# About the sequence satisfying $a_n=a_{n-1}a_{n+1}-1$

"Consider sequences of positive real numbers of the form x,2000,y,..., in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of x does the term 2001 appear somewhere in the sequence?

(A) 1 (B) 2 (C) 3 (D) 4 (E) More than 4"

Can anyone suggest a systematic way to solve this problem? Thanks!

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Perhaps a bit of a tedious solution but this is the most straight forward approach. By direct substitution it is not difficult to show that the sequence necessarily repeats itself after the $6$th term. We have $$a_1 = x$$ $$a_2 = 2000$$ $$a_3 = \frac{2001}{x}$$ $$a_4 = \frac{2001 + x}{2000x}$$ $$a_5 = \frac{1 + x}{2000}$$ $$a_6 = x$$ $$a_7 = 2000$$ and so on.... Clearly we can take $x=2001$ for a trivial appearance. For $a_3$ to be $2001$ we require $x=1$. For $a_4$ we require $x=0.0005$. For $a_5$ to be $2001$ we require $x = 4001999$. This gives a total of $4$ solutions for $x$.

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You have a sequence defined by $a_0=x,a_1=2000$, and the recurrence $$a_{n+1}=\frac{a_n+1}{a_{n-1}}\;.$$ Calculate the first few terms:

\begin{align*} a_0&=x\\\\ a_1&=2000\\\\ a_2&=\frac{2001}x\\\\ a_3&=\frac{2001+x}{2000x}\\\\ a_4&=\frac{\frac{2001(x+1)}{2000x}}{\frac{2001}{x}}=\frac{x+1}{2000}\\\\ a_5&=\frac{\frac{x+2001}{2000}}{\frac{x+2001}{2000x}}=x\\\\ a_6&=2000 \end{align*}

Clearly the sequence is period with period $5$, so $2001$ appears iff

$$2001\in\left\{x,\frac{2001}x,\frac{x+2001}{2000x},\frac{x+1}{2000}\right\}\;.$$

Thus, there are four solutions: $2001,1,\dfrac{4001999}{2001}$, and $4001999$.

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@Robert: Indeed, that becomes pretty apparent as soon as one looks closely at the specific calculations. –  Brian M. Scott Aug 15 '12 at 6:16

(This is basically EuYu's answer with the details of periodicity added; took a while to type up.)

Suppose that $a_0 , a_1 , \ldots$ is a generalised sequence of the type described, so that $a_i = a_{i-1} a_{i+1} - 1$ for all $i > 0$. Note that this condition is equivalent to demanding that $$a_{i+1} = \frac{ a_i + 1 }{a_{i-1}}.$$

Using this we find the following recurrences: $$a_2 = \frac{ a_1 + 1}{a_0}; \\ a_3 = \frac{ a_2 + 1}{a_1} = \frac{ \frac{ a_1 + 1}{a_0} }{a_1} = \frac{ a_0 + a_1 + 1 }{ a_0a_1 }; \\ a_4 = \frac{ a_3 + 1 }{a_2} = \frac{\frac{ a_0 + a_1 + 1 }{ a_0a_1 } + 1}{\frac{ a_1 + 1}{a_0}} = \frac{ ( a_0 + 1 )( a_1 + 1) }{ a_1 ( a_1 + 1 ) } = \frac{a_0 + 1}{a_1};\\ a_5 = \frac{ a_4 + 1 }{ a_3 } = \frac{ \frac{a_0 + 1}{a_1} + 1}{\frac{ a_0 + a_1 + 1 }{ a_0a_1 }} = \frac{ \left( \frac{a_0 + a_1 + 1}{a_1} \right) }{ \left( \frac{a_0+a_1+1}{a_0a_1} \right) } = a_0 \\ a_6 = \frac{ a_5 + 1 }{a_4} = \frac{ a_0 + 1}{ \left( \frac{ a_0 + 1 }{a_1} \right) } = a_1.$$

Thus every such sequence is periodic with period 5, so if 2001 appears, it must appear as either $a_0, a_1, a_2, a_3, a_4$.

1. Clearly if $a_0 = 2001$, we're done.
2. As we stipulate that $a_1 = 2000$, it is impossible for $a_1 = 2001$.
3. If $a_2 = 2001$, then it must be that $2001 = \frac{ 2000 + 1 }{a_0}$ and so $a_0 = 1$.
4. If $a_3 = 2001$, then it must be that $2001 = \frac{a_0 + 2000 + 1}{a_0 \cdot 2000}$, and it follows that $a_0 = \frac{2001}{2000 \cdot 2001 - 1}$.
5. If $a_4 = 2001$, then it must be that $2001 = \frac{ a_0 + 1 }{2000}$, and so $a_0 = 2001 \cdot 2000 - 1$.

There are thus exactly four values of $a_0$ such that 2001 appears in the sequence.

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We are given that $xy=2001$. Clearly if one of them is $1$ and the other is $2001$, then $2001$ appears. Try the other factorizations to see if $2001$ appears.