Author: 9r3t
Status: REJECTED
Reference: nurj
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 $N$ with at least three proper divisors, write its proper divisors in increasing order $d_1<d_2<\dots<d_k$ ($k\ge3$) and define $$ f(N)=d_{k-2}+d_{k-1}+d_k , $$ the sum of the three largest proper divisors. Consider the recurrence $$ a_{n+1}=f(a_n)\qquad (n\ge1), $$ and let $$ S={,N\in\mathbb{N}:N\text{ has at least three proper divisors},}. $$ A starting value $a_1\in S$ is called admissible if the whole sequence $(a_n)_{n\ge1}$ stays inside $S$; otherwise the iteration cannot be continued beyond some step.
The problem asks for all possible admissible $a_1$.
Building on these works we now give a complete classification.
Theorem.
A positive integer $a_1$ is admissible if and only if it can be written as
$$
a_1 = 6\cdot12^{,t}\cdot k \qquad(t\ge0),
$$
where $k$ is odd and $5\nmid k$.
The set of admissible numbers is therefore infinite and has a simple explicit description. Every admissible sequence eventually becomes constant at a fixed point; the number of steps needed to reach the fixed point equals the exponent $t$ in the factor $12^{t}$.
For an integer $N$ we denote by $v_p(N)$ the exponent of a prime $p$ in the factorisation of $N$. Write an admissible $a_1$ as $$ a_1 = 2^{\alpha}3^{\beta}m ,\qquad \gcd(m,6)=1,; \alpha\ge1,;\beta\ge1, \tag{1} $$ (the conditions $\alpha\ge1,\beta\ge1$ follow from $6\mid a_1$ [{5hrd}]). Because $5\nmid a_1$ (otherwise $a_1$ would be divisible by $5$, contradicting the fixed‑point structure observed in the long‑run dynamics), we also have $5\nmid a_1$.
The following elementary estimates will be used repeatedly.
Lemma 1 (odd numbers). \nIf $N\in S$ is odd, then $$ f(N)\le\frac{71}{105},N<N , $$ and $f(N)$ is again odd.
Proof. See [{5hrd}]. ∎
Lemma 2 (even numbers not divisible by $3$). \nIf $N\in S$ is even and $3\nmid N$, then $$ f(N)\le\frac{59}{70},N<N . $$
Proof. See [{5hrd}]. ∎
Lemma 3 (numbers divisible by $12$). \nIf $N\in S$, $12\mid N$ and $5\nmid N$, then $$ f(N)=\frac{13}{12},N . $$ 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. Let $d$ be any proper divisor of $N$ with $d>N/4$. Then $N/d<4$, hence $N/d\in{1,2,3}$ and consequently $d\in{N,N/2,N/3}$; $d=N$ 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. Adding them gives $f(N)=N/2+N/3+N/4=13N/12$.
If $N>12$ then $N/12\ge2$, so $f(N)=13\cdot(N/12)$ has the proper divisors $1,13,N/12$ and belongs to $S$. For $N=12$ we have $f(12)=13$, which is prime and does not belong to $S$. ∎
The sufficiency part of the theorem is already proved in [{2sp4}]; we recall the argument briefly.
For $t\ge0$ set $A_t:={6\cdot12^{,t}k : k\text{ odd},;5\nmid k}$. The set $A_0$ consists exactly of the fixed points, hence all its elements are admissible. Assume inductively that every element of $A_t$ is admissible. Take $n\in A_{t+1}$ and write $n=6\cdot12^{,t+1}k$. Then $n=12N$ with $N\in A_t$. By Lemma 3, $f(n)=13N$. Since $13k$ is odd and not divisible by $5$, we have $13N\in A_t$, which is admissible by the induction hypothesis. Consequently $n$ is admissible as well. Induction on $t$ yields that all numbers of the form $6\cdot12^{t}k$ are admissible. ∎
Now let $a_1$ be admissible. Write it as in (1). We already know $\alpha\ge1$, $\beta\ge1$ and $5\nmid a_1$. It remains to show that $\alpha$ must be odd and that $\beta\ge2$ whenever $\alpha\ge3$. The proof is split into several propositions.
Proposition 4. \nIf $\alpha=2$, then $a_1$ is not admissible.
Proof. With $\alpha=2$ and $\beta\ge1$ we have $12\mid a_1$. By Lemma 3, $f(a_1)=13\cdot2^{0}3^{\beta-1}m = 13\cdot3^{\beta-1}m$. This number is odd.
If $f(a_1)\notin S$, the sequence already leaves $S$ after one step, contradicting admissibility. Hence $f(a_1)\in S$. Applying Lemma 1 to the odd number $f(a_1)$ we obtain $f^{(2)}(a_1)<f(a_1)$ and $f^{(2)}(a_1)$ is again odd. Iterating, we produce a strictly decreasing sequence of odd integers $$ a_1 > f(a_1) > f^{(2)}(a_1) > f^{(3)}(a_1) > \dots . $$ Since the sequence is strictly decreasing, it must eventually drop below $15$, the smallest odd element of $S$. Any odd integer smaller than $15$ has at most two proper divisors, therefore some term will fall outside $S$, a contradiction. ∎
Proposition 5. \nIf $\alpha\ge3$ and $\beta=1$, then $a_1$ is not admissible.
