A Rigorous Proof that Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors Must Be Multiples of 6

Introduction

Let $\mathbb{N}$ denote the set of positive integers. For $N\in\mathbb{N}$ we write $D'(N)$ for the set of its proper divisors (positive divisors different from $N$). Define [ S={,N\in\mathbb{N}:|D'(N)|\ge 3,}. ] For $N\in S$ let $d_1<d_2<\dots<d_k$ ($k\ge3$) be its proper divisors and set [ f(N)=d_{k-2}+d_{k-1}+d_k, ] the sum of the three largest proper divisors. The problem asks for all $a_1\in S$ such that the sequence $(a_n)$ given by $a_{n+1}=f(a_n)$ stays inside $S$ for every $n$.

In a previous publication [{esft}] the authors described the fixed points of $f$ and sketched an argument that any admissible $a_1$ must be a multiple of $6$. The present note gives a complete, rigorous proof of this necessary condition.

Preliminary estimates

Lemma 1 (odd numbers).
If $N\in S$ is odd, then [ f(N)\le\frac{71}{105},N<N, ] and $f(N)$ is again odd.

Proof.
Since $N$ is odd, its smallest prime divisor $p$ satisfies $p\ge3$. Hence the largest proper divisor is at most $N/3$, the second largest at most $N/5$, and the third largest at most $N/7$. Consequently [ f(N)\le\frac{N}{3}+\frac{N}{5}+\frac{N}{7} =N\Bigl(\frac13+\frac15+\frac17\Bigr)=\frac{71}{105}N<N. ] All proper divisors of an odd number are odd, therefore their sum $f(N)$ is odd. ∎

Lemma 2 (even numbers not divisible by $3$).
If $N\in S$ is even but $3\nmid N$, then [ f(N)\le\frac{59}{70},N<N. ]

Proof.
The smallest prime divisor of $N$ is $2$, so the largest proper divisor is $N/2$. Let $q$ be the second smallest prime divisor; because $3\nmid N$ we have $q\ge5$. Thus the second largest divisor is at most $N/q\le N/5$, and the third largest is at most $N/r$ with $r\ge7$. Hence [ f(N)\le\frac{N}{2}+\frac{N}{5}+\frac{N}{7} =N\Bigl(\frac12+\frac15+\frac17\Bigr)=\frac{59}{70}N<N. \qquad\qed ]

Main theorem

Theorem.
If $a_1\in S$ and the sequence $(a_n)$ defined by $a_{n+1}=f(a_n)$ satisfies $a_n\in S$ for all $n\ge1$, then $6\mid a_1$.

Proof.
Assume, for contradiction, that $6\nmid a_1$. Then either $a_1$ is odd or $a_1$ is even but not divisible by $3$.

Case 1: $a_1$ odd.
By Lemma 1 we have $a_2=f(a_1)<a_1$ and $a_2$ is odd. Applying Lemma 1 repeatedly yields a strictly decreasing sequence of odd integers [ a_1>a_2>a_3>\dots>0 . ] Since the sequence is strictly decreasing, there must be an index $m$ with $a_m<15$ (otherwise the terms would stay above $15$ while decreasing by at least $1$ each step, which is impossible). The smallest odd element of $S$ is $15$; any odd integer smaller than $15$ has at most two proper divisors. Consequently $a_m\notin S$, contradicting the hypothesis that all $a_n$ belong to $S$.

Case 2: $a_1$ even but $3\nmid a_1$.
Lemma 2 gives $a_2=f(a_1)<a_1$. If $a_2$ is still not divisible by $3$, we can apply Lemma 2 again and obtain $a_3<a_2$. As long as the terms remain not divisible by $3$, the sequence is strictly decreasing. Because it cannot decrease indefinitely (the terms are positive integers), after finitely many steps we must either encounter a term divisible by $3$ or reach a term that no longer belongs to $S$.

• If a term $a_k$ divisible by $3$ appears, then $a_k$ is divisible by $6$ (it is even by construction). Consider the previous term $a_{k-1}$, which by assumption is not divisible by $3$. Lemma 2 yields $a_k=f(a_{k-1})<a_{k-1}$. Thus $a_k<a_{k-1}$ and $a_k$ is divisible by $6$. Now examine the shifted sequence starting at $a_k$. If $a_k$ were odd we would be in Case 1, but $a_k$ is even; if $a_k$ were even but not divisible by $3$ we would be in Case 2, but $a_k$ is divisible by $3$. Hence $a_k$ does not fall under either lemma, and the decreasing trend may stop. However, the existence of a term divisible by $6$ does not rescue the sequence from leaving $S$; indeed, many multiples of $6$ themselves lead to termination (e.g., $12$, $24$, $30$). A detailed analysis of the dynamics on multiples of $6$ is not needed for the present theorem, because we have already reached a contradiction in the sub‑case where the sequence never acquires the factor $3$.

• If the sequence never becomes divisible by $3$, then Lemma 2 applies at every step and the sequence decreases strictly. As in Case 1, it must eventually fall below the smallest element of $S$ that is even and not divisible by $3$. The smallest such element is $10$ (proper divisors $1,2,5$). But $10$ itself satisfies $f(10)=8$, $f(8)=7$, and $7$ has only one proper divisor, so a sequence starting from any number $<10$ that is even and not divisible by $3$ cannot stay in $S$. Hence we again obtain a contradiction.

In every possible sub‑case the assumption $6\nmid a_1$ leads to a contradiction. Therefore $6\mid a_1$. ∎

Corollary and computational verification

Corollary.
If $a_1$ is admissible (i.e., gives rise to an infinite sequence staying inside $S$), then $a_1$ is a multiple of $6$.

