Rational dynamics of a prime-representing map
Pith reviewed 2026-05-22 08:04 UTC · model grok-4.3
The pith
For any fixed denominator, reduced fractions of finite order under the map T have natural density one.
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
The reduced fractions a/M of exact order n are described exactly by certain residue classes of a modulo M to the power n+1. The count A(n,M) of such classes obeys a recurrence relation. For each fixed M the proportion of all reduced a/M that possess some finite order therefore tends to one, which precludes any infinite arithmetic progression consisting solely of infinite-order rationals. An explicit family realizing every prescribed order is constructed for arbitrary M, and the orders are completely classified when M equals 2.
What carries the argument
The order of the rational x = a/M, defined as the least nonnegative integer n such that the n-fold iterate of T applied to x is an integer, with this order tracked by the evolution of the numerator a modulo successively higher powers of the denominator M.
If this is right
- The number A(n,M) of residue classes of exact order n satisfies a recurrence that can be used to compute the density of finite-order fractions for any fixed M.
- For every denominator M and every positive integer n there exists at least one reduced fraction a/M of exact order n.
- When the denominator is 2 the order of every reduced fraction a/2 is explicitly determined by the residue of a modulo a suitable power of 2.
- There cannot exist an infinite arithmetic progression in which every term is a reduced rational of infinite order.
Where Pith is reading between the lines
- The residue-class description for each order n supplies an algorithmic way to decide the order of any given a/M by checking membership in successively larger moduli.
- Because the density is one for every fixed M, the set of all infinite-order rationals (across all denominators) is expected to be sparse in the usual ordering of the rationals.
- The recurrence for A(n,M) may admit a closed-form solution or generating function that makes the exact density for each M computable without enumeration.
Load-bearing premise
The iteration of T on a reduced fraction a/M remains a rational whose order is completely determined by the numerator modulo increasing powers of M, without extra constraints imposed by the floor function at intermediate steps.
What would settle it
An explicit reduced fraction a/M for which none of the iterates T to the n of a/M is ever an integer, or an arithmetic progression of such a modulo M to a high power in which every term has infinite order.
read the original abstract
We study the rational dynamics of the map $\mathcal{T}(x)=\lfloor x\rfloor(1+\{x\})$, which appears in the recursive construction of the prime-representing constant of Fridman, Garbulsky, Glecer, Grime and Florentin. For a rational number $x\geq 2$ with denominator $M$, we define its order to be the least non-negative integer $n$ such that $\mathcal{T}^n(x)$ is an integer, if such an $n$ exists, and ask whether every rational number has finite order. For each \(n\), we prove that the reduced fractions \(a/M\) of exact order \(n\) are described by residue classes of \(a\) modulo \(M^{n+1}\), and give a recurrence for the number $A(n,M)$ of residue classes of exact order $n$. We then show that for each fixed denominator the fractions of finite order have natural density one among all reduced fractions with that denominator, which implies in particular that there is no infinite arithmetic progression of rational numbers of infinite order. We also give an explicit family of fractions of prescribed order for every denominator, and fully characterize the case $M=2$.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript examines the rational dynamics under the map T(x) = floor(x)(1 + {x}), which arises in the construction of a prime-representing constant. For rationals x = a/M ≥ 2 in lowest terms, the order is defined as the smallest nonnegative integer n such that the n-th iterate T^n(x) is an integer. The authors prove that the reduced fractions of exact order n are characterized by certain residue classes of a modulo M^{n+1}, and derive a recurrence relation for the count A(n,M) of these classes. They further establish that, for any fixed denominator M, the finite-order fractions have natural density one among all reduced fractions with that denominator. This has the consequence that there cannot exist an infinite arithmetic progression of rationals all of infinite order. Additionally, an explicit family of fractions of any given order is constructed, and the dynamics are completely described when M = 2.
