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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:

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

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

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 $a_1$: $a_1.a_2a_3a_4\ldots$

For example, the Fibonacci sequence constructed from $\theta=2.956938891377988\ldots$ yields the concatenation $\tau=2.3581321345589\ldots$ Clearly, $\tau \neq \theta$ 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$a_n$：

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

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

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

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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