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Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?

The question extends as follows (January 16). Let

\begin{align*} r_{1,n} &= 1/c_n\\ r_{2,n} &= 1/r_{1,n}\\ r_{3,n} &= 1/r_{2,n}\\ &\vdots\\ s_{1,n} &= \lfloor{1/\{r_{1,n}\}}\rfloor, \text{ where }\{ \} \text{ denotes fractional part}\\ s_{2,n} &= \lfloor{r_{2,n}} \rfloor \\ s_{3,n} &= \lfloor{r_{3,n}} \rfloor \\ &\vdots\\ \end{align*}

Are $(s_{2,n})=(1,1,2,2,3,3, \ldots)$ and $(s_{3,n})=(2,1,3,1,4,1,6,1,7,1, \ldots)$, where the only numbers missing are $5+4h$ for $h \geq0$? In general, does the regularity of the continued fraction for $e$ imply some sort of regularity for $(s_{k,n})$ for $k \geq 5$?

Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?

The question extends as follows (January 16). Let

\begin{align*} r_{1,n} &= 1/c_n\\ r_{2,n} &= 1/r_{1,n}\\ r_{3,n} &= 1/r_{2,n}\\ &\vdots\\ s_{1,n} &= \lfloor{1/\{r_{1,n}\}}\rfloor, \text{ where }\{ \} \text{ denotes fractional part}\\ s_{2,n} &= \lfloor{r_{2,n}} \rfloor \\ s_{3,n} &= \lfloor{r_{3,n}} \rfloor \\ &\vdots\\ \end{align*}

Are $(s_{2,n})=(1,1,2,2,3,3, \ldots)$ and $(s_{3,n})=(2,1,3,1,4,1,6,1,7,1, \ldots)$, where the only numbers missing are $5+4h$ for $h \geq0$? In general, does the regularity of the continued fraction for $e$ imply some sort of regularity for $(s_{k,n})$ for $k \geq 5$?

Informally speaking, things look very irregular for higher $k$, and I wonder if the method in Answer 1 extends to these deeper cases.

Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?

Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?

The question extends as follows (January 16). Let

\begin{align*} r_{1,n} &= 1/c_n\\ r_{2,n} &= 1/r_{1,n}\\ r_{3,n} &= 1/r_{2,n}\\ &\vdots\\ s_{1,n} &= \lfloor{1/\{r_{1,n}\}}\rfloor, \text{ where }\{ \} \text{ denotes fractional part}\\ s_{2,n} &= \lfloor{r_{2,n}} \rfloor \\ s_{3,n} &= \lfloor{r_{3,n}} \rfloor \\ &\vdots\\ \end{align*}

Are $(s_{2,n})=(1,1,2,2,3,3, \ldots)$ and $(s_{3,n})=(2,1,3,1,4,1,6,1,7,1, \ldots)$, where the only numbers missing are $5+4h$ for $h \geq0$? In general, does the regularity of the continued fraction for $e$ imply some sort of regularity for $(s_{k,n})$ for $k \geq 5$?

Let

\begin{align*} c_n &= n!(e-\sum_{k=0}^n \frac{1}{k!}) \\ u_n &= \lfloor{\frac{1}{c_n} \rfloor} \\ v_n &= \lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?

Let

\begin{align*} c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\ \\ u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\ \\ v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}} \bigg\rfloor \end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$?