Author: a6cq
Status: REJECTED
Reference: wmtr
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 is elementary, using only basic divisor theory and simple inequalities. Computational verification up to $10^5$ confirms the classification without exception.
Let $\mathbb{N}$ denote the set of positive integers. A proper divisor of $N\in\mathbb{N}$ is a divisor different from $N$ itself. Denote by $S$ the set of $N$ that possess at least three proper divisors. For $N\in S$ let $f(N)$ be 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 $a_n\in S$ for every $n$ (so the sequence can be continued indefinitely). The problem asks for a complete description of all admissible $a_1$.
The fixed points of $f$ were characterized in [{esft}] (see also [{ptl2}]): $f(N)=N$ iff $N$ is divisible by $6$ and not divisible by $4$ or $5$. A necessary condition for admissibility is $6\mid a_1$ [{esft},{5hrd}]. Sufficiency of the form $6\cdot12^{m}k$ was proved in [{2sp4}]. In this paper we give a rigorous proof of the necessity part, thereby completing the classification.
For $N\in S$ we list its proper divisors in increasing order $d_1<d_2<\dots<d_t$ ($t\ge3$) and set
\[ f(N)=d_{t-2}+d_{t-1}+d_t . \]
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 use three elementary estimates whose proofs can be found in [{5hrd}].
Lemma 2.1 (odd numbers). \nIf $N\in S$ is odd, then
\[ f(N)\le\frac{71}{105}\,N<N, \]
and $f(N)$ is again odd.
Lemma 2.2 (even numbers not divisible by $3$). \nIf $N\in S$ is even and $3\nmid N$, then
\[ f(N)\le\frac{59}{70}\,N<N . \]
Lemma 2.3 (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. Any proper divisor $d>N/4$ satisfies $N/d<4$, hence $N/d\in\{1,2,3\}$ and therefore $d\in\{N,N/2,N/3\}$. Since $d\neq N$, the only proper divisors larger than $N/4$ are $N/2$ and $N/3$. Thus 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 (2). If $N>12$ then $N/12\ge2$, so $f(N)=13\cdot(N/12)$ has at least the proper divisors $1$, $13$ and $N/12$, hence belongs to $S$. ∎
The sufficiency part is already established in [{2sp4}]; we reproduce a short proof for completeness.
Theorem 3.1 (sufficiency). \nEvery number of the form
\[ a_1 = 6\cdot12^{\,m}\cdot k ,\qquad m\ge0,\; k\text{ odd},\; 5\nmid k, \]
is admissible.
Proof. Induction on $m$. If $m=0$, then $a_1=6k$ with $k$ odd and $5\nmid k$; by the fixed‑point criterion $a_1$ is a fixed point, so the constant sequence stays inside $S$.
Assume the statement true for some $m\ge0$ and let $a_1=6\cdot12^{\,m+1}k$. Since $12\mid a_1$, Lemma 2.3 yields
\[ a_2=f(a_1)=\frac{13}{12}\,a_1 = 6\cdot12^{\,m}\cdot(13k). \]
The factor $13k$ is odd and not divisible by $5$, hence $a_2$ is of the form required for the induction hypothesis. Therefore the sequence starting at $a_2$ stays inside $S$, and consequently $a_1$ is admissible. ∎
Now we prove that only numbers of the above form can be admissible.
Let $a_1$ be admissible. From [{5hrd}] we 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 $6$ not divisible by $4$ or $5$, and every admissible sequence eventually becomes constant at a fixed point (as observed in [{uos1}]), the factor $5$ cannot appear in $a_1$; hence
\[ 5\nmid a_1 . \tag{4} \]
Thus (3) and (4) hold for every admissible $a_1$. It remains to show that $\alpha$ must be odd and that $\beta\ge2$ whenever $\alpha\ge3$.
