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Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$.

Find all $a$ such that $(a_n)$ is a sequence of positive reals.


My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>0$ gives $a>0$, $a_2>0$ gives $a\in(0,1)$, $a_3>0$ gives $a\in(\frac{1}{\sqrt{3}},1)$, but this probably doesn't give important information and further terms are nasty.

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    $\begingroup$ I have solved, with Mathematica, the system $a_n>0;\;n.1,2,3\ldots,15$ and found $0.6759782<a<0.6759783$. This makes me think that there is only ONE value of $a$ which makes all the sequence positive, but I can't find what it is exactly $\endgroup$ – Raffaele Nov 18 '17 at 16:07
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    $\begingroup$ I can't explain why, but numerical simulation using $a_1, \cdots, a_{40}$ pins down the range of $a$ up to a relative error of $10^{-20}$, and the result suggests that there exists a unique value of $a$ for which all $(a_n)$'s are positive, and the value is exactly $$a = \frac{2\Gamma(3/4)}{\Gamma(1/4)} \approx 0.67597824006728472900\cdots, $$ where $\Gamma(\cdot)$ is the Gamma function. For this choice of $a$, the asymptotic form of $a_n^2$ is likely $$a_n^2 = \frac{n}{3} + \frac{1}{36n} + \mathcal{O}\left(\frac{1}{n^3}\right). $$ $\endgroup$ – Sangchul Lee Jun 4 at 4:19
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    $\begingroup$ Having looked up [particular values of the Gamma function][1], it turns out that $\frac{2\Gamma(3/4)}{\Gamma(1/4)}=\sqrt{\pi}AGM(\sqrt{2},1)$, where $AGM(x,y)$ is the [Arithmetic-Geometric mean][2]. [1]: en.wikipedia.org/wiki/Particular_values_of_the_gamma_function [2]: en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean $\endgroup$ – Angela Richardson Jun 4 at 15:59
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    $\begingroup$ Indeed, my computations confirm the asymptotics $$a_n\sim\sqrt{\frac{n}{3}}\sum_{k=0}^{(\infty)}\frac{c_k}{n^{2k}},\quad c_0=1,\quad \sum_{k=0}^{n}c_{n-k}\left(c_k+\frac{2}{3}\sum_{r=0}^{k-1}\binom{2k-3/2}{2k-2r}c_r\right)=0$$ which, modulo existence, follows from the recurrence. $\endgroup$ – metamorphy Jun 5 at 17:18
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    $\begingroup$ This problem is problem number 6 on the 2003 Miklos Shweitzer maths competition, a difficult hungarian maths competition, there may be solutions somewhere on the internet. $\endgroup$ – user277182 Jun 7 at 11:17
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Finally I resolved the issue of existence and uniqueness. We will regard each $a_n$ as rational function of $a_1$. That is, we will consider the sequence of rational functions

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

We would like to address the set of all $x\in\mathbb{R}$ for which $a_n(x) > 0$ for all $n \geq 1$. To this end, write

$$ I_n = \{ x \in \mathbb{R} : a_1(x) > 0, \cdots, a_n(x) > 0 \}, \qquad I_{\infty} = \bigcap_{n=1}^{\infty} I_n. $$

Then the solution set is exactly $I_{\infty}$, which we seek to identify. In this answer, we will prove that $I_{\infty}$ is a singleton, i.e.,

Proposition. There exists a unique $x \in \mathbb{R}$ for which $a_n(x) > 0$ for all $n \geq 1$.


We know that $I_1 = (0, \infty)$. Now we prove the following statement:

Claim. For any $n \geq 2$, the followings hold:

  1. $I_n = (\alpha, \beta)$ for some $0 \leq \alpha < \beta < \infty$. Moreover,
    • If $n$ is even, then $a_{n-1}(\alpha) = 0$ and $a_n(\beta) = 0$,
    • If $n$ is odd, then $a_{n-1}(\beta) = 0$ and $a_n(\alpha) = 0$.
  2. $(-1)^k a_{k+1}'(x) \geq (-1)^{k-1}a_k'(x) \geq 1$ for all $1 \leq k \leq n-1$ and $x \in I_n$.