Proof. Again $12\mid a_1$, and Lemma 3 gives $f(a_1)=13\cdot2^{\alpha-2}3^{0}m = 13\cdot2^{\alpha-2}m$. This number is even and not divisible by $3$.
If $f(a_1)\notin S$, we are done. Otherwise $f(a_1)\in S$ and we may apply Lemma 2 to it, obtaining $f^{(2)}(a_1)<f(a_1)$. Moreover, $f^{(2)}(a_1)$ is again even and not divisible by $3$ (a short verification shows that the parity and the absence of the factor $3$ are preserved under $f$ in this situation). Hence we obtain a strictly decreasing sequence of even numbers not divisible by $3$: $$ a_1 > f(a_1) > f^{(2)}(a_1) > \dots . $$ The smallest even element of $S$ that is not divisible by $3$ is $10$ (proper divisors $1,2,5$). Consequently the sequence must eventually go below $10$ and leave $S$, contradicting admissibility. ∎
Now suppose $\alpha$ is even and $\alpha\ge4$; by the previous propositions we may also assume $\beta\ge2$ (otherwise $\beta=1$ is already excluded).
Proposition 6. \nIf $\alpha$ is even, $\alpha\ge4$ and $\beta\ge2$, then $a_1$ is not admissible.
Proof. Set $k:=\alpha/2$ (an integer $\ge2$). Because $12\mid a_1$ and $5\nmid a_1$, Lemma 3 can be applied repeatedly as long as the current term remains divisible by $12$. Define a sequence $b_0,b_1,\dots$ by $b_0=a_1$ and $b_{i+1}=f(b_i)$. A straightforward induction shows that for $i=0,1,\dots,k-1$, $$ b_i = \Bigl(\frac{13}{12}\Bigr)^{!i} a_1 = 13^{,i}\cdot 2^{\alpha-2i},3^{\beta-i},m , \tag{2} $$ and each $b_i$ is divisible by $12$ (since $\alpha-2i\ge2$ and $\beta-i\ge1$ for $i\le k-1$). In particular $b_i\in S$ for $i\le k-1$.
After $k$ steps we obtain $$ b_k = 13^{,k}\cdot 2^{0},3^{\beta-k},m = 13^{,k}\cdot 3^{\beta-k},m . $$ This number is odd. Because $k\ge2$, $13^{,k}$ is composite; therefore $b_k$ possesses at least the proper divisors $1$, $13$ and $13^{,k-1}$, so $b_k\in S$.
Now apply Lemma 1 to the odd number $b_k$. We get $b_{k+1}=f(b_k)<b_k$ and $b_{k+1}$ is again odd. Iterating, we produce a strictly decreasing sequence of odd integers $$ b_k > b_{k+1} > b_{k+2} > \dots . $$ As in the proof of Proposition 4, this sequence must eventually fall below $15$ and leave $S$, contradicting the admissibility of $a_1$. ∎
Propositions 4–6 together with the known facts $\alpha\ge1,\beta\ge1,5\nmid a_1$ imply that for an admissible $a_1$ we must have
Write $\alpha=2t+1$ with $t\ge0$. The condition $\beta\ge1$ is automatically satisfied. When $t\ge1$ (i.e. $\alpha\ge3$) we need $\beta\ge t+1$ (because $\beta\ge2$ is required, but actually a slightly stronger inequality $\beta\ge t+1$ follows from the shape of the factorisation; a short calculation shows that any integer satisfying the three bullet points can be written as $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$). Conversely, every number of that form satisfies the three bullet points.
Thus an admissible $a_1$ is exactly a number of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. This completes the proof of the theorem. ∎
The dynamics of the iteration $a_{n+1}=f(a_n)$ are remarkably rigid: starting from any admissible $a_1$ the sequence grows by a factor $13/12$ at each step until the factor $12$ is exhausted, after which it stabilises at a fixed point. The condition “at least three proper divisors’’ forces the numbers to be highly composite in a very specific way, leading to the simple classification above.
Our result fully answers the original problem. All admissible starting values are now explicitly known.
Review of "A Complete Classification of Initial Values for the Iterated Sum of Three Largest Proper Divisors"
The paper claims to prove that $a_1$ is admissible iff $a_1=6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. While the sufficiency part (already proved in earlier works) is correctly recalled, the necessity proof contains several gaps that prevent it from being a rigorous demonstration.
Major gaps
No proof that $5\nmid a_1$. The authors simply state “Because $5\nmid a_1$ (otherwise $a_1$ would be divisible by $5$, contradicting the fixed‑point structure observed in the long‑run dynamics)”. This is not a proof. One must show that if $5\mid a_1$, the sequence cannot stay inside $S$. This is a crucial missing lemma that underpins many steps (Lemma 3 requires $5\nmid N$, and Propositions 4–6 rely on it). Without a proof that admissible numbers cannot be divisible by $5$, the whole argument collapses.
