Define an infinte graph $G$ whose vertices are $1,2,3,\ldots$ and connect $i\leftrightarrow j$ if and only if $a_i/i = a_j/j$. Note that $G$ is the disjoint union of cliques, each grouped on the common value of $a_i/i$. Assume for the sake of contradiction that no such clique has size $1$. Note that all cliques of $G$ have finite size; suppose a clique with vertices $(x_1,x_2,\ldots)$ has infinite size. Then \[ \frac{a_{x_1}}{x_1}=\frac{a_{x_2}}{x_2}=\cdots, \]but since $(x_i)$ is an unbounded sequence, $(a_{x_i})$ is also unbounded, contradiction. For each $n\ge 1$, define $n$ to be an ender if $n$ has no edge to an $n'>n$. Define $n$ to be a non-ender if $n$ is not an ender. Lemma 1. There are infinitely many non-enders. Proof. Assume for the sake of contradiction that for some finite $N$, all $n\ge N$ are non-enders. Note that no edge connects two numbers $N_1,N_2>N$; else, the lower is a non-ender. Also note that no two $N_1,N_2>N$ connect to the same $n'$, for any $n'<n$; else, $N_1 \leftrightarrow n' \leftrightarrow N_2$, so all are part of the same clique, contradicting $N_1,N_2$ not being an edge. Combining the above two facts implies that each of the values in $(N,\infty)$ has an edge to a unique value in $[1,N]$. This contradicts pigeonhole principle. $\blacksquare$ We construct an infinite sequence $(n_1,n_2,\ldots)$. Define $n_1=1$. Note that $1$ is a non-ender since it must have degree at least $1$ by assumption, and it must connect to a number greater than $1$. For each $i\ge 1$: Define $x_i$ to be the maximal integer for which $n_i\leftrightarrow n_i+x_i$ is an edge. We will show that $n_i$ is a non-ender later, which is necessary for this to be well-defined. Also note that $x_i$ exists since it is finite since cliques have finite size. Define $y_i$ to be the minimal nonnegative integer for which $n_i+x_i+y_i$ is a non-ender. This exists by Lemma 1. Now let $n_{i+1}=n_i+x_i+y_i$. Note that $n_k=n_1+(x_1+\cdots+x_{k-1})+(y_1+\cdots+y_{k-1})$ for all $k\ge 1$. Note that the definition of $x_i$ makes sense since we proved $n_1=1$ is a non-ender, and by construction, all later $n_i$ are non-enders. Lemma 2. [Key combinatorial estimate] We have $y_1+\cdots+y_{\ell} \le x_1+\cdots+x_\ell$ for all $\ell\ge 1$. Proof. Intuitive diagram: the bottom line is the number line, and the curved lines above are edges of $G$: Define $S_i=[n_i,n_i+x_i)$ and $E_i=[n_i+x_i,n_{i+1})$ for each $i\ge 1$. Everything in $E_i$ is an ender by construction, $n_{i+1}=n_i+x_i+y_i$ is the minimal non-ender after $n_i+x_i$. Consider some fixed $u\in E_i$, for some fixed $i$. We claim $u$ cannot have an edge to $n_j+x_j$, for any $j\le i$. (Except the trivial case of $j=i$ and $u=n_i+x_i$ so $u$ connects to itself.) Suppose it did. Then $u\leftrightarrow n_j+x+j\leftrightarrow n_j$, but since $u\ge n_i+x+i \ge n_j+x_j$, and one of these inequalities is strict since we threw out the trivial case, this means $n_j+x_j$ was not the maximal value that $n_j$ has an edge to, contradicting the construction of $x_j$. We further claim that $u$ does not have an edge to any element of $E_j$ for any $j\le i$. We showed above $u$ does not connect to the first element of any $E_j$. If $u$ has an edge to some $v\in E_j \setminus \{n_j+x_j\}$ for some $j\le i$, then $v$ is not an ender since it connects to $u>v$, contradiction. Finally, we claim no two $u,u' \in E_1\cup\cdots \cup E_i$ connect to the same element $s$ of $S_1\cup \cdots \cup S_i$. If they did, then $u\leftrightarrow s \leftrightarrow u'$, so $u$ and $u'$ are connected, so $\min\{u,u'\}$ is not an ender, contradiction. Combining the above facts implies that each element of $E_1\cup\cdots\cup E_i$ connects to a unique element of $S_1\cup \cdots \cup S_i$. Hence \[ |E_1\cup \cdots E_i| \le |S_1\cup \cdots \cup S_i| \implies y_1 + \cdots +y_i \le x_1+\cdots+x_i. \]$\blacksquare$ Now we use the $(n_i)$ sequence to prove that the corresponding subsequence of $(a_i)$ is unbounded. We have for all $i\ge 1$ that \[ \frac{a_{n_i}}{n_i} = \frac{a_{n_i+x_i}}{n_i+x_i} \implies a_{n_i+x_i} = \frac{n_i+x_i}{n_i}a_{n_i}, \]and since $(a_i)$ is increasing, we have \[ a_{n_{i+1}}=a_{n_i+x_i+y_i} > a_{n_i+x_i} = \frac{n_i+x_i}{n_i}a_{n_i}. \]Iterating the above procedure, we get that \begin{align*} \frac{a_{n_{i+1}}}{a_{n_1}} &> \frac{n_i+x_i}{n_i}\cdot \frac{n_{i-1}+x_{i-1}}{n_{i-1}} \cdot \cdots \cdot \frac{n_1+x_1}{n_1} \\ &=\prod_{\ell=1}^i \frac{n_\ell+x_\ell}{n_\ell} \\ &= \prod_{\ell=1}^i \frac{1+(x_1+\cdots+x_{\ell-1}+x_\ell)+(y_1+\cdots+y_{\ell-1})}{1+(x_1+\cdots+x_{\ell-1})+(y_1+\cdots+y_{\ell-1})} \\ &= \prod_{\ell=1}^i \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}}, \end{align*}where $X_k:=x_1+\cdots+x_k$ and $Y_k:=y_1+\cdots+y_k$ for all $k\ge 1$. We know $Y_\ell \le X_\ell$ for all $\ell$ by Lemma 2. We show this product is unboundedly large as $i$ increases. The function \[f(x)=\frac{1+X_\ell+x}{1+X_{\ell-1}+x}\]is decreasing for $x>0$; this can be verified by subtracting $1$ and noting that $X_\ell-X_{\ell-1}=x_{\ell}>0$. In particular, $f(Y_\ell) \ge f(X_\ell)$. Secondly, for $a\ge b \ge 1$, it can be verified that \[ \frac{\frac12+a}{\frac12+b} \ge \sqrt{\frac{a}{b}}. \]Now, using these two facts and AM-GM: \begin{align*} \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}} &\ge \frac{1+X_\ell+X_{\ell-1}}{1+X_{\ell-1} + X_{\ell-1}} =\frac{\frac12 + \frac{X_{\ell}+X_{\ell-1}}{2}}{\frac12 + X_{\ell-1}} \\ &> \sqrt{\frac{\frac{X_\ell+X_{\ell-1}}{2}}{X_{\ell-1}}} \ge \sqrt{\frac{\sqrt{X_\ell X_{\ell-1}}}{X_{\ell-1}}} = \left(\frac{X_\ell}{X_{\ell-1}}\right)^{1/4}. \end{align*}Finally, \begin{align*} \frac{a_{n_{i+1}}}{a_{n_1}}> \prod_{\ell=1}^i \frac{1+X_\ell+Y_{\ell-1}}{1+X_{\ell-1}+Y_{\ell-1}} > \prod_{\ell=2}^i \left(\frac{X_\ell}{X_{\ell-1}}\right)^{1/4} = \left(\frac{X_i}{X_1}\right)^{1/4}. \end{align*}(Note that we omitted one term -- it still works since each term is greater than $1$.) Now, we have a contradiction, since $X_i$ grows unboundedly large since $X_{i+1}-X_i = x_{i+1}$, so it grows by at least $1$ at each step.

