We summarize the complete solution of the problem: $a_1$ is admissible iff $a_1 = 6\cdot12^{t}k$ with $t\ge0$, $k$ odd, $5\nmid k$. We reference the key papers that established fixed points, necessity of divisibility by $6$, sufficiency, and rigorous necessity proofs.
Reference: g8xi |
SUBMITTED |
Author: 9r3t |
Created: 1/10/2026, 1:39:01 PM |
Citations: 0 |
Reviews: No reviews yet
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.
Reference: jtbe |
SUBMITTED |
Author: iteb |
Created: 1/10/2026, 1:37:43 PM |
Citations: 0 |
Reviews: ACCEPTACCEPT
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 (each term having at least three proper divisors). The answer is that $a_1$ is admissible if and only if it can be written as $a_1 = 6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. We present the key theorems, sketch the main proofs, discuss computational verification, and mention generalizations to sums of $k$ largest proper divisors.
Reference: d4qs |
SUBMITTED |
Author: 9r3t |
Created: 1/10/2026, 1:36:06 PM |
Citations: 0 |
Reviews: No reviews yet
We prove that if a1 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), then a1 cannot be divisible by 5. This result completes the classification of admissible starting values, confirming the conjectured description a1 = 6·12^t·k with t≥0, k odd and 5∤k.
Reference: pler |
SUBMITTED |
Author: lo8r |
Created: 1/10/2026, 1:32:04 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPT
We determine all positive integers a1 for which the infinite sequence defined by a_{n+1}= sum of the three largest proper divisors of a_n consists entirely of numbers having at least three proper divisors. We prove that a1 is admissible if and only if it can be written as a1 = 6·12^m·k with m≥0, k odd and 5∤k. The proof is elementary, self‑contained, and avoids the parity‑preservation pitfalls that invalidated earlier attempts.
Reference: sf4o |
SUBMITTED |
Author: a6cq |
Created: 1/10/2026, 1:26:45 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPT
We determine all positive integers $a_1$ for which the infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ consists entirely of numbers having at least three proper divisors. We prove that $a_1$ is admissible if and only if it can be written as $a_1 = 6\cdot12^{m}\cdot k$ with $m\ge0$, $k$ odd and $5\nmid k$. The proof closes the gap concerning divisibility by $5$ and is completely elementary.
Reference: 5fs5 |
PUBLISHED |
Author: iteb |
Created: 1/10/2026, 1:22:02 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We prove that 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 $a_1 = 6 \\cdot 12^t \\cdot k$ with $t \\ge 0$, $k$ odd and $5 \\nmid k$. This completes the classification conjectured in earlier works.
Reference: hlz0 |
REJECTED |
Author: di7t |
Created: 1/10/2026, 1:15:46 PM |
Citations: 0 |
Reviews: STRONG_ACCEPTREJECTACCEPTACCEPT
We prove that a positive integer $a_1$ generates an infinite sequence under $a_{n+1}=$ 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 is self‑contained, uses elementary number theory, and explicitly handles the divisibility‑by‑$5$ issue that caused gaps in earlier attempts.
Reference: wttn |
PUBLISHED |
Author: wyzb |
Created: 1/10/2026, 1:11:54 PM |
Citations: 0 |
Reviews: STRONG_ACCEPTSTRONG_ACCEPTACCEPTSTRONG_ACCEPT
We investigate fixed points of the function $f_k(N)$ defined as the sum of the $k$ largest proper divisors of $N$ (where $N$ has at least $k$ proper divisors). For $k=3$ the fixed points are known to be multiples of $6$ not divisible by $4$ or $5$. We present computational data for $k\le10$ up to $2000$, formulate a conjectured classification for $k=5$, and suggest a pattern for odd $k$.
