I was thinking about this question asked at Math.SE, when I came up with the following conjecture.

For every $q\in\mathbb Q$ consider a sequence $s_n^{(q)}$ (terms within the sequence are indexed by $n\in\mathbb N$): $$s_n^{(q)}=\left\lfloor\frac{n!\cdot q}e\right\rfloor-\,!n\cdot q$$ ($!n$ denotes the subfactorial).

Conjecture: For every $q$ the sequence $s_n^{(q)}$ is periodic. Furthemore, if $q$ is a positive integer is a reciprocal of a positive integer $m$, then the period is: $$p_{1/m}=\begin{cases}m,\quad m\text{ is even}\\2m,\quad m\text{ is odd}\end{cases}$$

Could you suggest any ideas how to prove this conjecture?

  • 3
    $\begingroup$ $1/e=\sum_{k=0}^\infty\frac{(-1)^k}{k!}$, so your formula is roughly the tail of the series multiplied by $n!\times q$. $\endgroup$ – Wadim Zudilin Nov 3 '15 at 5:21
  • $\begingroup$ Sorry, I made a mistake when I formulated conditions for $m$ or $2m$ period. It happens not when $q$ is an integer, but when it's reciprocal of an integer. $\endgroup$ – Vladimir Reshetnikov Nov 3 '15 at 17:41

Write $$ \frac{n!\,e^{-1}}m=x_n+\frac{(-1)^{n+m+1}y_n}m+\frac{(-1)^{n+1}\delta_n}m \qquad\text{and}\qquad \frac{!n}m=x_n+\frac{(-1)^{n+m+1}y_n}m, $$ where $$ x_n=\frac{n!}{m\cdot(n-m)!}\cdot(n-m)!\sum_{k=0}^{n-m}\frac{(-1)^k}{k!}\in\mathbb Z, \\ y_n=n!\sum_{k=n-m+1}^n\frac{(-1)^{k-(n-m+1)}}{k!}\in\mathbb Z, $$ and $$ \delta_n=\frac1{n+1}\biggl(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\dotsb\biggr) $$ satisfies $0<\delta_n<1$ (even this rough estimate is sufficient). Note that $$ y_n=n(n-1)\dotsb(n-m+1)-n(n-1)\dotsb(n-m+2)+\dotsb+(-1)^{m-1} $$ is $m$-periodic modulo $m$, because each factor in each term of the latter sum is; in other words, $\{y_n/m\}$ (the fractional part) is periodic with period $m$. It remains to observe that $$ \biggl\lfloor\frac{n!\,e^{-1}}m\biggr\rfloor-\biggl\lfloor\frac{!n}m\biggr\rfloor =\biggl\lfloor\frac{(-1)^{n+m+1}(y_n+(-1)^m\delta_n)}m\biggr\rfloor-\frac{(-1)^{n+m+1}y_n}m $$ and that $$ \biggl\lfloor\frac{\pm(y_n+(-1)^m\delta_n)}m\biggr\rfloor =\biggl\lfloor\frac{\pm y_n}m\biggr\rfloor \quad\text{or}\quad \biggl\lfloor\frac{\pm(y_n-1)}m\biggr\rfloor, $$ respectively, depending on whether $m$ is even or odd. The above argument works for $1/m$ replaced with $\ell/m$, though to guarantee the estimate for $\delta_n$ one needs $|\ell|\ge n$, so that the periodicity will be eventual; the period can double if $\ell<0$.

Proving that $m$ or $2m$ is the minimal period is equivalent to showing that the $m$ residues $(-1)^{m-1}y_0,(-1)^{m-1}y_1,\dots,(-1)^{m-1}y_{m-1}$ are not periodic modulo $m$: $$ (-1)^{m-1}y_0=1, \quad (-1)^{m-1}y_1=1-1, \quad \ldots, \\ (-1)^{m-1}y_k=1-k+k(k-1)-k(k-1)(k-2)+\dots+(-1)^kk!\,, \quad \ldots $$ This seems to be correct but the distribution of those residues modulo $m$ is quite chaotic.


Since $$ \frac1e=e^{-1}=\sum_{k=0}^\infty\frac{(-1)^k}{k!}, $$ we can write $$ q\,n!\,e^{-1}=q\,n!\sum_{k=0}^n\frac{(-1)^k}{k!} +(-1)^{n+1}\frac q{n+1}\biggl(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\cdots\biggr) =t_n+(-1)^{n+1}\epsilon_n, $$ where $t_n\in\mathbb Z$ and $0<\epsilon_n<1$ for $n\ge q$. Then $\lfloor q\,n!\,e^{-1}\rfloor=t_n$ if $n$ is odd and $=t_n-1$ if $n$ is even. Furthermore, $t_n={}!n\cdot q$. In other words, the sequence $s_n^{(q)}$ alternates with period 2 between $0$ and $-1$ for $n\ge q$ (that is, eventually).

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    $\begingroup$ In the last sentence, did you mean that $q$ is an integer? $\endgroup$ – Vladimir Reshetnikov Nov 3 '15 at 17:42
  • 1
    $\begingroup$ Yes, of course, as it was a positive integer at the time the answer was posted. $\endgroup$ – Wadim Zudilin Nov 11 '15 at 2:09

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