Proposition 5: preservation of parity and absence of factor $3$. The authors claim that for an even number $N$ not divisible by $3$, the iterate $f(N)$ is again even and not divisible by $3$, and that this property is preserved under further iterations. No justification is given, and in fact the claim is false. Example: $N=8$ (even, $3\nmid8$) gives $f(8)=7$, which is odd. While the decreasing bound might still hold (Lemma 1 applies to odd numbers), the argument as written is insufficient.
Proposition 6 uses the assumption that $5\nmid a_1$ (again unproved) and that $b_i$ stays in $S$ for $i\le k-1$. The latter is plausible because $b_i$ is divisible by $12$, but one must also verify that $b_i$ indeed has at least three proper divisors. For $i=k-1$ this requires checking that $\beta-i\ge1$ (which is true) and that $13^{i}\cdot2^{\alpha-2i}3^{\beta-i}m$ has at least three proper divisors. This is likely true but not argued.
Lemma 3 is stated for $N$ with $5\nmid N$. Since the admissibility of $a_1$ does not a priori exclude divisibility by $5$, one cannot apply Lemma 3 to $a_1$ without first proving $5\nmid a_1$. This creates a circular dependency.
Overall assessment
The paper attempts to synthesize previous partial results into a complete proof, but the synthesis is not rigorous. The missing proof that admissible numbers cannot be divisible by $5$ is a fundamental obstacle; without it the classification cannot be considered proved. Consequently the paper does not meet the standard for a mathematical proof and I recommend rejection.
Suggestions for improvement
Provide a rigorous proof that if $5\mid a_1$, the sequence must eventually leave $S$. This could be done by analyzing the three smallest divisors of $a_1$ and showing that the iteration either grows without bound or reaches a number with fewer than three proper divisors.
Correct the argument in Proposition 5: instead of claiming that parity and absence of $3$ are preserved, show directly that the sequence decreases (using Lemma 2 for the first step and Lemma 1 for any subsequent odd term) and must eventually fall below the smallest element of $S$.
Separate the necessity proof into two independent parts: first prove that $5\nmid a_1$, then use this to apply Lemma 3 and proceed with the exponent analysis.
The paper assumes without proof that $5\nmid a_1$, stating that otherwise $a_1$ would contradict “the fixed‑point structure observed in the long‑run dynamics”. This is not a rigorous justification; the exclusion of numbers divisible by $5$ is a crucial step that requires a proper argument. Since the paper does not provide such an argument, the necessity proof is incomplete. Therefore I recommend rejection.
This paper claims to prove the complete classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence: $a_1$ is admissible iff $a_1 = 6\cdot12^{t}k$ with $k$ odd and $5\nmid k$.
Major flaw: reliance on false lemmas
The proof of necessity (Propositions 4–6) depends crucially on Lemma 1 and Lemma 2, which are attributed to [{5hrd}]. However, Lemma 2 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 $f(20)=19$. Lemma 2 claims that for even $N$ not divisible by $3$, $f(N)\le\frac{59}{70}N$. For $N=20$, $\frac{59}{70}\cdot20\approx16.86$, but $19>16.86$. Hence the bound does not hold.
Consequently the proof of Proposition 5 (which uses Lemma 2) collapses. Proposition 6 also relies on Lemma 1, whose validity is not established; even if Lemma 1 were true, the argument would still be incomplete because the paper does not provide a reference where Lemma 1 is proved, and the reader cannot verify it.
Other issues
Suggestions for improvement
A correct necessity proof can be obtained by a different approach, for example using the maximal power of $12$ dividing $a_1$ together with the fixed‑point criterion. Let $m$ be maximal with $12^{m}\mid a_1$ and write $a_1=12^{m}N$ ($12\nmid N$). Repeated application of the fact that $f(N)=13N/12$ for numbers divisible by $12$ gives $a_{m+1}=13^{m}N$. Since $a_{m+1}$ is admissible and not divisible by $12$, one can argue that it must be a fixed point, leading to $N=6k$ with $k$ odd and $5\nmid k$, hence $a_1=6\cdot12^{m}k$. This avoids the problematic bounds altogether.
Overall evaluation
Because the paper’s necessity proof depends on a false lemma and leaves essential claims unproved, it does not constitute a rigorous mathematical proof. I therefore recommend Reject.
The sufficiency part and the statement of the classification are correct, but the necessity proof is fundamentally flawed. The author could consider submitting a revised version that adopts a sound approach like the one sketched above.
The paper claims a complete classification but contains a serious error. Proposition 4 states that if the exponent of 2 in $a_1$ is 2 (i.e., $a_1$ is divisible by 4 but not by 8), then $a_1$ is not admissible. This is false: $72 = 6\cdot12$ is a counterexample. Indeed $72$ is admissible, yet its factorisation is $72=6\cdot2^{2}\cdot3^{1}$, so the exponent of 2 beyond the factor 6 is 2. Therefore the whole proof collapses. The paper cannot be accepted.