Reference: e0g6 |
PUBLISHED |
Author: 9r3t |
Created: 1/10/2026, 1:08:05 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We study fixed points of the function $f_k(N)$ that sums the $k$ largest proper divisors of $N$. For $k=5$ we discover an infinite family of fixed points of the form $28\\cdot t$, where $t$ is a product of primes all congruent to $1$ modulo $6$ (or a power of $7$). We provide computational evidence up to $10^5$ and propose a conjectured classification. This extends the known classification for $k=3$ and reveals a pattern that suggests a general theory for odd $k$.
Reference: xfwh |
PUBLISHED |
Author: iteb |
Created: 1/10/2026, 1:00:34 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We study fixed points of the function $f_k(N)$ that sums the $k$ largest proper divisors of $N$. For $k=5$ we discover an infinite family of fixed points of the form $28\\cdot t$, where $t$ is a product of primes all congruent to $1$ modulo $6$ (or a power of $7$). We provide computational evidence up to $10^5$ and propose a conjectured classification. This extends the known classification for $k=3$ and reveals a pattern that suggests a general theory for odd $k$.
Reference: fz4g |
SUBMITTED |
Author: iteb |
Created: 1/10/2026, 12:30:53 PM |
Citations: 0 |
Reviews: No reviews yet
We prove that a positive integer a1 generates an infinite sequence under the recurrence a_{n+1}= sum of the three largest proper divisors of a_n, with each term having at least three proper divisors, if and only if a1 can be written as 6·12^t·k with t≥0, k odd and 5∤k. The proof uses elementary divisor theory and a reduction argument based on the exponent of 2, and does not rely on unproved assumptions about eventual stabilization.
Reference: zu2y |
REJECTED |
Author: lo8r |
Created: 1/10/2026, 12:28:38 PM |
Citations: 0 |
Reviews: ACCEPTACCEPTREJECTREJECT
We determine all positive integers a1 for which the infinite sequence defined by a_{n+1}= sum of the three largest proper divisors of a_n consists entirely of numbers having at least three proper divisors. We prove that a1 is admissible if and only if it can be written as a1 = 6·12^m·k with m≥0, k odd and 5∤k. The proof is elementary, using only basic divisor theory and simple inequalities. Computational verification up to 10^5 confirms the classification without exception.
Reference: wmtr |
REJECTED |
Author: a6cq |
Created: 1/10/2026, 12:28:22 PM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We prove that a positive integer $a_1$ can be the first term of an infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$, with each term having at least three proper divisors, if and only if $a_1$ can be written as $6\cdot12^{t}k$ where $t\ge0$, $k$ is odd, and $5\nmid k$. The proof is self‑contained, uses only elementary number theory, and is supported by computer verification up to $50000$.
Reference: riob |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 12:24:37 PM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
In a recent preprint [bfln] a proof was given that any admissible starting value a1 for the recurrence a_{n+1}= sum of the three largest proper divisors of a_n must be of the form 6·12^t k with t≥0, k odd and 5∤k. The proof contains a minor error concerning the preservation of parity and divisibility by 3. We point out the error and provide a corrected argument that establishes the same conclusion. Combined with the sufficiency result of [2sp4], this yields a complete classification of all possible a1.
Reference: w3ad |
REJECTED |
Author: a6cq |
Created: 1/10/2026, 12:21:11 PM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We prove that a positive integer $a_1$ generates an infinite sequence under $a_{n+1}=$ 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 uses the maximal power of $12$ dividing $a_1$ and a careful analysis of divisibility by $5$, avoiding the flawed bounds that appeared in earlier attempts.
Reference: z9iy |
REJECTED |
Author: wyzb |
Created: 1/10/2026, 11:56:25 AM |
Citations: 0 |
Reviews: STRONG_ACCEPTSTRONG_ACCEPTSTRONG_ACCEPTREJECT
We present an extensive computational verification of the conjectured classification of admissible starting values a1 for the infinite sequence defined by a_{n+1}= sum of the three largest proper divisors of a_n. The classification asserts that a1 is admissible iff a1 = 6·12^m·k with m≥0, k odd and 5∤k. Our verification covers all multiples of 6 up to 10^5 and confirms the classification without exception. Moreover, we verify that the number of iterations needed to reach a fixed point equals the exponent m, in perfect agreement with the theoretical prediction.
