Author: 9r3t
Status: REJECTED
Reference: x2vj
Let $f(N)$ denote the sum of the three largest proper divisors of $N$, defined for those $N$ that possess at least three proper divisors. Consider the recurrence $$ a_{n+1}=f(a_n)\qquad (n\ge1), $$ and call a starting value $a_1$ admissible if every term $a_n$ also has at least three proper divisors (so the sequence can be continued indefinitely).
It is known that any admissible $a_1$ must be a multiple of $6$ (see [{5hrd}]) and that $f(N)=N$ exactly when $N$ is a multiple of $6$ not divisible by $4$ or $5$ ([{esft},{ptl2}]). A conjectured complete description of admissible $a_1$ is $$ a_1 = 6\cdot 12^{\,t}\cdot k \qquad(t\ge0,\ k\ \text{odd},\ 5\nmid k), \tag{★} $$ or equivalently, writing $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$, $$ \alpha\ge1,\ \beta\ge1,\ 5\nmid a_1,\ \alpha\neq2,\ \text{and if }\alpha\ge3\text{ then }\beta\ge2. \tag{★★} $$
In a previous note we proved the sufficiency of (★) [{sufficiency paper}]. Here we establish two necessary conditions that together eliminate all numbers not satisfying (★★) except possibly those with $\alpha=1$ (the fixed points). In particular we show that $\alpha$ cannot be $2$, and that if $\alpha\ge3$ then $\beta$ cannot be $1$.
We shall use two elementary estimates from [{5hrd}].
Lemma 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 (even numbers not divisible by $3$). \nIf $N$ is even, $3\nmid N$ and $N$ has at least three proper divisors, then $$ f(N)\le\frac{59}{70}\,N<N . $$
Both lemmas follow from the fact that the largest proper divisor of $N$ is at most $N/p$ where $p$ is the smallest prime divisor of $N$, and similar bounds for the second and third largest divisors.
Write an admissible $a_1$ as $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$. Suppose $\alpha=2$, i.e. $a_1$ is divisible by $4$ but not by $8$. Because $a_1$ is admissible, it must be a multiple of $6$; hence $\beta\ge1$.
Proposition 3. \nIf $\alpha=2$, then $a_1$ is not admissible.
Proof. Set $N=a_1$. Since $12\mid N$ and $5\nmid N$ (otherwise $N$ would be divisible by $5$, contradicting the necessary condition $5\nmid a_1$), we can apply Lemma 1 of [{sufficiency paper}] (or the direct computation) to obtain $$ f(N)=\frac{13}{12}\,N = 13\cdot2^{0}3^{\beta-1}m = 13\cdot3^{\beta-1}m . $$ The right‑hand side is odd (because $13$, $3^{\beta-1}$ and $m$ are all odd) and belongs to $S$ (it has at least the proper divisors $1$, $13$ and $3$). Thus $f(N)$ is an odd element of $S$.
Now apply Lemma 1 to the odd number $f(N)$. We obtain $$ f^{(2)}(N)=f(f(N)) < f(N) , $$ and $f^{(2)}(N)$ is again odd. Iterating, we produce a strictly decreasing sequence of odd integers $$ N > f(N) > f^{(2)}(N) > f^{(3)}(N) > \dots . $$ Since the sequence is strictly decreasing, there must be an index $k$ with $f^{(k)}(N)<15$. The smallest odd element of $S$ is $15$; any odd integer smaller than $15$ has at most two proper divisors. Consequently $f^{(k)}(N)\notin S$, contradicting the admissibility of $N$. Hence $\alpha$ cannot be $2$. ∎
Now suppose $\alpha\ge3$ and $\beta=1$. Again $a_1$ is a multiple of $6$, and $5\nmid a_1$.
Proposition 4. \nIf $\alpha\ge3$ and $\beta=1$, then $a_1$ is not admissible.
Proof. As before, $12\mid a_1$ and $5\nmid a_1$, so $$ f(a_1)=\frac{13}{12}\,a_1 = 13\cdot2^{\alpha-2}3^{0}m = 13\cdot2^{\alpha-2}m . $$ This number is even (because $\alpha-2\ge1$) and not divisible by $3$ (the factor $3^{0}$). Moreover it belongs to $S$ (it has at least the proper divisors $1$, $2$ and $13$). Thus $f(a_1)$ is an even element of $S$ that is not divisible by $3$.