FTSoC suppose numbers appear at least twice. Partition $\mathbb{N}$ into groups $S_1, S_2, \ldots$ so that for every group $S_i$, the value of $\tfrac{a_k}{k}$ is constant across all $k \in S_i$. By the assumption of the opposite of the problem, $|S_i| \geq 2$ for all $i$. Claim: All groups have finite size. Proof: Suppose there is an group with infinite size, and let $m$ be its smallest element. Then, we can choose arbitrarily large $M$ in the group for which\[\frac{a_m}{m} = \frac{a_M}{M} \implies a_M = \frac{Ma_m}{m} \geq 1\]which is no good. Call a number an apex predator if it is the greatest element of some group. Let $M_i$ be the sequence of all apex predators in increasing order. We prove the seemingly random claim: Claim:\[\frac{M_1}{1}\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{n+1}}{M_n + 1}\right) \geq \frac{2n+2}{\sqrt{2n+1}}\]Proof: Rewrite this as\[\frac{M_1}{1}\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{n+1}}{M_n + 1}\right) = M_{n+1} \left(\prod_{i = 1}^n \frac{M_i}{M_i + 1}\right)\]where since each group has size at least $2$, we can conclude that among any first $n$ positive integers there are at most $\tfrac{n}{2}$ apex predators. This also tells us that the $i$-th apex predator must be $\geq 2i$, else there are too many apex predators among the first $M_i$ positive integers. Therefore, it follows that\[M_{n+1} \left(\prod_{i = 1}^n \frac{M_i}{M_i + 1}\right) \geq (2n+2) \left(\prod_{i = 1}^n \frac{2i}{2i + 1}\right) = \frac{(2n+2)!!}{(2n+1)!!}\]which we can induct to prove is $\geq \tfrac{2n+2}{\sqrt{2n+1}}$. Now with this, we will be able to construct some $a_M \geq 1$. Let $y_1 = 1$ and $M_{r_1}$ be the apex predator in its group. Let $y_2$ be the least number $> M_{r_1}$ that is not an apex predator, then let $M_{r_2}$ be the apex predator in $y_2$'s group, and so on defining the sequence $y_i$ and $r_i$. For any $k$, note\[a_{M_{r_k}} = \left(\frac{M_{r_k}}{y_k}\right)a_{y_k} > \left(\frac{M_{r_k}}{y_k}\right)a_{M_{r_{k-1}}} > \ldots > \left[\left(\frac{M_{r_k}}{y_k}\right)\ldots \left(\frac{M_{r_1}}{y_1}\right)\right]a_1.\]It is actually not hard to see that\[\left(\frac{M_{r_k}}{y_k}\right)\ldots \left(\frac{M_{r_1}}{y_1}\right) > \left(\frac{M_1}{1}\right)\left(\frac{M_2}{M_1 + 1}\right)\ldots \left(\frac{M_{r_k}}{M_{r_{k-1}} + 1}\right)\]mainly due to two things: Note that we actually do get the $y_i$'s successfully on the RHS because if we have consecutive apex predators $M_i, M_{i+1}$, then $\tfrac{M_{i+1}}{M_i + 1}$ is $1$ so it does not matter, so in effect we are skipping past all consecutive apex predators to get the least number that is not an apex predator. If the $r_i$'s skip any apex predators, say $M_{r_i} < M_{\ell_1} < \ldots < M_{\ell_t} < M_{r_{i+1}}$, then\[\frac{M_{r_{i+1}}}{y_{i+1}} > \prod_{i = 1}^{t + 1} \left(\frac{M_{\ell_i}}{M_{\ell_{i-1}} + 1}\right)\]is true. Hence, by the lemma, we know\[a_{M_{r_k}} > \frac{2r_k}{\sqrt{2r_k - 1}} a_1\]so continuously going to further terms in the sequence we can get arbitrarily large $r_k$ so we can create an arbitrarily large constant $c$ for which $a_{something} = ca_1$, breaking $(0, 1)$, a contradiction so we are done.

