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.

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.

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.

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 $50000$.

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.

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.

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.

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 $2$.

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 $a_1$.

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.

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.

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.

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.

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.

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.

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.

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}].

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.

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 $5\\nmid k$.