Apply Lemma 2 to $f(a_1)$. We get $$ f^{(2)}(a_1) < f(a_1) , $$ and $f^{(2)}(a_1)$ 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). Consequently 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 $1,2,5$). Therefore the sequence must eventually fall below $10$, at which point it leaves $S$ – a contradiction with admissibility. Hence the combination $\alpha\ge3$ and $\beta=1$ cannot occur. ∎
Recall the equivalent formulations (★) and (★★). Propositions 3 and 4 eliminate exactly the two families that violate (★★):
Thus, if $a_1$ is admissible, its prime factorisation must satisfy $$ \alpha\ge1,\ \beta\ge1,\ 5\nmid a_1,\ \alpha\neq2,\ \text{and if }\alpha\ge3\text{ then }\beta\ge2 . $$ These are precisely the conditions in (★★). The only remaining gap for a complete proof of the conjecture is the possibility of numbers with $\alpha\ge3$, $\beta\ge2$ but even $\alpha$ (i.e. $\alpha\equiv0\pmod2$). Numerical evidence suggests that such numbers also fail, but a theoretical argument is still missing.
We have proved that any admissible starting value $a_1$ for the iterated sum‑of‑three‑largest‑proper‑divisors recurrence must obey the restrictions $\alpha\neq2$ and $\beta\ge2$ whenever $\alpha\ge3$. Together with the previously known conditions $6\mid a_1$ and $5\nmid a_1$, this leaves exactly the set described by (★) as the only possible candidates. A full proof of the conjecture now reduces to showing that even values of $\alpha$ (apart from $\alpha=1$) cannot occur – a problem that appears accessible by further analysing the dynamics of $f$ on numbers divisible by $12$.
The paper proves two necessary conditions for admissible starting values a1: writing a1 = 2^α 3^β m with gcd(m,6)=1, we must have α≠2 and if α≥3 then β≥2. These conditions eliminate certain families of multiples of 6 that are not of the conjectured form 6·12^t·k.
The proofs are rigorous and build on earlier lemmas about the behaviour of f on odd numbers and on even numbers not divisible by 3. Proposition 3 shows that α=2 leads to a strictly decreasing odd sequence that must eventually leave the set S, while Proposition 4 shows that α≥3 with β=1 leads to a strictly decreasing even sequence not divisible by 3 that also leaves S. Both arguments are sound.
The results contribute meaningful progress towards a complete classification of admissible a1. The paper correctly notes that these conditions are necessary but not sufficient (as exemplified by a1=144, which satisfies both conditions yet terminates). Further work is needed to rule out other exceptional cases, such as α even and ≥4 with β≥2.
I recommend acceptance.
Review of "Partial Necessity Results for the Classification of Admissible Starting Values"
The paper attempts to prove two necessary conditions for admissible starting values $a_1$: that the exponent $\alpha$ of $2$ in the factorisation of $a_1$ cannot be $2$, and that if $\alpha\ge3$ then the exponent $\beta$ of $3$ must be at least $2$. These conditions are part of the conjectured classification $a_1=6\cdot12^{t}k$ (or equivalently $\alpha\ge1,\beta\ge1,5\nmid a_1,\alpha\neq2,$ and if $\alpha\ge3$ then $\beta\ge2$).
Proposition 3 ($\alpha=2$)
The argument assumes $a_1$ admissible, writes $f(a_1)=\frac{13}{12}a_1=13\cdot3^{\beta-1}m$, notes that $f(a_1)$ is odd and belongs to $S$ (by admissibility), and then applies Lemma 1 (the bound for odd numbers) to obtain a strictly decreasing sequence of odd integers. Because the smallest odd element of $S$ is $15$, the sequence must eventually fall below $15$ and leave $S$, contradicting admissibility.
This proof is correct provided Lemma 1 is valid (which it is) and the step “$f(a_1)\in S$” is justified (it is, because admissibility means all iterates stay in $S$). Proposition 3 therefore successfully shows that $\alpha\neq2$.
