A Very Average Bunch Of Numbers

Relevant wiki: Induction - Recurrence Relations

The recursion relation implies that $x_n$ is an arithmetic progression. I choose not to use this fact but go for a more elementary approach:

Lemma : $2x_n = x_{n-a} + x_{n+a}$ .

Proof by induction on $a$ . The case $a = 0$ is trivial, and $a = 1$ is the given recursion relation. Now assume the relation is true up to some value of $a \geq 1$ . Then $x_{n-(a+1)} + x_{n+(a+1)} \\ = 2x_{n-a} - x_{n-(a-1)} + 2x_{n+a} - x_{n+(a-1)} \\ = 2(x_{n-a}+x_{n+a}) - (x_{n-(a-1)} + x_{n+(a-1)}) \\ = 4x_n - 2x_n = 2x_n,$ Q.E.D.

Taking $n = 343434$ and $a = 171717$ we conclude that $x_{171717} + x_{515151} = 2x_{343434}\ \ \ \therefore\ \ \ x_{171717} = 2x_{343434} - x_{515151};$ similarly, $x_0 + x_{343434} = 2x_{171717}\ \ \ \therefore\ \ \ x_0 = 2x_{171717} - x_{343434} = 3x_{343434} - 2x_{515151},$ and since $x_{343434} + x_{515151} = 5$ , we can write this as $x_0 = 5x_{343434} - 10\ \ \ \text{or}\ \ \ x_0 = 15 - 5x_{515151}.$ Since $x_{343434} \geq 0$ we have $x_0 \geq -10$ ; since $x_{515151} \geq 0$ we have $x_0 \leq 15$ .

Thus $x_0$ can take all integer values in the range $x_0 \in \{ -10, -9, \dots, 14, 15 \},$ showing that there are $15 - (-10) + 1 = \boxed{26}$ possible values for $x_0$ .

The first recursion relation implies an arithmetic progression.

• $x_{0,min} = -10$ for $x_{343434} = 0$ and $x_{515151}=5$
• $x_{0,max} = 15$ for $x_{343434} = 5$ and $x_{515151}=0$

So, $x_0$ has the range from $-10$ to $15$ .

So it has $\boxed{26}$ possible integral values.

Although @Pi Han Goh pointed out that this solution is a little unsatisfying and somewhat incomplete. Can anyone think of a more straightforward way to explain the solution?

Perhaps it helps to consider that the numbers $343434$ and $515151$ have little relevance other than that $343434$ is $\frac{2}{3}$ of $515151$ . So, the problem is essentially the same as if you said $x_2$ and $x_3$ , a much simpler problem to visualize/solve.

$\begin{aligned} x_n & =\dfrac{{x}_{n-1} + {x}_{n+1}}{2} \\ 2{x}_{n} & ={x}_{n-1} + {x}_{n+1} \\ {x}_{n}+{x}_{n} & ={x}_{n-1} + {x}_{n+1} \\ {x}_{n}-{x}_{n-1} & ={x}_{n+1} - {x}_{n} \end{aligned}$

Let ${x}_{n}-{x}_{n-1}={x}_{n+1} - {x}_{n}=k$ , then ${x}_{n-1}={x}_{n}-k$ and ${x}_{n+1}={x}_{n}+k$

${x}_{n-1},{x}_{n},{x}_{n+1} \Leftrightarrow {x}_{n}\color{#3D99F6}{-k}, {x}_{n}, {x}_{n}\color{#3D99F6}{+k}$ (it shows us that ${x}_{n}$ follow aritmetical progression)

Then, we can get that ${x}_{n}={x}_{0}+nk$

$n=343434 \implies {x}_{343434}={x}_{0}+343434k$

$n=515151 \implies {x}_{515151}={x}_{0}+515151k$

$\begin{aligned} {x}_{343434}+{x}_{515151} & =({x}_{0}+343434k)+({x}_{0}+515151k) \\ 5 & ={x}_{0}+343434k+{x}_{0}+515151k \\ 5 & =2{x}_{0}+858585k \\ 5-2{x}_{0} & =858585k \\ k & =\dfrac{5-2{x}_{0}}{858585} \end{aligned}$

$\begin{aligned} {x}_{343434} & \geq 0 \\ {x}_{0}+343434k & \geq 0 \\ {x}_{0}+343434 \times \dfrac{5-2{x}_{0}}{858585} & \geq 0 \\ {x}_{0}+\dfrac{2}{5}(5-2{x}_{0}) & \geq 0 \\ {x}_{0}+2-\dfrac{4}{5} {x}_{0} & \geq 0 \\ \dfrac{1}{5}{x}_{0} & \geq -2 \\ {x}_{0} & \geq -10 \end{aligned}$

$\begin{aligned} {x}_{515151} & \geq 0 \\ {x}_{0}+515151k & \geq 0 \\ {x}_{0}+515151 \times \dfrac{5-2{x}_{0}}{858585} & \geq 0 \\ {x}_{0}+\dfrac{3}{5}(5-2{x}_{0}) & \geq 0 \\ {x}_{0}+3-\dfrac{6}{5} {x}_{0} & \geq 0 \\ 3 & \geq \dfrac{1}{5}{x}_{0} \\ 15 & \geq {x}_{0} \\ {x}_{0} & \leq 15 \end{aligned}$

$\implies -10 \leq {x}_{0} \leq 15$

So, the number possible integral values of ${x}_{0}$ is $\boxed{26}$