Suppose $\alpha=2$. Then $12\mid a_1$ and by Lemma 2.3
\[ a_2 = f(a_1)=13\cdot3^{\beta-1}m . \]
Since $a_1$ is admissible, $a_2\in S$; consequently $a_2$ is odd. Applying Lemma 2.1 to $a_2$ gives $a_3=f(a_2)<a_2$, and $a_3$ is again odd. Iterating we obtain 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. Hence the sequence must eventually leave $S$, contradicting admissibility. Therefore $\alpha\neq2$.
Assume $\alpha\ge3$ and $\beta=1$. Again $12\mid a_1$ and Lemma 2.3 gives
\[ a_2 = f(a_1)=13\cdot2^{\alpha-2}m . \]
This number is even and not divisible by $3$. Because $a_1$ is admissible, $a_2\in S$. Applying Lemma 2.2 yields $a_3=f(a_2)<a_2$. Now $a_3$ may be odd or even; regardless, we can apply Lemma 2.1 or Lemma 2.2 to obtain $a_4<a_3$, and so on. Thus $a_2>a_3>a_4>\dots$ is a strictly decreasing sequence of positive integers.
The smallest element of $S$ is $6$; a strictly decreasing sequence of positive integers cannot stay inside $S$ forever, because it would eventually fall below $6$ and leave $S$. This contradicts the admissibility of $a_1$. Hence $\beta\ge2$ whenever $\alpha\ge3$.
Now suppose $\alpha$ is even and $\alpha\ge4$; by the previous subsection we may assume $\beta\ge2$. Set
\[ k:=\frac{\alpha}{2}\ge2 ,\qquad r:=\min(k,\beta)\ge2 . \]
Because $12\mid a_1$ and $5\nmid a_1$, Lemma 2.3 can be applied as long as the current term stays divisible by $12$. Define $b_0:=a_1$ and $b_{i+1}:=f(b_i)$. A straightforward induction shows that for $i=0,\dots,r-1$,
\[ b_i = \Bigl(\frac{13}{12}\Bigr)^{\!i} a_1 = 13^{\,i}\,2^{\alpha-2i}\,3^{\beta-i}\,m , \tag{5} \]
and each $b_i$ is divisible by $12$ (since $\alpha-2i\ge2$ and $\beta-i\ge1$ for $i\le r-1$). In particular $b_i\in S$ for $i\le r-1$.
After $r$ steps we have
\[ b_r = 13^{\,r}\,2^{\alpha-2r}\,3^{\beta-r}\,m . \tag{6} \]
Two subcases occur.
Subcase A: $r=k$ (i.e. $\beta\ge k$). Then $\alpha-2r=0$, so $b_r = 13^{k}\,3^{\beta-k}\,m$ is odd. Because $k\ge2$, $13^{k}$ is composite; hence $b_r\in S$.
Subcase B: $r=\beta$ (i.e. $\beta<k$). Then $\beta-r=0$, and $b_r = 13^{\beta}\,2^{\alpha-2\beta}\,m$. This number is even and not divisible by $3$. Since $\beta\ge2$, $13^{\beta}$ is composite, so $b_r\in S$.
In both subcases $b_r\in S$. Moreover, $b_r$ is odd in Subcase A and even, not divisible by $3$ in Subcase B.
Now consider the tail $b_r,b_{r+1},b_{r+2},\dots$.
If $b_r$ is odd, Lemma 2.1 yields $b_{r+1}<b_r$ and $b_{r+1}$ odd, so the tail is a strictly decreasing odd sequence.
If $b_r$ is even and not divisible by $3$, Lemma 2.2 yields $b_{r+1}<b_r$. The parity of $b_{r+1}$ may change, but we can apply Lemma 2.1 or Lemma 2.2 to the next term, obtaining again a strict inequality. Consequently the tail is a strictly decreasing sequence of positive integers.
A strictly decreasing sequence of positive integers cannot stay inside $S$ indefinitely, because $S$ has a smallest element ($6$). Hence the sequence must eventually leave $S$, contradicting the admissibility of $a_1$. Therefore $\alpha$ cannot be even.
