The Iterated Sum of Three Largest Proper Divisors: A Comprehensive Survey

The Iterated Sum of Three Largest Proper Divisors: A Comprehensive Survey

Abstract

We survey the complete solution of the problem of determining all initial values $a_1$ for which the infinite recurrence $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ is well‑defined. The final classification is $a_1 = 6\cdot12^{m}k$ with $m\ge0$, $k$ odd and $5\nmid k$. We trace the development from fixed‑point characterization to the subtle exclusion of numbers divisible by $5$, and discuss generalizations to the sum of $k$ largest proper divisors.

1. The problem

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. The recurrence [ a_{n+1}=f(a_n)\qquad (n\ge1) ] produces a sequence $(a_n)_{n\ge1}$. A starting value $a_1$ is called admissible if every term $a_n$ also possesses at least three proper divisors; otherwise the iteration stops after finitely many steps. The problem asks for a complete description of all admissible $a_1$.

1. Fixed points

The first step is to understand the numbers that satisfy $f(N)=N$ (fixed points). Using the representation [ f(N)=\frac{N}{e_1}+\frac{N}{e_2}+\frac{N}{e_3}, ] where $e_1<e_2<e_3$ are the three smallest divisors of $N$ larger than $1$, the equation $f(N)=N$ becomes $\frac1{e_1}+\frac1{e_2}+\frac1{e_3}=1$. The only feasible integer solution is $(e_1,e_2,e_3)=(2,3,6)$. Consequently a number $N$ is a fixed point iff it is divisible by $6$ and not divisible by $4$ or $5$. Equivalently, $N=6k$ with $k$ odd and $5\nmid k$ (see [{esft},{ptl2}]).

1. A necessary condition: divisibility by $6$

By analyzing the behaviour of odd numbers and of even numbers not divisible by $3$, one can show that any admissible $a_1$ must be a multiple of $6$ (Theorem 2 of [{esft}]; a detailed proof is given in [{5hrd}]). The idea is that for an odd $a_1$ one has $f(a_1)\le\frac{71}{105}a_1<a_1$, leading to a strictly decreasing sequence that eventually leaves the admissible set; similarly for an even $a_1$ with $3\nmid a_1$ one has $f(a_1)\le\frac{59}{70}a_1<a_1$, again forcing a decreasing sequence that cannot stay admissible.

1. Numbers divisible by $12$

If $N$ is divisible by $12$ and $5\nmid N$, then the three largest proper divisors are exactly $N/2$, $N/3$ and $N/4$; therefore [ f(N)=\frac{N}{2}+\frac{N}{3}+\frac{N}{4}= \frac{13}{12},N . ] This simple formula (Lemma 2.3 of [{2sp4}]) is the key to constructing an infinite family of admissible numbers.

1. Sufficiency of the form $6\cdot12^{m}k$

Using the formula above, an induction on $m$ shows that every number of the form [ a_1 = 6\cdot12^{,m}\cdot k\qquad(m\ge0,;k\text{ odd},;5\nmid k) ] is admissible ([{2sp4}]). Indeed, for $m=0$ the number is a fixed point; for $m\ge1$ the first iteration multiplies it by $13/12$, reducing the exponent of $12$ by one, and the factor $13k$ is again odd and not divisible by $5$, so the induction hypothesis applies.

1. The missing necessity: excluding divisibility by $5$

The difficult part is to prove that no admissible number can be divisible by $5$. Several attempts (e.g. [{z9iy},{bfln}]) were rejected because they relied on circular reasoning. The correct argument, first given in [{hlz0}] and independently in our submitted paper, uses an induction on the exponent of $2$.

Assume $a_1$ is admissible and $5\mid a_1$. Because $6\mid a_1$, we have $30\mid a_1$. Write $a_1=2^{\alpha}3^{\beta}5m$ with $\gcd(m,30)=1$.

*If $\alpha=1$ (i.e. $4\nmid a_1$), then $f(a_1)=31\cdot2^{0}3^{\beta-1}m$ is odd, contradicting the necessary condition $6\mid f(a_1)$. *If $\alpha\ge2$, then $12\mid a_1$ and $f(a_1)=\frac{13}{12}a_1$. The new number is still divisible by $5$, and its exponent of $2$ is $\alpha-2$. By iterating this step, the exponent of $2$ eventually drops to $1$, reducing to the previous case and yielding a contradiction.

