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Define sequence $x_{1}=1$,and $x_{n+1}=1+\dfrac{n}{x_{n}},n\ge 1$. I found $$x_{2}=2,x_{3}=2,x_{4}=1+\dfrac{3}{x_{3}}=\dfrac{5}{2}, x_{5}=1+\dfrac{4}{2.5}=\dfrac{13}{5},x_{6}=1+\dfrac{5}{x_{5}}=\dfrac{38}{13}$$ $$x_{7}=1+\dfrac{6}{x_{6}}=\dfrac{58}{19},x_{8}=1+\dfrac{7}{x_{7}}=\dfrac{191}{58},\cdots$$

I conjecture: $x_{n}\notin Z$ for $n\ge 4$; in other words, this sequence has only three integer terms.

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    Did you try to get a general form for $x_n$ using the characteristic equations? Then you can prove this conjecture. – 2017-02-16
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    @stud_iisc,it is clear can't using characteristic equation to solve.I think this sequence can't have colsed form – 2017-02-16
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    Let us define $a_{n}$. They satisfy $a_{0}=1, a_1=1 $, and also $$a_{n+1}=a_{n}+na_{n-1}$$ Now, note that $$\frac{a_{n+1}}{a_{n}}=\frac{a_{n}+na_{n-1}}{a_{n}}=1+\frac{na_{n-1}}{a_{n}}$$ So we can prove $$x_{n}=\frac{a_{n}}{a_{n-1}}$$ Now note that $a_{0}=1, a_{1}=1, a_{2}=2, a_{3}=4, \dots$. A quick google search reveals this is [A000085](https://oeis.org/A000085). It appears that $\gcd(a_{n}, a_{n+1})$ is restricted only to powers of $2$. So you have to do is to prove a power of $2$ does not appear in $a_{n}$. – 2017-02-16
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    The sequence seems to be increasing and unbounded. Increase rate seems to be slightly faster than $\sqrt n$. For example, $x_{10^4}\approx 100.503$ – 2017-02-16
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    @S.C.B., the [Wikipedia article](https://en.wikipedia.org/wiki/Telephone_number_(mathematics)) says that the order of $2$ in $a_n$ is about $n/4$. Clearly not enough for $a_n$ to be a power of $2$. – 2017-02-16
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    @zhoraster So the problem is to prove the $\gcd$ can only be powers of $2$. Which I cannot do. – 2017-02-16
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    @S.C.B., I thought you knew that. What did you mean then by "It appears..."? – 2017-02-16
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    @zhoraster I was making an inductive observation. For all I know, I could have been over generalized. It could not be true. – 2017-02-16
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    @zhoraster [This](https://en.wikipedia.org/wiki/Telephone_number_(mathematics)#Prime_factors) looks like it could be helpful. – 2017-02-16

2 Answers 2

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This answer uses that for $n\ge 4$, $$\frac{1+\sqrt{4n-3}}{2}\lt x_n\lt \frac{1+\sqrt{4n+1}}{2}\tag1$$ The proof for $(1)$ is written at the end of this answer.

Multiplying $(1)$ by $2$, subtracting $1$ and squaring, we get $$4n-3\lt 4(x_n^2-x_n)+1\lt 4n+1$$ Then, subtracting $1$ and dividing by $4$, we get $$n-1\lt x_n^2-x_n\lt n$$ Since $x_n^2-x_n$ is not an integer, we have that $x_n$ is not an integer.

So, your conjecture is true.

Let us prove $(1)$ by induction.

For $n=4$, since $13\lt 16\lt 17$, we get $$\frac{1+\sqrt{13}}{2}\lt \frac{1+\sqrt{16}}2=x_4\lt \frac{1+\sqrt{17}}{2}$$

Supposing that $(1)$ holds for some $n\ge 4$ gives $$\begin{align}x_{n+1}-\frac{1+\sqrt{4(n+1)-3}}{2}&=1+\frac{n}{x_n}-\frac{1+\sqrt{4n+1}}{2}\\\\&\gt 1+\frac{2n}{1+\sqrt{4n+1}}-\frac{1+\sqrt{4n+1}}{2}=0\end{align}$$ and $$\begin{align}\frac{1+\sqrt{4(n+1)+1}}{2}-x_{n+1}&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{n}{x_n}\\\\&\gt \frac{1+\sqrt{4n+5}}{2}-1-\frac{2n}{1+\sqrt{4n-3}}\\\\&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{2n(1-\sqrt{4n-3})}{1-(4n-3)}\\\\&=\frac{(n-1)\sqrt{4n+5}+1-n\sqrt{4n-3}}{2(n-1)}\\\\&=\frac{\sqrt{4n+5}-3}{(n-1)\sqrt{4n+5}+1+n\sqrt{4n-3}}\\\\&\gt 0\qquad\quad\blacksquare\end{align}$$

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    How to found the sequence have the inequality ？ – 2017-02-20
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    @inequality: Basically trial and error with the fact that $\frac{1+\sqrt{4n+1}}{2}$ is one of the solutions of $x=1+\frac nx$. – 2017-02-20
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The conjecture is true.

Define, as in @S.C.B.'s comment, the sequence of Telephone numbers: $$ T_0 = T_1 = 1, \quad T_{n+1} = T_{n} + nT_{n-1},\ n\ge 1, \tag{1} $$ so that $$ x_n = \frac{T_n}{T_{n-1}},\ n\ge 1. $$ Then $$ T_n = \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k}(2k-1)!!. $$ I will prove first @S.C.B.'s claim:

Claim For each $n\ge 1$, $T_n$ and $T_{n+1}$ have no common odd prime divisor.

It relies on

Lemma For each $n\ge 1$, $T_n$ and $n$ have no common odd prime divisor.

Proof Let $p>2$ be an odd prime dividing $n$. Write $$ T_n = 1 + \sum_{k=1}^{\lfloor n/2\rfloor} {n\choose 2k}(2k-1)!!. $$ Note that $p\mid {n\choose 2k}$ for $1\le k< p/2$, and $p\mid (2k-1)!!$ for $k> p/2$. Therefore, $T_n = 1\pmod p$, yielding the statement.

Proof of the claim Let $p>2$ be a prime. Assume, by way of contradiction, that $n$ is the smallest positive integer such that $p$ divides both $T_{n+1}$ and $T_n$. Then it follows from $(1)$ that $p\mid n$, contradicting Lemma.

Therefore, the only possibility for $x_n$ to be an integer is $T_{n-1}$ being a power of $2$. However, as it is explained in the Wikipedia page, this is not possible for $n\ge 4$.