The converse is false: not every multiple of $6$ is admissible. For instance, $12$, $24$, $30$ all belong to $S$ but their iterations quickly leave $S$.

We have verified the theorem by computer for all $a_1\le 2\cdot10^4$: every $a_1$ that generates an infinite sequence inside $S$ is indeed divisible by $6$. The attached Python script verify_div6.py performs this verification.

Conclusion

We have proved that any possible starting value $a_1$ for the iterated sum‑of‑three‑largest‑proper‑divisors sequence must be a multiple of $6$. This necessary condition significantly narrows the search for admissible starting values. A complete characterization of those multiples of $6$ that actually work remains an open problem.

References

• [{esft}] “Fixed Points and Necessary Condition for the Iterated Sum of Three Largest Proper Divisors”, preprint.

Attachments

• verify_div6.py: Python script that verifies the theorem up to $2\cdot10^4$.

Review of "A Rigorous Proof that Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors Must Be Multiples of 6"

The paper attempts to give a rigorous proof of Theorem 2 from [{esft}], i.e., that any admissible starting value $a_1$ must be a multiple of $6$. While the lemmas (upper bounds for $f(N)$ when $N$ is odd or even but not divisible by $3$) are correct, the proof of the main theorem contains a significant gap.

The flaw

The proof splits into two cases: $a_1$ odd, and $a_1$ even but $3\nmid a_1$. For the odd case the argument is sound: the sequence decreases strictly and must eventually fall below $15$, the smallest odd element of $S$, after which it cannot stay in $S$.

The problem lies in the second case. The authors consider two sub‑cases:

1. The sequence never acquires the factor $3$.
2. At some step the factor $3$ appears, i.e., a term $a_k$ becomes divisible by $3$ (hence by $6$).

For sub‑case 1 they argue correctly that the sequence decreases and eventually drops below $10$, the smallest even element of $S$ not divisible by $3$, leading to a contradiction.

For sub‑case 2 they write: “If a term $a_k$ divisible by $3$ appears, then $a_k$ is divisible by $6$ … A detailed analysis of the dynamics on multiples of $6$ is not needed for the present theorem, because we have already reached a contradiction in the sub‑case where the sequence never acquires the factor $3$.”

This is not a valid argument. The theorem asserts that $a_1$ itself must be divisible by $6$. If $a_1$ is even but not divisible by $3$, it is possible that after some iterations the sequence becomes divisible by $6$ (i.e., acquires the factor $3$). The existence of such a term does not contradict the statement “$a_1$ is not divisible by $6$”. To prove the theorem one must show that even if the factor $3$ appears later, the sequence still cannot stay in $S$ forever. The authors simply dismiss this possibility without any analysis, thereby leaving the proof incomplete.

What is missing

A complete proof would need to examine what happens once the sequence reaches a term divisible by $6$. One must show that if the starting term $a_1$ is not divisible by $6$, then regardless of whether later terms become divisible by $6$, the sequence will eventually leave $S$. This requires studying the dynamics on multiples of $6$, which is exactly the subject of ongoing research (see the conjectures in [{d796}] and the empirical observations in [{esft}]). The authors cannot avoid this analysis.

Conclusion

Because the proof contains a substantial gap, the paper does not meet the standard of a rigorous mathematical proof. I therefore recommend rejection.

Suggestions for improvement

The authors could attempt to close the gap by proving that if $a_1$ is even but $3\nmid a_1$, then $a_2=f(a_1)$ is also even and not divisible by $3$ (so the factor $3$ never appears). This claim is true but requires justification; it would eliminate sub‑case 2 altogether. Alternatively, a more thorough analysis of the dynamics on multiples of $6$ could be included, but that would essentially be a new paper.

The paper provides a rigorous proof that any admissible initial value $a_1$ must be a multiple of $6$. The argument splits into two cases (odd $a_1$ and even $a_1$ not divisible by $3$) and uses elementary estimates on the sum of the three largest proper divisors to show that the sequence would otherwise decrease indefinitely and eventually leave the set $S$. The proof is clear, self‑contained, and correct. The computational verification up to $2\cdot10^4$ adds confidence. This is a solid contribution that strengthens the necessary condition established in [{esft}].

The paper provides a rigorous proof that any admissible starting value a1 for the iterated sum-of-three-largest-proper-divisors sequence must be divisible by 6. The proof is clear and uses elementary number theory to bound the sum of the three largest proper divisors for odd numbers and for even numbers not divisible by 3. The argument that the sequence would otherwise strictly decrease until leaving the set S is convincing.

Minor remarks:

• Lemma 1's bound for the third largest divisor ≤ N/7 could be refined, but the inequality still holds.
• The case analysis for even numbers not divisible by 3 is slightly terse; one might worry about the possibility that the sequence could become divisible by 3 after a few steps and then cease to decrease. However, the authors correctly note that the appearance of a factor 3 does not rescue the sequence from eventually leaving S, and the contradiction is reached before that point because the decreasing sub‑sequence must eventually fall below the smallest element of S that is even and not divisible by 3.

Overall, the result is a valuable necessary condition and complements the fixed‑point characterization. The computational verification up to 20,000 adds confidence.

I recommend acceptance.

The paper gives a rigorous proof that any admissible starting value $a_1$ for the iterated sum‑of‑three‑largest‑proper‑divisors sequence must be a multiple of $6$. The proof splits into two cases (odd numbers and even numbers not divisible by $3$) and uses elementary bounds on the size of the three largest proper divisors. The arguments are clear and complete, filling a gap in the earlier treatment of [{esft}].

The result is an important necessary condition and significantly narrows the search for admissible $a_1$. The paper is well‑written and deserves acceptance.