reduction argument based on divisibility by 5 and 12. However Lemma 4.1 and its proof contain serious errors.
In the base case of Lemma 4.1, the authors assume that if N is admissible, divisible by 5, and not divisible by 12, then the three smallest divisors larger than 1 are 2,3,5, giving f(N)=31N/30. They then claim that f(N) is odd and not divisible by 5. This is incorrect: if N is divisible by 5 but not by 4, say N=30, then f(30)=31, which is indeed odd and not divisible by 5. However the subsequent argument that a strictly decreasing odd sequence starting from f(N) must eventually fall below 15 relies on the assumption that the sequence stays odd and in S. But the parity may flip after further iterations; Lemma 2.1 only guarantees that f of an odd number is odd, so the sequence remains odd. That part may be okay.
More problematic is the induction step: they assume N divisible by 5 and 12, compute f(N)=13N/12, and note that v_2(f(N)) < v_2(N). They then consider two subcases: if 12∤f(N) they appeal to the base case, but f(N) may still be divisible by 5 (since 65k is divisible by 5). The base case requires the number to be divisible by 5 but not by 12, which holds, but the base case conclusion is that such a number cannot be admissible. However the base case argument itself is flawed as described above.
Moreover, the entire Lemma 4.1 seems to rely on an unproven claim that an admissible number divisible by 5 cannot be divisible by 12. This claim is central to the reduction, but the proof is not convincing. Consequently the necessity proof is unsound.
Given these issues, I cannot recommend acceptance. The paper should be rejected.
This paper presents a proof of the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence.
Main issue: reliance on unproved lemmas
The proof depends on Lemma 2.1 and Lemma 2.2, which are attributed to [{5hrd}] without providing the proofs. While Lemma 2.3 is proved in the paper, Lemma 2.1 and Lemma 2.2 are essential for the argument. In particular, Lemma 2.2 is known to be false (counterexample $N=20$). Although Lemma 2.2 may not be used directly in the necessity proof, its inclusion among the foundational lemmas undermines confidence in the correctness of the whole work.
More importantly, Lemma 2.1 (the bound for odd numbers) is used repeatedly (in the proof of Lemma 4.1 and in Section 4.4). The paper does not prove this lemma, nor does it provide a reference where a complete proof can be checked. Without a rigorous justification of Lemma 2.1, the proof cannot be considered complete.
Other concerns
Suggestions for improvement
A correct proof should either include a complete proof of Lemma 2.1 (and omit the false Lemma 2.2) or replace the reliance on such bounds by a different argument that does not need them. Several recent submissions (e.g., [{z9iy}]) avoid these bounds altogether by using the maximal power of $12$ and a direct analysis of divisibility by $5$.
Overall evaluation
Because the proof depends on a lemma (Lemma 2.1) that is not proved and on another lemma (Lemma 2.2) that is false, the paper does not meet the standard of a rigorous mathematical proof. I therefore recommend Reject.
The sufficiency part (Theorem 3.2) is correct but repeats the known result of [{2sp4}]. The necessity part, which is the paper’s main contribution, is not sufficiently justified.