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

From papers

For p0p\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,n0(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 p2p\ne2 and p=2p=2.

Generalized Collatz convergence and cycle conjecture. For all p0p\geq0 with p2p\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)=(12222p+11),\Omega(1)=\bigl(1\rightarrow2\rightarrow2^2\rightarrow\cdots\rightarrow2^{p+1}\rightarrow1\bigr),

and for every integer n1n\geq1 there exists k0k\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)=(12481)\Omega(1)=(1\rightarrow2\rightarrow4\rightarrow8\rightarrow1) and Ω(67)=(6710215623635653667)\Omega(67)=(67\rightarrow102\rightarrow156\rightarrow236\rightarrow356\rightarrow536\rightarrow67), and for every n1n\geq1 there exists k0k\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.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

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

Solutions 0

No solutions have been posted yet.