… is a mathematical problem that is very easy to state, but not easy to solve. It is described here:
https://en.wikipedia.org/wiki/Collatz_conjecture
As long as no AI has taken on this problem, we humans can continue to think about it. Here come…
Some new Facts on the Collatz Conjecture (2026)
We introduce the „accelerated“ Collatz function $\bar{c}(n)$, which combines several steps of the iteration and
thus accelerates it significantly compared to the classic Collatz function $C(n)$ and the Terras variant $T(n)$.
For this purpose, instead of the initial value $n$, we consider the number $n+1$ and terminate the iteration as soon as a smaller number is reached. This yields several interesting results:
For this purpose, instead of the initial value $n$, we consider the number $n+1$ and terminate the iteration as soon as a smaller number is reached. This yields several interesting results:
1. Definition & Elementary Properties
- $2$-adic Transformation: For odd numbers $n = r \cdot 2^k – 1$, $\bar{c}(n) = (3^k r – 1)/2^q$ combines all divisions by $2$ into a single step.
For even numbers $n = r \cdot 2^k$, it is analogously $\bar{c}(n) = r$.
The parameter pairs $(k_i, q_i)$ play a central role for further analysis. - Always odd: For all $n \in \mathbb{N}$, $\bar{c}(n)$ is always odd.
- Fixed point at the end: $\bar{c}(1) = 1$ is a true fixed point of the iteration.
- 75% Immediate Stop: All even numbers $n \equiv 0 \pmod 2$ as well as all numbers in the residue class $n \equiv 1 \pmod 4$ drop below the initial value in exactly $1$ step ($\bar{c}(n) < n$). Only $n \equiv 3 \pmod 4$ remains for deeper analysis.
- Extreme values: Only 5 known initial values satisfy the maximum excursion rate $\rho_{\bar{c}}(n_0) > 2$ and temporarily reach a value $n_t > n_0^2$ during iteration: $n_0 \in \{27, 3716509988199, 1980976057694848447, 2971464086542272671, 3961952115389696895\}$.
$75\%$
Stopping in $1$ step
$6\times \text{ to } 10\times$
Speedup vs. $C(n)$
$\approx 2.5$
Average stopping time
2. Acceleration Factors & Algebraic Structure
- Shortening: For $n_0 = 27$, $\bar{c}(n)$ requires only $15$ steps to drop (compared to $59$ for $T(n)$ and $96$ for $C(n)$). For $n_0 = 1023$, $2$ steps are sufficient ($1023 \to 7381 \to 173$).
- Speedup: A single step under $\bar{c}(n)$ corresponds on average to $2k+q$ steps of the standard function $C(n)$. The expected values are $6$ for $n \equiv 3 \pmod 8$ and $10$ for $n \equiv 7 \pmod 8$, respectively.
- Linearity: $t$ steps map the initial value to $n_t = (3^{K_t} n_0 + B_t)/2^{S_t}$ with $K_t = \sum_{i=1}^t k_i$ and $S_t = K_t + \sum_{i=1}^t q_i$; $B_t$ is a recursively defined sum. Thus, a formal relation exists between the initial and final value.
- Inversion: Inversion forces residue classes of the form $n_0 \equiv 3^{-K_t}(2^{S_t} – B_t) \pmod{2^{S_t+1}}$ as initial values. To invert $27 \to 31 \to 121$, we have $K_t=7$, $S_t=9$, and $B_t=2903$. This yields $n_0 \equiv 27 \pmod{1024}$.
3. Asymptotic Behavior & Prefix Stabilization
- One-Step Reduction: For $n \equiv 3 \pmod 4$, the probability of dropping in the first step is exactly $\mathbb{P}(n_1 < n_0) = \sum_{k=2}^{\infty} 2^{1-k – \lfloor \alpha k \rfloor} \approx 42.745\%$ (with $\alpha = \log_2 3 – 1 \approx 0.58496$).
- Drift at $n = 2^m – 1$: The number of steps follows on average the linear regression $\mathcal{L}(\bar{c}, 2^m-1) \approx 0.7047\,m – 1.2047\,q_1 + 1$.
- Prefix Stabilization: For families of initial values $10^m + j$, all parameter pairs $(k_i, q_i)$ of the first $t$ steps match those of $j$ as soon as $m \ge S_t + 1$ holds. Example: $10^{60} + 31$ requires exactly as many steps as $31$ itself.
- Stopping Time Constancy: There even exists a bound $M(j)$ beyond which the stopping time $\mathcal{L}(\bar{c}, 10^m+j)$ is absolutely constant (e.g., exactly $14$ steps for $10^m + 31$ for all $m \ge M(31)=56$).
- Average Stopping Time: For all $n \equiv 3 \pmod 4$, the average number of steps for $\bar{c}(n)$ remains constant at approx. $2.5$ and does not increase further like in $C(n)$ and $T(n)$.