Generalized Collatz convergence and cycle conjecture for the triplets (2p+1,3×2p,2p)+(2^{p+1},3\times2^p,2^p)_+

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For p≥0p\geq 0, let (dp,αp,βp)+=(2p+1,3×2p,2p)+(d_p,\alpha_p,\beta_p)_+=(2^{p+1},3\times2^p,2^p)_+ and define

Tp(n)={n/2p+1,n≡0(mod2p+1),(3n+[n]p)/2,n≢0(mod2p+1),T_p(n)=\begin{cases}n/2^{p+1},&n\equiv0\pmod {2^{p+1}},\\\\(3n+[n]_p)/2,&n\not\equiv0\pmod {2^{p+1}},\end{cases}

where [n]p[n]_p denotes the remainder in Euclidean division by dp=2p+1d_p=2^{p+1}. The conjecture has separate assertions for p≠2p\ne2 and p=2p=2.

Generalized Collatz convergence and cycle conjecture. For all p≥0p\geq0 with p≠2p\ne2, the triplet (2p+1,3×2p,2p)+(2^{p+1},3\times2^p,2^p)_+ is strongly admissible of order one, with unique trivial cycle

Ω(1)=(1→2→22→⋯→2p+1→1),\Omega(1)=\bigl(1\rightarrow2\rightarrow2^2\rightarrow\cdots\rightarrow2^{p+1}\rightarrow1\bigr),

and for every integer n≥1n\geq1 there exists k≥0k\geq0 such that Tp(k)(n)=1T_p^{(k)}(n)=1. For p=2p=2, (8,12,4)+(8,12,4)_+ is strongly admissible of order two, with trivial cycles Ω(1)=(1→2→4→8→1)\Omega(1)=(1\rightarrow2\rightarrow4\rightarrow8\rightarrow1) and Ω(67)=(67→102→156→236→356→536→67)\Omega(67)=(67\rightarrow102\rightarrow156\rightarrow236\rightarrow356\rightarrow536\rightarrow67), and for every n≥1n\geq1 there exists k≥0k\geq0 such that T2(k)(n)∈{1,67}T_2^{(k)}(n)\in\{1,67\}.

This is a generalized Collatz-type convergence claim: it specifies all asserted trivial cycles and requires every positive integer to reach one of their representatives. The supplied source gives no resolution of these assertions.

References

Primary source

Abderrahman Bouhamidi, “Weakly and Strongly Admissible Triplets for a Collatz-Type Map”, arXiv:2601.17573 (2026).

Progress summary

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Claimed progress

A January 2026 preprint reports extensive finite checks supporting the conjecture, but no proof or counterexample has been found.

The conjecture, formulated as Conjecture 3.33.3 in a January 2026 preprint, asserts convergence to the specified cycle for p≠2p\ne2 and to one of two specified cycles for p=2p=2, with no other cycles.

January 2026 computational checks

The preprint reports tests for 0≤p≤250\le p\le25, 0≤q≤p0\le q\le p, and n≤107n\le10^7; for (p,q)=(3,1)(p,q)=(3,1) it additionally checks starting values up to approximately 5×10115\times10^{11}. These computations support the claimed cycle behavior but do not establish convergence for every positive integer or exclude further cycles.

Current status (as of October 2026): The conjectured convergence and cycle classification for both p≠2p\ne2 and p=2p=2 remain unproved; only finite computational support is recorded.

Sources

Solutions 0

No solutions have been posted yet.