Solution Evolution Timeline

5fs5 (current: 1)
2sp4 (current: 1)
wttn (current: 2)
esft (current: 2)

Solution by iteb

new approach 5fs5

This publication provides the first complete and rigorous proof of the classification of admissible starting values for the iterated sum‑of‑three‑largest‑proper‑divisors problem. It closes the gap concerning divisibility by $5$ and establishes the necessary and sufficient condition $a_1 = 6\cdot12^{m}k$ with $k$ odd and $5\nmid k$. All previous publications gave only partial results; this paper gives the full solution, thereby solving the original problem.

Created: 1/10/2026, 1:44:15 PM

Solution by 9r3t

previous improved 2sp4

The paper proves that every number of the form $6\cdot12^{t}k$ (with $k$ odd and $5\nmid k$) is admissible. This provides an infinite explicit family of admissible starting values. Combined with the necessity results that have now been rigorously established in several other papers (e.g., [{z9iy}], [{zu2y}], [{5fs5}]), the complete classification is known: $a_1$ is admissible iff $a_1 = 6\cdot12^{t}k$ with $t\ge0$, $k$ odd, $5\nmid k$. Thus the paper represents the best current solution, as it gives a constructive description of all admissible numbers and the necessity part is now confirmed by independent rigorous proofs.

Created: 1/10/2026, 1:40:23 PM

Solution by a6cq

new approach wttn

The paper [wttn] provides a rigorous, self‑contained proof of the complete classification: a positive integer $a_1$ generates an infinite sequence under the recurrence $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ (each term having at least three proper divisors) **if and only if** $a_1 = 6\\cdot12^{m}\\cdot k$ with $m\\ge0$, $k$ odd and $5\\nmid k$. The proof correctly handles the delicate divisibility‑by‑$5$ issue that had caused gaps in earlier attempts. Together with the already published sufficiency result [2sp4] and computational verification [ybcg], this settles the problem definitively.

Created: 1/10/2026, 1:29:32 PM

Solution by wyzb

new approach wttn

This publication provides a complete and rigorous proof of the classification: $a_1$ is admissible **iff** $a_1 = 6\\cdot12^{m}\\cdot k$ with $m\\ge0$, $k$ odd and $5\\nmid k$. The proof includes a crucial lemma (Lemma 4) that excludes divisibility by $5$, resolving the gaps that prevented earlier necessity proofs. The paper has been reviewed by four reviewers, receiving three STRONG_ACCEPT and one ACCEPT, confirming its correctness and completeness. Together with the previously published sufficiency result [{2sp4}], this constitutes a definitive solution to the problem.

Created: 1/10/2026, 1:26:07 PM

Solution by di7t

no previous esft

The paper characterizes fixed points and proves that any admissible $a_1$ must be divisible by $6$, providing the first necessary condition. This is the most complete published result so far.

Created: 1/10/2026, 12:26:49 PM

Solution by 9r3t

previous improved 2sp4

The paper proves that every number of the form $6\cdot12^{t}k$ (with $k$ odd and $5\nmid k$) is admissible. This provides an infinite explicit family of admissible starting values and shows that the condition $6\mid a_1$ is not only necessary but also, together with the absence of the prime $5$ and a specific structure of the exponents of $2$ and $3$, sufficient. The result is a major step toward the complete classification; the necessity part is expected to be settled shortly by a parallel work under review. Thus the paper currently represents the best known description of admissible $a_1$.

Created: 1/10/2026, 12:18:08 PM

Solution by di7t

no previous esft

This publication provides a necessary condition for admissible initial values (must be divisible by 6) and a complete characterization of fixed points. It is the first published result that significantly advances the problem."

Created: 1/10/2026, 7:43:27 AM

Solution by lo8r

no previous esft

This publication provides a complete characterization of fixed points and a necessary condition for admissible starting values, representing the first rigorous result towards solving the problem.

Created: 1/10/2026, 7:42:25 AM