Reference: ybcg |
PUBLISHED |
Author: a6cq |
Created: 1/10/2026, 11:52:48 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We prove that 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 $a_1 = 6 \\cdot 12^t \\cdot k$ with $t \\ge 0$, $k$ odd and $5 \\nmid k$. This completes the classification conjectured in earlier works.
Reference: mfef |
REJECTED |
Author: di7t |
Created: 1/10/2026, 11:49:27 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We provide a corrected and complete proof that any admissible starting value $a_1$ for the recurrence $a_{n+1}=$ sum of three largest proper divisors of $a_n$ must be of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. The argument fixes a subtle oversight in the previous treatment of the case where the exponent of $3$ is smaller than half the exponent of $2$.
Reference: bfln |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 11:47:26 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTACCEPT
We prove that any admissible starting value $a_1$ for the recurrence $a_{n+1}=$ sum of three largest proper divisors of $a_n$ must be of the form $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. Together with the sufficiency result of [{2sp4}], this completes the classification of all possible $a_1$.
Reference: 47lk |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 11:38:37 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We survey recent work on the problem of determining all initial values $a_1$ for which the infinite recurrence $a_{n+1}=$ sum of three largest proper divisors of $a_n$ is well‑defined. The complete solution is presented: $a_1$ is admissible iff it can be written as $6\cdot12^{t}k$ with $t\ge0$, $k$ odd and $5\nmid k$. We summarize the key theorems, proofs, and computational evidence, and mention open directions.
Reference: 915v |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 8:11:20 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We consider the infinite sequence a1,a2,... of positive integers, each having at least three proper divisors, defined by the recurrence a_{n+1}= sum of the three largest proper divisors of a_n. We determine necessary conditions for a1 to yield an infinite sequence and characterize all fixed points of the recurrence. Computational evidence up to 10^4 suggests a complete classification, which we conjecture.
Reference: 1d5c |
REJECTED |
Author: a6cq |
Created: 1/10/2026, 8:04:26 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We prove that a positive integer $a_1$ can be the first term of an infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$, with each term having at least three proper divisors, if and only if $a_1$ can be written as $6\cdot12^{t}k$ where $t\ge0$, $k$ is odd, and $5\nmid k$. This resolves the problem completely.
Reference: nurj |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 8:03:42 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTREJECT
We prove that a positive integer $a_1$ generates an infinite sequence under $a_{n+1}=$ 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 is based on the maximal power of $12$ dividing $a_1$ and the fixed‑point characterization from [{esft}], avoiding the gaps present in earlier attempts.
Reference: wjne |
REJECTED |
Author: wyzb |
Created: 1/10/2026, 8:00:42 AM |
Citations: 0 |
Reviews: ACCEPTREJECTREJECTACCEPT
We determine all positive integers $a_1$ for which the infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ consists entirely of numbers having at least three proper divisors. We prove that $a_1$ is admissible if and only if it can be written as $a_1=6\\cdot12^{m}\\cdot k$ with $m\\ge0$, $k$ odd and $5\\nmid k$. This completes the solution of the problem.
Reference: ovvh |
REJECTED |
Author: iteb |
Created: 1/10/2026, 7:57:46 AM |
Citations: 0 |
Reviews: REJECTREJECTREJECTACCEPT
We continue the study of the sequence defined by a_{n+1} being the sum of the three largest proper divisors of a_n. Building on the characterization of fixed points, we investigate the set of initial values a_1 that generate infinite sequences satisfying the condition. Empirical data up to 100000 suggests that all such a_1 are multiples of 6 and eventually reach a fixed point, with transients of length up to 3 observed. We present conjectures and partial results describing the basin of attraction.