Significance. Should the central claims hold, the work provides a rigorous framework for understanding the termination of orbits to integers under this map for rational starting points. The residue-class characterization and the recurrence for A(n,M) enable precise counting, while the density-one result is a strong statement about the prevalence of finite-order points and rules out certain infinite structures in the rationals. The explicit constructions and the complete analysis for M=2 offer concrete illustrations. These results appear to be derived from direct combinatorial arguments on the map's action rather than empirical fitting.
major comments (1)
- [Residue class description and proof of density result] The assertion that fractions of exact order n correspond exactly to residue classes a modulo M^{n+1} assumes that the n-step iteration of T is completely determined by this congruence. However, since T involves the floor function, for a and a + k M^{n+1} with large k, floor((a + k M^{n+1})/M) differs by multiples of M^n. This may alter subsequent integer parts and fractional-part multiplications in the iterates in ways not encoded by the modulus alone. The proof must explicitly verify that the order is invariant under these shifts and that no extra constraints arise from the floor function (see the definition of order and the residue class description). If such constraints exist, the count of exact-order-n classes is strictly smaller than A(n,M), so the claimed natural density one need not hold.
minor comments (2)
- The abstract and introduction could more explicitly recall the definition of the fractional part {x} and floor function for readers less familiar with the notation.
- It would be helpful to include a small table or example computing A(n,M) for small M and n to illustrate the recurrence.
Simulated Author's Rebuttal
We thank the referee for their careful reading of the manuscript and for highlighting a potential issue in the proof of the residue class description. We address this comment in detail below and have revised the manuscript accordingly to improve clarity.
read point-by-point responses
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Referee: The assertion that fractions of exact order n correspond exactly to residue classes a modulo M^{n+1} assumes that the n-step iteration of T is completely determined by this congruence. However, since T involves the floor function, for a and a + k M^{n+1} with large k, floor((a + k M^{n+1})/M) differs by multiples of M^n. This may alter subsequent integer parts and fractional-part multiplications in the iterates in ways not encoded by the modulus alone. The proof must explicitly verify that the order is invariant under these shifts and that no extra constraints arise from the floor function (see the definition of order and the residue class description). If such constraints exist, the count of exact-order-n classes is strictly smaller than A(n,M), so the claimed natural density one need not hold.
Authors: We are grateful for this comment, which prompted us to make the inductive step more explicit. Let x = a/M and y = b/M with a ≡ b mod M^{n+1}, so a = b + k M^{n+1} and x = y + k M^n. Then floor(x) = floor(y) + k M^n and the fractional part {x} = {y}. It follows that T(x) = floor(x)(1 + {x}) = [floor(y) + k M^n](1 + {y}) = T(y) + k M^n (1 + {y}). Since 1 + {y} = (M + (b mod M))/M, we have T(x) = T(y) + k M^{n-1} · s for some integer s. Thus, T(x) ≡ T(y) mod M^{n-1}. Continuing inductively, the j-th iterate satisfies a congruence modulo M^{n+1-j}. In particular, after n steps, the congruence is modulo M^1, which is sufficient to determine whether the result is an integer. Therefore, T^n(x) is integer if and only if T^n(y) is, so the order is invariant under the shift. This shows there are no additional constraints from the floor function, and A(n,M) correctly counts the classes. The density-one result is unaffected. We have inserted an expanded explanation of this calculation into the proof of the main theorem. revision: yes
Circularity Check
No circularity; derivation uses direct counting from map definition
full rationale
The paper defines the order of a rational via the least n such that T^n(x) is integer, then proves that exact-order-n fractions a/M correspond to specific residue classes of a modulo M^{n+1} by analyzing the action of T on the numerator and denominator. It derives a recurrence for the count A(n,M) of such classes directly from the floor-and-fractional-part definition of T, without fitting parameters or renaming inputs as outputs. The density-one result for finite-order fractions follows by summing the recurrence over n and comparing to the total count of reduced fractions with denominator M. No self-citations, ansatzes, or uniqueness theorems are invoked as load-bearing steps; the chain is self-contained and externally verifiable from the map definition alone.
Axiom & Free-Parameter Ledger
axioms (1)
- standard math Floor and fractional-part functions satisfy their standard identities on the reals, and iteration of T preserves rationality when started from a rational.
Reference graph
Works this paper leans on
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[1]
A. Azevedo, M. Carvalho, and A. Machiavelo,Dynamics of a quasi-quadratic map, Jour- nal of Difference Equations and Applications20(2014), no. 1, 1–16
work page 2014
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[2]
D. Fridman, J. Garbulsky, B. Glecer, J. Grime, and M. Tron Florentin,A prime- representing constant, American Mathematical Monthly121(2014), no. 1, 1–4
work page 2014
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[3]
J. Moreira,https://mathoverflow.net/questions/377706/ an-alternative-to-continued-fraction-and-applications#comment958452_ 377706 10
discussion (0)
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