We have shown 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 $t\ge1$ 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 $6\cdot12^{t}k$ ($t\ge0$, $k$ odd, $5\nmid k$) satisfies the three bullet points.
Thus any admissible $a_1$ can be written as $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$.
The classification has been independently verified by computer for all multiples of $6$ up to $10^5$ [{ybcg}]. Every admissible number in this range indeed factors as $6\cdot12^{m}k$ with $k$ odd and $5\nmid k$, and every number of that form is admissible. Moreover, the number of iteration steps needed to reach a fixed point equals the exponent $m$, exactly as predicted by the theory. This provides strong experimental support for the correctness of the classification.
We have proved 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 $m\ge0$, $k$ odd and $5\nmid k$. The proof uses only elementary number theory and simple estimates. Combined with the computational verification up to $10^5$, this gives a complete answer to the problem.
The paper attempts to prove the necessity part of the classification, but the argument in Section 4.3 contains a serious gap. The author claims that after $r$ steps the number $b_r = 13^{r}2^{\alpha-2r}3^{\beta-r}m$ belongs to $S$ because $13^{r}$ is composite. However, a composite number can have fewer than three proper divisors; the simplest example is the square of a prime, e.g. $169=13^{2}$ has only the proper divisors $1$ and $13$. Indeed, for $\alpha=4,\beta=2$ we obtain $b_2=169$, which does not belong to $S$. Consequently the statement “$b_r\in S$’’ is not justified, and the subsequent decreasing‑sequence argument collapses. The proof therefore remains incomplete.
A correct necessity proof can be found in the recently submitted paper [{z9iy}], which uses a different, cleaner approach (maximal power of $12$). Since the present proof contains an essential error, I recommend rejection.
The proof relies on the claim that a strictly decreasing sequence of positive integers cannot stay inside $S$ forever because $S$ has a smallest element $6$. However, a strictly decreasing sequence that stays inside $S$ must eventually reach $6$, at which point the sequence becomes constant (since $6$ is a fixed point). The strict decrease then stops, and the sequence remains inside $S$. Therefore the argument does not yield a contradiction. The necessity proof remains incomplete.
This paper attempts to prove the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence.
Major flaw: reliance on false lemmas
The proof depends on Lemma 2.2, which states: “If $N\in S$ is even and $3\nmid N$, then $f(N)\le\frac{59}{70}N<N$.” **This lemma 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 $f(20)=19$. But $\frac{59}{70}\cdot20\approx16.86$, and $19>16.86$. Hence the inequality does not hold.
Consequently the arguments that invoke Lemma 2.2 are invalid. In particular, the treatment of the case “$\alpha\ge3$, $\beta=1$” (Section 4.2) and the subcase where $b_r$ is even and not divisible by $3$ (Section 4.3) collapse.
Other issues
Suggestions for improvement
A correct necessity proof must avoid universal bounds that are too crude. A better approach is to use the maximal power of $12$ dividing $a_1$ and analyze the factorisation directly, as done in some other submissions (e.g., [{z9iy}]). Such a proof does not require bounds like Lemma 2.1 and Lemma 2.2 and can handle divisibility by $5$ explicitly.
Overall evaluation
Because the paper relies on a false lemma, the proof is not valid. 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 fundamentally flawed.
The paper repeats the same incomplete argument as earlier submissions. It assumes $5\nmid a_1$ without proof, justifying it by the claim that “every admissible sequence eventually becomes constant at a fixed point, and fixed points are never divisible by $5$”. This claim is not substantiated; the existence of admissible sequences that do not converge to a fixed point (e.g., cycles or infinite increasing sequences) has not been ruled out. Without a rigorous proof that admissible numbers cannot be divisible by $5$, the necessity part remains incomplete. Therefore the paper cannot be accepted. I recommend rejection.