REVIEW 3 major objections 4 minor 7 references
On a Recursive Integer Sequence Implying the Nonexistence of Odd Perfect Numbers
T0 review · 3 major / 4 minor · reviewed 2026-08-07 · deepseek-v4-flash
Pith's one-line read A halve-or-sum-divisors iteration is conjectured to always reach 1, which would rule out odd perfect numbers.
desk verdict A new recursive prime family with a correct OPN connection, but the main theorem's proof has an unclosed product-multiplicity gap and the conjectures outrun the evidence. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The engine is the map $R$ itself, together with a factorization lemma: $\sigma(p^\alpha)$ is a power of 2 only when $p$ is a Mersenne prime and $\alpha=1$. That gives Corollary 3.2, that $R(n)$ is a power of 2 exactly when $n$ is a square-free product of distinct Mersenne primes. The recursive sets $P_i$ are then built so that a prime $p\in P_i$ satisfies $p+1=2^r q$ with $q\in P_{i-1}$; applying $R$ once and halving $r+1$ times sends $p$ to $q$, reducing the index, and induction on $i$ carries the argument for square-free products. The proof invokes multiplicativity of $\sigma$ to pass from primes to products, which is the step that needs a separate justification.
What would settle it
Compute the orbit of every square-free product of primes from $P_1\cup P_2\cup P_3$ up to $10^6$; the theorem asserts each reaches 1, so a single orbit that cycles without reaching 1 (for example an odd perfect number would create the 2-cycle $N\leftrightarrow 2N$) or that runs beyond $10^5$ steps without reaching 1 would falsify Theorem 3.3. Already visible is the failure of the proof's reduction for $15$: $R^2(15)=12$ while $R^2(3)R^2(5)=9$, so the induction needs a separate argument for products before the theorem follows.
Extended reading notes
Core claim
On its own terms, the paper's central claim is that $R(n)=\sigma(n)$ for odd $n$ and $R(n)=n/2$ for even $n$ has the property that $R^k(n)=1$ for some $k$ for every positive integer $n$ (Conjecture 3.5), and that this property is already provable for every square-free product of primes belonging to the recursively defined sets $P_1=\{3,7,31,127,\ldots\}$ (the Mersenne primes) and $P_i=\{p\text{ prime}: p=2^\alpha p_{i-1}-1\text{ for some }p_{i-1}\in P_{i-1},\ \alpha\in\mathbb{N}\}$ (Theorem 3.3). The link to odd perfect numbers is a two-line argument: if $N$ were odd and perfect, then $\sigma(N)=2N$, so $R(N)=2N$ and $R(2N)=N$, and the orbit would cycle forever instead of reaching 1. The paper also notes that the conjecture would forbid odd $2k$-perfect numbers and odd superperfect numbers.
Load-bearing premise
The argument for the main theorem assumes that proving the orbit reaches 1 for each prime in the recursively defined sets automatically proves it for every square-free product of those primes; this reduction is valid only if the product orbit factors into the prime orbits, which the paper does not show and which already fails at the second step for the product $3\cdot 5$.
Editorial extensions
If this is right
- If Conjecture 3.5 holds, no odd perfect number exists, because such an $N$ would satisfy $R(N)=2N$ and $R(2N)=N$, a cycle that never reaches 1.
- The same conjecture would rule out odd $2k$-perfect numbers for every $k\in\mathbb{N}$, containing odd perfect numbers as the case $k=2$.
- It would also rule out odd superperfect numbers: such an $N$ must be an odd square, making $\sigma(N)$ odd, so the orbit runs $N\to\sigma(N)\to 2N\to N$ instead of reaching 1.
- The theorem supplies a concrete infinite family for which the conjecture is true, extending beyond the Mersenne primes themselves to primes constructed as $2^\alpha p_{i-1}-1$.
Reading between the lines
- The proof of Theorem 3.3 relies on reducing square-free products to their prime factors because $\sigma$ is multiplicative; a direct check on $n=3\cdot 5$ shows the factorization is lost after the first halving step ($R^2(15)=12$ while $R^2(3)R^2(5)=9$), so the induction as written needs an additional closure argument for products before the theorem can be considered established.
- One testable extension is to search computationally for orbits that enter a cycle other than the trivial one; such a cycle would refute Conjecture 3.5. The odd-perfect case would be the special two-cycle $N\leftrightarrow 2N$, so the conjecture is a sufficient condition for the nonexistence of odd perfect numbers, not a necessary one.
- Remark 3.4 suggests the family can be widened by adding primes whose divisor sum factors entirely into primes from the existing sets; systematically closing this family under that rule would produce a larger provable set and could be compared numerically with the sets defined in the paper.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper defines a recursive map R on positive integers by R(n)=sigma(n) if n is odd and R(n)=n/2 if n is even, conjectures that every trajectory eventually reaches 1, and notes that this conjecture would imply the nonexistence of odd perfect numbers. The main unconditional contribution is a family of square-free integers, built from recursively defined sets P_i of primes, for which Theorem 3.3 claims the trajectory reaches 1. The proof attempts to reduce the statement to the case of a single prime via multiplicativity of sigma and an induction on i. Corollary 3.2 and Lemma 3.1 are used to justify the base case.
