In this paper we introduce and discuss the sequence of \emph{real numbers} defined as $u_0 \in \mathbb R$ and $u_{n+1} = \Delta(u_n)$ where \begin{equation*} \Delta(x) = \begin{cases} \frac{x}{2} &\text{if } \operatorname{frac}(x)<\frac{1}{2} \\[4px] \frac{3x+1}{2} & \text{if } \operatorname{frac}(x)\geq\frac{1}{2} \end{cases} \end{equation*} This sequence is reminiscent of the famous Collatz sequence, and seems to exhibit an interesting behaviour. Indeed, we conjecture that iterating $\Delta$ will eventually either converge to zero, or loop over sequences of real numbers with integer parts $1,2,4,7,11,18,9,4,7,3,5,9,4,7,11,18,9,4,7,3,6,3,1,2,4,7,3,6,3$. We prove this conjecture for $u_0 \in [0, 100]$. Extending the proof to larger fixed values seems to be a matter of computing power. The authors pledge to offer a reward to the first person who proves or refutes the conjecture completely -- with a proof published in a serious refereed mathematical conference or journal.

中文翻译：

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分数 $3n+1$ 猜想

在本文中，我们介绍和讨论定义为 $u_0 \in \mathbb R$ 和 $u_{n+1} = \Delta(u_n)$ where \begin{equation*} \Delta (x) = \begin{cases} \frac{x}{2} &\text{if } \operatorname{frac}(x)<\frac{1}{2} \\[4px] \frac{3x+ 1}{2} & \text{if } \operatorname{frac}(x)\geq\frac{1}{2} \end{cases} \end{equation*} 这个序列让人想起著名的 Collat​​z 序列，并且似乎表现出一种有趣的行为。事实上，我们推测迭代 $\Delta$ 最终要么收敛到零，要么遍历整数部分 $1,2,4,7,11,18,9,4,7,3,5,9 的实数序列,4,7,11,18,9,4,7,3,6,3,1,2,4,7,3,6,3$。我们为 $u_0 \in [0, 100]$ 证明了这个猜想。将证明扩展到更大的固定值似乎是计算能力的问题。