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Problem 751

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Concatenation Coincidence

A non-decreasing sequence of integers $a_n$ can be generated from any positive real value $\theta$ by the following procedure:

$$b_1=\theta$$
$$b_n=\lfloor b_{n-1}\rfloor (b_{n-1}-\lfloor b_{n-1}\rfloor +1)\mathrm{\forall }n\ge 2$$
$$a_n=\lfloor b_n\rfloor$$
Where $\lfloor .\rfloor$ is the floor function.

For example, $\theta =2.956938891377988\dots$ generates the Fibonacci sequence: $2,3,5,8,13,21,34,55,89,\dots$

The concatenation of a sequence of positive integers $a_n$ is a real value denoted $\tau$ constructed by concatenating the elements of the sequence after the decimal point, starting at $a_1$: $a_1.a_2{a}_3{a}_4\dots$

For example, the Fibonacci sequence constructed from $\theta =2.956938891377988\dots$ yields the concatenation $\tau =2.3581321345589\dots$ Clearly, $\tau \ne \theta$ for this value of $\theta$.

Find the only value of $\theta$ for which the generated sequence starts at $a_1=2$ and the concatenation of the generated sequence equals the original value: $\tau =\theta$. Give your answer rounded to $24$ places after the decimal point.

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拼接巧合

给定任意正实数$\theta$，根据以下过程可以生成一个不下降的整数序列$a_n$：

$$b_1=\theta$$
$$b_n=\lfloor b_{n-1}\rfloor (b_{n-1}-\lfloor b_{n-1}\rfloor +1)\mathrm{\forall }n\ge 2$$
$$a_n=\lfloor b_n\rfloor$$
其中$\lfloor .\rfloor$表示下取整函数。

例如，$\theta =2.956938891377988\dots$能够生成斐波那契数列：$2,3,5,8,13,21,34,55,89,\dots$

数列$a_n$的拼接是指将$a_1$作为整数部分、将序列的其它元素拼接作为小数部分所生成的实数，记为$\tau$。

例如，由$\theta =2.956938891377988\dots$构造的斐波那契数列，其拼接即为$\tau =2.3581321345589\dots$。显然，对于$\theta$的这一取值，有$\tau \ne \theta$。

对于所有生成序列中$a_1=2$的$\theta$，求出唯一使得生成序列的拼接恰好等于原实数，也即使得$\tau =\theta$的$\theta$的取值，并将你的答案保留小数点后$24$位。

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