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.

  • 5
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 7
    $\begingroup$ 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

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 $n\to\infty$.

  • $\begingroup$ 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

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


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.


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...


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.