This posting consists of several mildly-related questions, motivated from this posting. The main object is the following sequence.

$$a_0 = 0, \qquad a_1 = x, \qquad a_{n+1} = \frac{n}{a_n} - a_n - a_{n-1}. \tag{*}$$

Question 1. Numerical experiment suggests that there is a unique value of $x$ for which $a_n > 0$ for all $n \geq 1$. Can we prove/disprove this?

If we write $I_n = \{ x \in \mathbb{R} : a_1 > 0, \cdots, a_n > 0\}$, then obviously $I_n$ is a nested sequence of open sets that begins with $I_1 = (0, \infty)$. Moreover, the experiment suggests that $I_n$ are all intervals, and the endpoints of $I_n$ are adjacent poles of $a_{n+1}$ and $a_{n+1}$ is strictly monotone on $I_n$. Provided this is correct, we easily see that there is a unique zero of $a_{n+1}$ on $I_{n+1}$, which then determines $I_{n+1}$.

Plots of a_{n+1} on I_n

Question 2. The same experiment also suggests that the value of such unique $x$ is

$$ \frac{\operatorname{AGM}(1,\sqrt{2})}{\sqrt{\pi}} = \frac{2\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \approx 0.675978240067284728995\cdots.$$

At this point, I completely have no idea why this value arises, but I have checked that this is correct up to hundreds of digits. (I progressively refined the range of $x$ so that $a_n$ stays positive for a longer time.) Again, will it ever have a chance to be proved?

My original suspicion was that we may rearrange the recurrence relation to obtain continued fraction, but it was of no avail. To be honest, I have never seen this type of problem, and will be glad if I can learn anything new about it.

Question 3. Given that the above question seems to bold to answer, perhaps we may consider its variants:

  1. (Variant 1) $a_0 = 0$, $a_1 = x$, and $a_{n+1} = \frac{n}{a_n} - a_n - p a_{n-1}$, where $p \in \mathbb{R}$.

  2. (Variant 2) $a_0 = 0$, $a_1 = x$, and $a_{n+1} = \frac{1}{a_n} - a_n - a_{n-1}$.

  3. (Variant 3) $a_0 = 0$, $a_1 = x$, and $a_{n+1} = \frac{n^2}{a_n} - a_n - a_{n-1}$.

Again, in each case, numerical experiment suggests that there is a unique $x$ for which $(a_n)_{n\geq 1}$ stays positive. Moreover,

  • For Variant 1, it seems that $x = 1/\sqrt{3}$ for $p = -2$ but I have no guess for general $p$, even when it is an integer.

  • For Variant 2, it is conjectured that $x = 4/3\sqrt{3}$.

  • For Variant 3, we can check that $x = 1/\sqrt{3}$ is such one. Indeed, we find that $a_n = n/\sqrt{3}$ solves the recurrence relation.

Then we may ask whether the version of Question 1-2 can be proved for these variants.


  1. I managed to answer Question 1. Check this answer.
  • 3
    $\begingroup$ This. Is. Hard. $\endgroup$ – DinosaurEgg Jun 5 at 2:36
  • $\begingroup$ Another variant is $a_0=w$, $a_1=x$, $a_{n+1} = n/a_n - a_n - a_{n-1}$. (So the original problem is $w=0$.) Numerical experimentation suggests that there is a curve of values $(w,x)$ for which the recurrence stays positive, and this curve slants gently down from roughly $(0, 0.7)$ to roughly $(1,0.4)$. $\endgroup$ – David E Speyer Jun 6 at 13:55
  • 2
    $\begingroup$ Continuing the experiments, if $x_n$ is the unique zero of $a_{n+1}$ on $\color{red}{I_n}$ (typo?), and $x=2\Gamma(3/4)/\Gamma(1/4)$ is the (conjectured) limit, then computations suggest $$\lim_{n\to\infty}\frac{x_n-x}{x-x_{n+1}}=2+\sqrt{3}.$$ $\endgroup$ – metamorphy Jun 7 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.