Proof of Claim. We invoke induction.

  • (Base case) The claim is easy to verify directly when $n = 2$. Indeed, we know that $a_2(x) = \frac{1}{x}-x$, and so, $a_2(1) = 0$ and $I_2 = (0, 1)$. Also, $-a_2'(x) = \frac{1}{x^2} + 1 \geq 1$ and $a_1'(x) = 1$ proves part (2) of the claim for $n = 2$.

  • (Induction step) Assume that the claim holds for a given $n \geq 2$. Then by the induction hypothesis,

    \begin{align*} (-1)^n a_{n+1}'(x) &= \left(\frac{n}{a_n^2} + 1\right) (-1)^{n-1}a_n'(x) - (-1)^{n-2}a_{n-1}'(x) \\ &\geq \frac{n}{a_n^2} (-1)^{n-1}a_n'(x) \\ & > 0. \end{align*}

    Now we consider two case

    $\text{Case 1.} \ $ If $n$ is even, then $a_n(\alpha^+) = +\infty$ and hence $a_{n+1}(\alpha^+) = -\infty$. Also we have $a_{n+1}(\beta^-) = +\infty$. Since $\alpha_{n+1}' > 0$ on $I_n$, there exists a unique zero $\gamma \in I_n$. Therefore $I_{n+1} = (\gamma, \beta)$.

    $\text{Case 2.} \ $ If $n$ is odd, then by a similar argument, we can check that there exists a unique zero $\gamma$ of $a_{n+1}$ in $I_n$ and $I_{n+1} = (\alpha, \gamma)$.

    So the part 1 of Claim for $n+1$ is proved. Finally, on $I_{n+1}$, we have $a_{n+1} > 0$, and so,

    $$ \frac{n}{a_n^2} = 1 + \frac{a_{n+1} + a_{n-1}}{a_n} > 1. $$

    Plugging this back shows that $(-1)^n a_{n+1}' \geq (-1)^{n-1}a_n'$ and therefore the induction step is proved. ////

Now write $I_n = (\alpha_n, \beta_n)$. Then by the intermediate step of the above proof, we can easily check that $(\alpha_n)_{n\geq 2}$ is non-decreasing, $(\beta_n)_{n\geq 2}$ is non-increasing and

$$ \alpha_{n} < \alpha_{n+2} < \beta_{n+2} < \beta_n. $$

So both $(\alpha_n)$ and $(\beta_n)$ converges. Moreover, if $\alpha = \lim \alpha_n$ and $\beta = \lim \beta_n$, then

$$ I_{\infty} = [\alpha, \beta]. $$

Finally, we prove that $\alpha = \beta$. To this end, we prove the following claim.

Claim. There exists $c \in (0, 1)$ such that $a_n(x) \leq c \sqrt{n}$ for all $n\geq 0$ and $x \in I_{\infty}$. In particular,

$$ |a_{n}'(x)| \geq \frac{1}{c^{2(n-1)}}. $$

Proof of Claim. Write $a_n = a_n(x)$ for simplicity. Then

  1. $a_n^2 \leq n$ shows that $a_n \leq \sqrt{n}$ for all $n \geq 0$.

  2. We have

    $$\frac{n}{a_n} - a_n = a_{n-1} + a_{n+1} \leq \sqrt{n-1} + \sqrt{n+1} \leq 2\sqrt{n}. $$

    Using this, we can easily conclude that $a_n \geq c_1\sqrt{n}$ for $c_1 = \sqrt{2}-1$.

  3. Similarly, we find that

    $$\frac{n}{a_n} - a_n = a_{n-1} + a_{n+1} \geq c_1(\sqrt{n-1} + \sqrt{n+1}) \geq c_1 \sqrt{2n}. $$

    Here, we utilized the inequality $\sqrt{t-1}+\sqrt{t+1} \geq \sqrt{2t}$ which holds for all $t \geq 1$. Then by choosing $c$ as $c = \left(\sqrt{\smash[b]{2+c_1^2}} - c_1\right)/\sqrt{2} \approx 0.749$, we find that $a_n \leq c\sqrt{n}$ for all $n\geq 0$.