Significance. If Conjecture 3.5 were true, the argued implication for odd perfect numbers would be correct and significant. The recursive family in Theorem 3.3 is a concrete and potentially useful partial result, and the paper names a specific route by which a resolution of the conjecture would resolve a centuries-old problem. However, the paper's only unconditional theorem is not established by the given proof, and the statement of Lemma 3.1 is at least ambiguous and, on the literal reading, false. The work does not include machine-checked proofs or reproducible code, and the conjectural part is conditional on an unproved statement that is not supported by new evidence beyond the (unproved) family classification.
major comments (3)
- [Theorem 3.3 proof] The first sentence of the proof, 'Since sigma is a multiplicative function, it suffices to prove the statement for prime numbers,' is not justified and is false as a method. For a square-free product n = p_1 ... p_m, one application of R gives sigma(n) = product_j (p_j + 1) = 2^T product_j q_j, where q_j is the odd part of p_j + 1. The q_j need not be distinct, so after T halvings the state is a product with repeated prime factors and is not square-free; sigma can no longer be computed factorwise. For example, n = 5*23 gives R(115) = sigma(115) = 144 = 2^4 * 3^2, so R^4(115) = 9, and R(9) = 13, whereas the product of the individual prime trajectories would give R(3)*R(3) = 16. The induction on i tracks only a single prime p -> q, and no closure statement is proved for products with multiplicities such as q^2. Thus the proof does not establish Theorem 3.3, which is the paper's central positive result.
- [Lemma 3.1] Lemma 3.1 as typeset reads 'sigma(p^alpha) = 2m for some m in N if and only if p is a Mersenne prime and alpha = 1.' If '2m' means twice m, the lemma is false: for p = 3 and alpha = 3, sigma(3^3) = 40 is even but alpha is not 1. If the intended statement is sigma(p^alpha) = 2^m, a power of two, then the typesetting must be corrected throughout the proof, where '2r', '2^beta', and '2m' appear without superscripts. Corollary 3.2 inherits this issue, and since the base case of the main theorem is said to follow from Corollary 3.2, the ambiguity affects the proof of Theorem 3.3.
- [Remark 3.7(i)] The claim that Conjecture 3.5 rules out odd 2k-perfect numbers for every positive integer k is not supported. For k = 1, an odd perfect N gives R(N) = 2N and R(2N) = N, a 2-cycle, contradicting eventual convergence to 1. For k > 1, an odd 2k-perfect N gives R(N) = 2kN and R(2kN) = kN, which need not return to N, so no cycle is exhibited and the stated implication does not follow from the conjecture alone.
minor comments (4)
- [Conjecture 3.5] In the displayed definition of R^{k+1}, the condition 'if xn is odd' should read 'if R^k(n) is odd'; the symbol x appears without definition.
- [Theorem 3.3 proof] The proof refers to an arbitrary i >= 0, but the sets P_i are defined only for i >= 1; the initial set P_1 is used as the base case, so the indexing should be adjusted for consistency.
- [Example 2.2] The computation of sigma(n) is displayed as an unformatted string of factors; rewriting it as a product of (p^{alpha+1}-1)/(p-1) terms would make the verification readable.
- [Introduction] The sentence 'Touchard [Tou53] showed that an odd perfect number must be of one of the forms 36m + 1, 9, 13, 25' is imprecise and appears garbled; the subsequent statement about the refined form 12m + 1 or 36m + 9 should be reconciled with it.
Circularity Check
No circular derivation: the recursive construction and conditional odd-perfect-number implication are not equivalent to their inputs; the invalid product-to-prime reduction in Theorem 3.3 is a correctness gap, not a circular step.
full rationale
The paper contains no step in which a derived result is equivalent to its inputs by construction. The sets P_i are defined recursively and the induction in Theorem 3.3 uses exactly that recursive definition; the main theorem is not obtained by fitting a parameter to the data it then 'predicts'. The odd-perfect-number consequence is explicitly conditional: Remark 3.7 says 'If Conjecture 3.5 is true...', and the chain R(N)=2N, R(2N)=N is a direct consequence of sigma(N)=2N, not an imported theorem. All references are classical external results (Euler, Touchard, Ochem-Rao, etc.), with no load-bearing self-citation. The only serious issue is in the proof of Theorem 3.3, where the sentence 'Since sigma is a multiplicative function, it suffices to prove the statement for prime numbers' is not justified: after halving, the odd part of sigma(n) can have repeated prime factors (e.g. 5*23 gives 3^2), so later applications of sigma are not products of independent prime trajectories. That is a correctness gap, not a circularity: it does not assume Theorem 3.3, and it cannot be exhibited as an equality between the theorem and its inputs. Hence no circular step meets the quoting standard, and the score is 0. The proof gap should be reported as a correctness risk.
Assumptions & free parameters
assumptions (2)
- standard math Multiplicativity of sigma for coprime factors
- domain assumption Conjecture 3.5, that R eventually reaches 1 for every n
Cite this review
Pith. "Pith review of On a Recursive Integer Sequence Implying the Nonexistence of Odd Perfect Numbers." pith.science (2026). https://pith.science/paper/XDKBPYV2
@misc{pith2026250601830,
author = {Pith},
title = {Pith review of: On a Recursive Integer Sequence Implying the Nonexistence of Odd Perfect Numbers},
year = {2026},
howpublished = {\url{https://pith.science/paper/XDKBPYV2}},
note = {Machine review of arXiv:2506.01830}
}
read the original abstract
We define a sequence of positive integers recursively, where each term is determined as follows: starting with a given positive integer, if the term is odd, the next is the sum of its positive divisors; if the term is even, the subsequent term is half the term. In this paper, we conjecture that this sequence eventually reaches one for all initial values. Furthermore, we classify a family of integers for which this conjecture holds.
Reference graph
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Reviewed August 7, 2026 · model on record in the stance chip above.
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