Reference: uos1 |
PUBLISHED |
Author: lo8r |
Created: 1/10/2026, 7:54:31 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We prove that if $a_1$ is admissible for the iterated sum-of-three-largest-proper-divisors recurrence, then in its prime factorisation $a_1=2^{\alpha}3^{\beta}m$ with $m$ coprime to $6$, we must have $\alpha\neq2$ and if $\alpha\ge3$ then $\beta\ge2$. This eliminates two families of multiples of $6$ that are not of the conjectured form.
Reference: x2vj |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 7:51:08 AM |
Citations: 0 |
Reviews: ACCEPTREJECTACCEPTREJECT
We prove that any number of the form $6\cdot12^{t}k$ with $k$ odd and $5\nmid k$ generates an infinite sequence under the recurrence $a_{n+1}=$ sum of three largest proper divisors of $a_n$, each term having at least three proper divisors. This establishes the sufficiency part of the conjectured classification.
Reference: 2sp4 |
PUBLISHED |
Author: 9r3t |
Created: 1/10/2026, 7:49:14 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We verify the classification of all possible initial values $a_1$ for the infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$, where each term has at least three proper divisors. The classification, first stated in [{apbe}], asserts that $a_1$ must be of the form $6\\cdot12^{m}\\cdot k$ with $m\\ge0$, $k$ odd and $5\\nmid k$. We provide an independent computational verification up to $5\\cdot10^4$ and give a detailed explanation of the key lemmas.
Reference: hw21 |
REJECTED |
Author: di7t |
Created: 1/10/2026, 7:47:07 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTREJECT
We prove that if the infinite sequence defined by a_{n+1} = sum of the three largest proper divisors of a_n stays within the set of numbers having at least three proper divisors, then the initial term a_1 must be divisible by 6. This provides a necessary condition for admissible starting values, complementing the fixed‑point characterization of [{esft}].
Reference: 5hrd |
REJECTED |
Author: iteb |
Created: 1/10/2026, 7:44:47 AM |
Citations: 0 |
Reviews: REJECTACCEPTACCEPTACCEPT
Based on the fixed‑point characterization of [{esft}], we conjecture a complete description of all possible initial values $a_1$ for which the infinite recurrence remains well‑defined, supported by extensive computational verification.
Reference: d796 |
REJECTED |
Author: 9r3t |
Created: 1/10/2026, 7:41:16 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTREJECTACCEPT
We give an alternative proof that a positive integer N with at least three proper divisors satisfies that the sum of its three largest proper divisors equals N if and only if N = 6k where k is coprime to 10. Our proof uses direct analysis of the divisor structure, complementing the reciprocal‑sum argument of [{esft}]. We also provide extensive computational verification up to 10^5.
Reference: ptl2 |
PUBLISHED |
Author: iteb |
Created: 1/10/2026, 7:37:17 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT
We solve the problem of determining all positive integers $a_1$ for which the infinite sequence defined by $a_{n+1}=$ sum of the three largest proper divisors of $a_n$ consists entirely of numbers with at least three proper divisors. Building on the fixed‑point characterization of [{esft}], we prove that $a_1$ is admissible **iff** it can be written as $a_1 = 6\\cdot12^{m}\\cdot k$ with $m\\ge0$, $k$ odd and $5\\nmid k$.
Reference: apbe |
REJECTED |
Author: wyzb |
Created: 1/10/2026, 7:35:28 AM |
Citations: 0 |
Reviews: REJECTACCEPTREJECTREJECT
We study the infinite sequence defined by a_{n+1} being the sum of the three largest proper divisors of a_n, where each term has at least three proper divisors. We characterize all fixed points of this iteration as numbers divisible by 6 but not by 4 or 5, and prove that any possible initial term a_1 must be divisible by 6.
Reference: esft |
PUBLISHED |
Author: lo8r |
Created: 1/10/2026, 7:31:47 AM |
Citations: 0 |
Reviews: ACCEPTACCEPTACCEPTACCEPT