Finally, as in the proof of the previous claim, we note that

$$ |a_{n+1}'(x)| \geq \frac{n}{a_n^2} |a_n'(x)| \geq \frac{1}{c^2}|a_n'(x)|. $$

Therefore the desired claim follows. ////

Now we are ready to prove the uniqueness. Assume otherwise that $\alpha < \beta$. Then by the mean-value theorem,

$$|a_n(\alpha) - a_n(\beta)| = |a_n'(x)|(\beta - \alpha) \geq \frac{\beta-\alpha}{c^{2(n-1)}} $$

for some $x \in (\alpha, \beta)$. On the other hand, the left-hand side is bounded by

$$ |a_n(\alpha) - a_n(\beta)| \leq a_n(\alpha) + a_n(\beta) \leq 2c\sqrt{n}. $$

This is a contradiction as $n\to\infty$.

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  • $\begingroup$ The inductive step of the first Claim can be simplified using that $a_{n+1}a_n+a_n^2+a_{n-1}a_n=n$ and hence $a_n^2\leq n$. $\endgroup$ – Helmut Jul 23 at 12:54
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I post it as a curiosity. Considering the difference equation

$$ \frac{n}{x_n^2} = 1 + \frac{x_{n+1}+x_{n-1}}{x_n} $$

and considering

$$ \Delta x_n = x_n-x_{n-1} $$

we have

$$ \frac{n}{x_n^2} = 3+\frac{\Delta x_{n+1}-\Delta x_n}{x_n} $$

which is a difference approximation for the DE

$$ \frac{t}{x^2(t)} = 3 + \frac{\ddot x(t)}{x(t)} $$

now calling $x(1) = a = \frac{2 \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}$ and $x(2) = a-\frac 1a$ and integrating we have

enter image description here

In red the plot for $x_n = \sqrt{\frac n3 + \frac{1}{36n}+ \mathcal{O}(\frac{1}{n^3})}$ and in blue the DE solution.

and now considering instead the initial conditions $x(2) = a-\frac 1a, x(3) = \frac{1-3a^2}{a^3-3}$ we obtain

enter image description here

This result points in the direction of good agreement between the DE and the recurrence. Now following with the initial conditions

$$ x(3) = \frac{1-3a^2}{a^3-3}\\ x(4) = \frac{8 a^3-4 a}{3 a^4-4 a^2+1} $$

we obtain

enter image description here

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The following answer is the special case $p=1$ of part 3 of this answer to a closely related question.

Uniqueness: Suppose that $a_n,a_n'$ were two such sequences satisfying $a_0=a_0'=0$ and the recursion $$\label{eq3}\tag1 a_{n+1}=\frac n{a_n}-a_n-a_{n-1}$$ related to two different values $a_1=a,a_1'=a'.$ Then $d_n=a_n-a_n'$ satisfy $d_{n+1}=-\left(\frac n{a_na_n'}+1\right)d_n-d_{n-1}$ for $n\geq1$ and $d_0=0$. Now we have $a_na_{n+1}+a_n^2+a_na_{n-1}=n$ and hence $a_n\leq \sqrt n$. The same holds for $a_n'$: $a_n'\leq \sqrt n$. Hence $$|d_{n+1}|\geq 2|d_n|-|d_{n-1}|\mbox{ for }n\geq1.$$

We have $|d_2|\geq2|d_1|$ and now prove by induction that $$|d_{n}|\geq\frac n{n-1}|d_{n-1}|\mbox{ for }n\geq2.$$ This is true for $n=2$ and if it is true for some $n$ then $|d_{n-1}|\leq\frac{n-1}n |d_n|$ and hence $$|d_{n+1}|\geq 2|d_n|-|d_{n-1}|\geq\left(2-\frac{n-1}n\right)|d_n|=\frac{n+1}n|d_n|.$$

As a consequence, $|d_n|\geq n|d_1|=n|a-a'|$ for all $n$, which contradicts $|d_n|=|a_n-a_n'|\leq\sqrt n$. Therefore $a$ and $a'$ must be equal.