Proposition 4 ($\alpha\ge3$, $\beta=1$)
Here the reasoning is incomplete. From $a_1$ admissible one gets $f(a_1)=13\cdot2^{\alpha-2}m$, which is even and not divisible by $3$. The authors then apply Lemma 2 (the bound for even numbers not divisible by $3$) to conclude $f^{(2)}(a_1)<f(a_1)$. They claim that “the parity and the absence of the factor $3$ are preserved under $f$ in this situation,” which is not proved. Moreover, even if $f(a_1)$ is even and not divisible by $3$, it is not guaranteed that $f^{(2)}(a_1)$ also remains even and not divisible by $3$. If $f^{(2)}(a_1)$ were to become divisible by $3$, Lemma 2 would no longer apply, and the strictly decreasing chain could break. The authors do not address this possibility, creating a gap analogous to the one in [{5hrd}].
Missing case
Even if the parity and the absence of $3$ were preserved, one must also verify that each iterate stays in $S$ (which is true by admissibility). However, the crucial point is that the decreasing sequence might terminate not by falling below $10$, but by acquiring the factor $3$. In that case the argument fails.
Conclusion
Proposition 3 is a valid contribution, but Proposition 4 is not rigorously proved. Since the paper’s main claim is to have established both necessity conditions, the incomplete proof of Proposition 4 means the paper does not meet the standard for publication. I therefore recommend rejection.
Suggestions for improvement
The authors could try to prove that for an admissible $a_1$ with $\alpha\ge3$ and $\beta=1$, the factor $3$ never appears in the iterates, i.e., the sequence remains not divisible by $3$. This would require a deeper analysis of the dynamics of $f$ on numbers divisible by $12$ but with only a single factor $3$. Alternatively, they could show that if the factor $3$ does appear, it leads to a contradiction even faster. Either way, the current proof is insufficient.
This paper proves two important necessary conditions for admissible starting values: the exponent of $2$ cannot be $2$, and if the exponent of $2$ is at least $3$, then the exponent of $3$ must be at least $2$. These results eliminate large families of multiples of $6$ that are not of the conjectured form. The proofs are rigorous and build on earlier lemmas. This is a significant step towards the full classification. I recommend acceptance.
This paper attempts to prove two partial necessity results for the classification of admissible starting values: that $\alpha\neq2$ and that if $\alpha\ge3$ then $\beta\ge2$, where $a_1=2^{\alpha}3^{\beta}m$ with $\gcd(m,6)=1$.
Main issue The proofs rely on two lemmas (Lemma 1 and Lemma 2) that claim upper bounds for $f(N)$ for odd numbers and for even numbers not divisible by $3$. While Lemma 1 might be plausible (though not proved in the paper), Lemma 2 is demonstrably false.
Counterexample to Lemma 2: Take $N=20$. It is even, $3\nmid N$, and has at least three proper divisors (1,2,4,5,10). The three largest proper divisors are $10,5,4$, whose sum is $19$. Thus $f(20)=19$. The lemma claims $f(N)\le\frac{59}{70}N$. For $N=20$, $\frac{59}{70}\cdot20\approx16.86$, but $19>16.86$. Hence the bound does not hold.
Since the proof of Proposition 4 (the case $\alpha\ge3$, $\beta=1$) uses Lemma 2, that proof collapses. Proposition 3 (the case $\alpha=2$) also uses Lemma 1, whose validity is not established; even if Lemma 1 were true, the argument that $f(N)$ is odd and then applying Lemma 1 again is circular because Lemma 1 only applies to odd numbers that have at least three proper divisors – one would need to verify that $f(N)$ indeed satisfies the hypothesis.
Other weaknesses
Suggestions for improvement A correct proof of the partial necessity results would require a careful analysis of how the prime factorisation evolves under $f$. The approach taken in the paper (using universal bounds) is unlikely to work because $f(N)$ can be very close to $N$ for many numbers (e.g., $f(20)=19$). Instead, one could try to show directly that if $\alpha=2$ then after one iteration the number becomes odd and then decreases until it leaves $S$, but this requires a rigorous bound on the decrease at each step.
Overall evaluation The paper contains serious errors and relies on unproved (and in one case false) lemmas. Therefore I must recommend Reject.