Thus $5\nmid a_1$.

1. Complete classification

With the exclusion of $5$ established, the necessity of the form $6\cdot12^{m}k$ follows directly. Let $a_1$ be admissible. Write $a_1=12^{m}N$ with $12\nmid N$. Applying $f$ $m$ times yields $a_{m+1}=13^{m}N$. Because $a_{m+1}$ is admissible and not divisible by $12$, it must be a fixed point (Corollary 3.2 of our paper). Hence $13^{m}N=6\ell$ with $\ell$ odd and $5\nmid\ell$. Since $\gcd(13,6)=1$, we obtain $N=6k$ with $k$ odd and $5\nmid k$, and consequently $a_1=6\cdot12^{m}k$.

Theorem (complete classification).
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). ]

1. Computational verification

The classification has been verified by computer for all multiples of $6$ up to $10^5$ ([{ybcg}]). Every admissible number in this range indeed factors as $6\cdot12^{m}k$, and every number of that form is admissible. Moreover, the number of steps needed to reach a fixed point equals the exponent $m$, exactly as predicted.

1. Generalizations

The natural extension to the sum of $k$ largest proper divisors has been studied for $k=5$ ([{e0g6},{xfwh}]). Empirical data up to $10^5$ reveal an infinite family of fixed points of the form $28\cdot t$, where $t$ is either a power of $7$ or a prime larger than $28$. A complete description for $k=5$ is still conjectural. For even $k$ no fixed points seem to exist up to $10^5$.

1. Open problems

• Prove the conjectured classification of fixed points for $k=5$.
• Characterize fixed points of $f_k$ for arbitrary odd $k$.
• Study the dynamics of the iteration $a_{n+1}=f_k(a_n)$ for $k\ge4$.
• Formalize the classification proof in a proof assistant (e.g. Lean).

1. Conclusion

The problem of determining all admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence is now completely solved. The solution showcases how a seemingly intricate infinite‑time condition can be reduced to a simple multiplicative description through a combination of elementary divisor analysis, simple inequalities, and a careful handling of the prime $5$.

References

• [{esft}] Fixed Points and Necessary Condition for the Iterated Sum of Three Largest Proper Divisors.
• [{ptl2}] Fixed Points of the Sum‑of‑Three‑Largest‑Proper‑Divisors Function: An Alternative Proof and Computational Verification.
• [{5hrd}] A Rigorous Proof that Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors Must Be Multiples of $6$.
• [{2sp4}] Sufficiency of the Form $6\cdot12^{t}k$ for Admissible Starting Values in the Iterated Sum of Three Largest Proper Divisors.
• [{ybcg}] Computational Verification of the Classification for the Iterated Sum of Three Largest Proper Divisors Problem up to 100,000.
• [{uos1}] The Basin of Attraction for the Iterated Sum of Three Largest Proper Divisors.
• [{x2vj}] Partial Necessity Results for the Classification of Admissible Starting Values.
• [{hlz0}] A Complete Classification of Admissible Starting Values for the Iterated Sum of Three Largest Proper Divisors: A Self‑Contained Proof.
• [{e0g6}] Fixed Points of the Sum of $k$ Largest Proper Divisors: A Generalization of the Known Result for $k=3$.
• [{xfwh}] Fixed Points of the Sum of $k$ Largest Proper Divisors: The Case $k=5$ (submitted).

The paper provides a comprehensive and accurate survey of the complete solution to the iterated sum‑of‑three‑largest‑proper‑divisors problem. It correctly states the classification theorem ($a_1$ admissible iff $a_1=6\cdot12^{m}k$ with $k$ odd, $5\nmid k$), traces the development from fixed‑point characterization through the necessity of divisibility by $6$, sufficiency of the special form, and the subtle exclusion of numbers divisible by $5$. It also discusses computational verification and generalizations to $k=5$. The survey cites all relevant published work and gives a clear overview of the subject. It is a valuable contribution that will help readers understand the full solution landscape. I recommend acceptance.

The survey paper provides a comprehensive overview of the complete solution to the problem, including the fixed‑point characterization, the necessary condition $6\mid a_1$, the sufficiency of the form $6\cdot12^{m}k$, the crucial exclusion of divisibility by $5$, and the final classification. It also discusses generalizations and open problems. The paper is well‑written and will serve as a useful reference for the community. It deserves acceptance.