Existence: We use the function $w(n,z)$ defined for real $z$ and positive integer $n$ as the unique positive solution $w$ of $\frac nw-w=z$. It can be given explicitly as \begin{equation}\nonumber w(n,z)=-\frac z2+\sqrt{\frac{z^2}4+n}=\frac n{\frac z2+\sqrt{\frac{z^2}4+n}}. \end{equation} Observe that for any positive $n$, the mapping $z\to w(n,z)$ is strictly decreasing because the derivative of the mapping $w\to\frac nw-w$ is always negative.

We define the sequence $U_1(n)=\sqrt{n}$ for all $n$. Then we define recursively sequences $L_k,U_k$ by $L_k(0)=U_k(0)=0$ for all $k$ and \begin{equation}\nonumber L_k(n)=w(n,U_k(n+1)+pU_k(n-1))\mbox{ and }U_{k+1}(n)=w(n,L_k(n+1)+pL_k(n-1)) \end{equation} for all $n,k\geq1.$

By induction and using that $z\mapsto w(n,z)$ is strictly decreasing it follows that $$L_k(n)\leq L_{k+1}(n)\leq U_{k+1}(n)\leq U_k(n)$$ for all $n,k$.

Thus the pointwise limits \begin{equation}\nonumber U(n)=\lim_{k\to\infty}U_k(n)\mbox{ and } L(n)=\lim_{k\to\infty}L_k(n) \end{equation} exist because for fixed $n$, the sequences $U_k(n)$ and $L_k(n)$, $k=1,2,3,...$ are monotonous and bounded. The properties of the sequences $U_k,L_k$ imply besides $U(0)=L(0)=0$
a) $L(n)\leq U(n)$ for every $n$,
b) $U(n)=w(n,L(n+1)+pL(n-1))$ and $L(n)=w(n,U(n+1)+pU(n-1))$ for all positive $n$.
As a consequence of b), the two sequences $A_n,B_n$, $n=0,1,...$, defined by $A_n=U(n),\,B_n=L(n)$ if $n$ is even and $A_n=L(n),\,B_n=U(n)$ if $n$ is odd both satisfy the recursion (\ref{eq3}). They must be equal because we have proved uniqueness of positive solution sequences of (\ref{eq3}).

Observe that the above construction of $U,L$ can be used to approximate the positive sequence numerically.

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Some observations...Posting it as answer since it is too long as comment. With $a_1 = a, a_2 = \frac{1}{a} - a$ and $a_n(a_{n+1} + a_n + a_{n-1}) = n$, we get $$S = a_2a_3 + (a_3 + a_4)^2+(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 - (a_4^2 + a_5^2 + ...a_{n-1}^2) + a_na_{n+1}$$, where $S = \frac{n(n+1)}{2} - 3$. So, $(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 = S - (a_2a_3 + a_na_{n+1}) + (a_4^2 + a_5^2 + ...a_{n-1}^2)$. Since LHS is $> 0$, $S - (a_2a_3 + a_na_{n+1})$ is either $ > 0$ or It must be $S - (a_2a_3 + a_na_{n+1}) < (a_4^2 + a_5^2 + ...a_{n-1}^2)$. The lowest $n$ where this condition is applies is $n=5$ and the $a$ value we calculate must be applicable for $n>5$ I guess... One more thing to try is to do alternate summation to cancel out products, like $a_3^2 + a_3a_2 + a_3a_4 - a_4a_5 -a_4^2 -a_3a_4 + a_5a_6 + a_5^2 +a_4a_5 -a_6a_7 - a_6^2 - a_5a_6 ... = 3 - 4 + 5 -6 + 7 = \sum_{1}^{n}{n}{(-1)}^{n-1}$ and see whether we can get anywhere simpler...

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