The diagonal generalized Collatz conjecture

For $p\geq0$ , let $(d_p,\alpha_p,\beta_p)_+=(2^{p+1},3\cdot2^p,2^p)_+$ and let $T_p$ be its associated mapping. Diagonal generalized Collatz conjecture. If $p\geq0$ and $p\neq2$ , the triplet is strongly admissible of order one, with unique cycle

\Omega(1)=(1\to2\to2^2\to\cdots\to2^{p+1}\to1),

of length $p+2$ , and every $n\geq1$ satisfies $T_p^{(k)}(n)=1$ for some $k\geq0$ . For $p=2$ , the triplet $(8,12,4)_+$ is strongly admissible of order two, with cycles $\Omega(1)=(1\to2\to4\to8\to1)$ and $\Omega(67)=(67\to102\to156\to236\to356\to536\to67)$ , and every $n\geq1$ satisfies $T_2^{(k)}(n)\in\{1,67\}$ for some $k\geq0$ . This is a special case of the generalized conjecture above and includes the classical Collatz mapping when $p=0$ ; the paper offers computational support but no proof.

The conjecture, formulated by Abderrahman Bouhamidi in January 2026, predicts that every trajectory in this generalized Collatz family reaches the specified cycle or cycles. It includes the classical case $p=0$ .

Known results

Bouhamidi reports tests for $0\leq p\leq25$ and initial values $n\leq10^7$ , supporting the predicted cycles but proving nothing exhaustive.
For the broader family, additional large finite checks are reported, including a case with $p=3$ and $q=1$ ; these do not establish the diagonal conjecture.

January 2026 computational study

Bouhamidi’s paper states the diagonal assertions as conjectures and explicitly offers numerical support rather than a proof. No retrieved source reports a counterexample, proof, independent verification, withdrawal, or retraction.

Current status (as of October 2026): The conjecture remains open; only finite computational support is recorded, including the classical case $p=0$ .

The diagonal generalized Collatz conjecture

Primary source

Progress summary

Solutions 0