Assume for contradiction none of \(\frac{a_1}1\), \(\frac{a_2}2\), \(\ldots\) are unique. Say \(i\) is a peak if for all \(j>i\), we have \(\frac{a_j}j\ne\frac{a_i}i\). Then for each \(j\), there is a peak \(i\ge j\) with \(\frac{a_i}i=\frac{a_j}j\); moreover, for each \(k\), there are at most \(k/2\) peaks among 1, 2, \(\ldots\), \(k\). Claim: For each \(n\), there is an \(N>n\) with \(a_N\ge1.99a_n\). Proof. Let \(x_1=n\), and consider the \((x_i)\) and \((y_i)\) defined as follows: Let \(y_i>x_i\) be the peak with \(\frac{a_{y_i}}{y_i}=\frac{a_{x_i}}{x_i}\). Let \(x_{i+1}\) be the smallest non-peak with \(x_{i+1}>y_i\). Since \(a_{x_{i+1}}>a_{y_i}\), we always have \[a_{y_{i+1}}=\frac{y_{i+1}}{x_{i+1}}\cdot a_{x_{i+1}}>\frac{y_{i+1}}{x_{i+1}}\cdot a_{y_i}.\]It follows that for each \(k\), \[a_{y_k}\ge\frac{y_k}{x_k}\cdot\frac{y_{k-1}}{x_{k-1}}\cdots\frac{y_1}{x_1}\cdot a_n.\] Let there be \(m\) peaks \(j\le n\). By construction, for each \(i\), each of \(y_i\), \(y_i+1\), \(\ldots\), \(x_{i+1}-1\) is a peak. Thus the number of peaks among 1, \(\ldots\), \(y_k\) is \[m+(x_2-y_1)+(x_3-y_2)+\cdots+(x_k-y_{k-1})+1.\]Recall there are at most \(y_k/2\) peaks among 1, \(\ldots\), \(y_k\), so \begin{align*} m+(x_2-y_1)+\cdots+(x_k-y_{k-1})+1&\le\tfrac12y_k\\ \implies (y_k-x_k)+\cdots+(y_1-x_1) &=y_k-x_1-\big[(x_2-y_1)+\cdots+(x_k-y_{k-1})\big]\\ &\ge y_k-n-\big[\tfrac12y_k-m-1\big]\\ &=\tfrac12y_k-(n-m-1). \end{align*} Now notice that \begin{align*} \frac{a_{y_k}}{a_1}&\ge\frac{y_k}{x_k}\cdot\frac{y_{k-1}}{x_{k-1}}\cdots\frac{y_1}{x_1}\\ &=\underbrace{ \left(\frac{y_k}{y_k-1}\cdot\frac{y_k-1}{y_k-2}\cdots\cdot\frac{x_k+1}{x_k}\right) \cdots \left(\frac{y_1}{y_1-1}\cdots\frac{x_1+1}{x_1}\right) }_{\ge\frac12y_k-(n-m-1)\text{ fractions}}\\ &\ge\frac{y_k}{y_k-1}\cdot\frac{y_k-1}{y_k-2}\cdots \frac{y_k-(\frac12y_k-(n-m-1)-1)}{y_k-(\frac12y_k-(n-m-1))}\\ &=\frac{y_k}{\frac12y_k+(n-m-1)}>1.99 \end{align*}for sufficiently large \(y_k\) (i.e.\ sufficiently large \(k\)). \(\blacksquare\) Now inductively, for each \(k\) there is an \(N_k\) with \[a_{N_k}\ge1.99^k\cdot a_1.\]Taking \(1.99^k>\frac1{a_1}\) gives \(a_{N_k}>1\), contradiction.

Wait, I'm probably being really really dumb right now (sorry if this is the case), but @above can you tell me why there are no peak indices between $x_i$ and $y_i$ ($x_i<y_i$) for $i=1,2,..., k$? Or I should assume that $m+(x_2-y_1)+(x_3-y_2)+\cdots+(x_k-y_{k-1})+1.$ is just a lower bound of the peak indices in $1,2,..., y_k$ (well, probably it would be better if it is clarified, if this is the case, because now one assumes that this is the exact count; not that it changes anything though)

Lemma. No real number can occur infinitely many times in the sequence. Proof. Suppose such a number $c$ exists. Then $\frac{a_k}{k}=c\Rightarrow k=\frac{a_k}{c}<\frac{1}{c}$ is a contradiction. Let $b_k=\frac{a_k}{k}$. Define a sequence $1=p_1<q_1<p_2<q_2<\dots$ recursively. For each $k\geq 1$, let $q_k$ be the largest positive integer with $b_{q_k}=b_{p_k}$, and let $p_{k+1}$ be the smallest positive integer greater than $q_k$ such that there exists some positive integer $q>p_{k+1}$ with $b_q=b_{p_{k+1}}$. Now $a_{q_k}=\frac{q_k}{p_k}\cdot a_{p_k}>\frac{q_k}{p_k}\cdot a_{q_{k-1}}>\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot a_{q_{k-2}}>\dots>\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot\dots\cdot\frac{q_2}{p_2}\cdot a_{q_1}=\frac{q_k}{p_k}\cdot\frac{q_{k-1}}{p_{k-1}}\cdot\dots\cdot\frac{q_1}{p_1}\cdot a_1$. It suffices to show that $\prod\limits_{i=1}^{\infty}\frac{q_i}{p_i}$ diverges as there will be some large