# Showing that $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ with $a_0 = 0$ and $a_1=2\Gamma(\frac34)\big/\Gamma(\frac14)$ stays positive for $n\geq1$.

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 $$I_{n+1}$$.

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.

Progress.

• 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 $(1,0.4)$. – David E Speyer Jun 6 at 13:55
• 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 $$\lim_{n\to\infty}\frac{x_n-x}{x-x_{n+1}}=2+\sqrt{3}.$$ – metamorphy Jun 7 at 7:28