Marcus’s Tent-Map Cycle Length

Abstract

Marcus’s tent-map cycle length equals the signed suborder of two.

The tent map on [0,1] sends x to 1-|2x-1|. The symbols n, N, k, and m denote natural numbers; N=2n+1 in the theorem. For N>0 and k<=N, tent(N,k) is its numerator action at the rational point k/N. Subtraction is natural subtraction, with no truncation in either orbit branch. The function minimalPeriod gives the least positive return time of a point, or zero when there is no positive return. The signed suborder of 2 modulo 2n+1 is the least m>0 with 2^m congruent to 1 or -1; the latter residue is written 2n. Congruence means that the modulus divides the integer difference, and IsLeast asserts membership in the displayed set and a lower bound for every member. Only Marcus’s Jul 16 2025 cycle-length sentence in A003558 is settled here. Kaprekar cycles, x^2-2 iteration, and the Kappraff-Adamson base conjecture are not claimed. The hypothesis n>0 excludes n=0, where N=1 and 2/N is outside [0,1].

Definition 1.1 (The numerator tent map).

$$\forall N\in \mathbb{N},k\in \mathbb{N},\mathrm{tent}\left(N,k\right)=\{\begin{array}{ll}2\cdot k& (2\cdot k\le N)\\ 2\cdot N-2\cdot k& (N<2\cdot k)\end{array}$$

Formalization. D5/S1/Recurrence/MarcusTentMapSuborder.tent (✓ std3).

Citation. N. J. A. Sloane; Michel Marcus (2025). OEIS A003558, least m with 2^m = ±1 mod 2n+1, with Marcus’s tent-map cycle-length conjecture. URL: https://oeis.org/A003558.

Commentary.

If 2k<=N, the absolute-value expression equals 2k/N. If N<2k and k<=N, it equals (2N-2k)/N, and the nonnegative numerator agrees with natural subtraction. The displayed definition is total on natural N and k; its interpretation as a rational tent-map numerator uses N>0 and k<=N.

Theorem 1.2 (The cycle-length conjecture).

$$\forall n\in \mathbb{N},(0<n)\Rightarrow (\mathrm{IsLeast}\left(\left\{m\in \mathbb{N}\mid (0<m)\wedge ((2^m\equiv 1(\mathrm{mod}2\cdot n+1))\vee (2^m\equiv 2\cdot n(\mathrm{mod}2\cdot n+1)))\right\},\mathrm{minimalPeriod}\left(\mathrm{tent}\left(2\cdot n+1\right),2\right)\right))$$

Proof. Machine-checked in Lean as D5/S1/Recurrence/MarcusTentMapSuborder.result (✓ std3). ∎

Resolves. Problems/oeis-a003558-marcus-tent-map-suborder (proved) by D5/S1/Recurrence/MarcusTentMapSuborder.result.

Citation. N. J. A. Sloane; Michel Marcus (2025). OEIS A003558, least m with 2^m = ±1 mod 2n+1, with Marcus’s tent-map cycle-length conjecture. URL: https://oeis.org/A003558.

Commentary.

For N=2n+1, induction shows that every iterate from numerator 2 stays even and at most N, and represents either sign of 2^(m+1) modulo N. Since N is odd, the even representative of either sign of 2 in this interval must be 2. Cancelling the unit 2 therefore makes a return at time m equivalent to 2^m being congruent to 1 or -1. Euler’s theorem supplies a positive return, and minimalPeriod gives the least one. Division by the fixed positive N preserves equality of numerators, so these are also the return times of 2/N.

References

• Truth anchor: D5/S1/Recurrence/MarcusTentMapSuborder.result
• Truth anchor: D5/S1/Recurrence/MarcusTentMapSuborder.tent