enough $k$ where $a_{q_k}$ will be forced to be greater than $1$. We now colour the positive integers as such: $p_i$ are green, $q_i$ are red, any $r$ with $p_i<r<q_i$ are yellow, any $r$ with $q_i<r<p_{i+1}$ are blue. Consider a blue integer $q_i<r<p_{i+1}$. By condition $\exists s\neq r, b_s=b_r$. If $s>r$, the minimality of $p_{i+1}$ is contradicted. If $s=p_j$ for some $j\leq i$, the maximality of $q_j$ is contradicted. If $s=q_j$, we can pick $s=p_j$ instead. If $s<r$ is blue, where $q_j<s<p_{j+1}$, the minimality of $p_{j+1}$ is contradicted. Thus $s$ must be a yellow number smaller than $r$. No two $r_1<r_2$ can map to the same $s$ since then $r_2$ can map to $r_1$, a contradiction. In other words, there is an injection mapping each blue number to a smaller yellow number. Let $x_1<x_2<\dots$ be the sequence of green and yellow numbers. We have $$\prod\limits_{i=1}^{\infty}\frac{q_i}{p_i}=\prod\limits_{i=1}^{\infty}\left(1+\frac{q_i-p_i}{p_i}\right)\geq 1+\sum\limits_{i=1}^{\infty}\frac{q_i-p_i}{p_i}\geq 1+\sum\limits_{i=1}^{\infty}\left(\frac{1}{p_i}+\frac{1}{p_i+1}+\dots+\frac{1}{q_i-1}\right)=1+\sum\limits_{i=1}^{\infty}\frac{1}{x_i}.$$But now consider the positive integers less than some $x_i$. There are exactly $i-1$ green and yellow numbers. There are not more blue numbers than yellow numbers by the above injection. There are also not more red numbers than green numbers. Thus there are at most $2i-2$ positive integers less than $x_i$, so $x_i\leq 2i-1$, so the above series diverges. $\Box$

My solution seems to be easier than the rest of the ones in this thread, but I think it's still valid, so here it is... Claim: Every value of $\frac{a_k}{k}$ only occurs finitely many times. Proof: Notice that $\frac{a_n}{n} < \frac{1}{n}$ and $\frac{1}{n}$ always decreases and gets arbitrarily close to $0$, so for any fixed value of $k$ there exists some constant $M$ so that for every $n > M$ we have $\frac{a_k}{k} > \frac{1}{n}$, so $\frac{a_k}{k} = \frac{a_n}{n}$ can only be possible for integers $n$ up to $M$, which is finite, as desired $\square$ Now suppose FTSoC every distinct value of $\frac{a_k}{k}$ appears at least twice. Now let $k_1$ be the largest integer such that $\frac{a_1}{1} = \frac{a_{k_1}}{k_1}$. Then it follows that $a_{k_1} = a_1 \cdot \frac{k_1}{1}$ Now let $k_2$ be the largest integer such that $\frac{a_{k_1+1}}{k_1+1} = \frac{a_{k_2}}{k_2}$. Then we must have $$a_{k_2} = a_{k_1 + 1} \cdot \frac{k_2}{k_1 + 1} > a_{k_1} \cdot \frac{k_2}{k_1 + 1} = a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1}$$ We can continue this inductively: suppose $$a_{k_n} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_n}{k_{n-1} + 1}$$and let ${k_{n+1}}$ be the largest integer such that $\frac{a_{k_n+1}}{k_n+1} = \frac{a_{k_{n+1}}}{k_{n+1}}$. Then we must have $$a_{k_{n+1}} = a_{k_n+1} \cdot \frac{k_{n+1}}{k_n+1} > a_{k_n} \cdot \frac{k_{n+1}}{k_n+1} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_{n+1}}{k_{n} + 1}$$as desired. Now note that the last expression can be rewritten as $$a_1 \cdot \frac{k_1}{k_1 + 1} \cdot \frac{k_2}{k_2 + 1} \cdot ... \cdot \frac{k_n}{k_n + 1} \cdot k_{n+1}$$and the expression $\frac{n}{n+1} = 1 - \frac{1}{n+1}$ is minimized when $n$ itself is minimized, so we are left to minimize each of $k_1$ through $k_{n+1}$. Notice that the minimum possible value of $k_1$ is $2$ otherwise the value of $\frac{a_1}{1}$ would be unique. This forces $k_2$ to be at least $4$ otherwise the value of $\frac{a_3}{3}$ would be unique as $\frac{a_1}{1} = \frac{a_2}{2} \neq \frac{a_3}{3}$. Then we see that the minimum possible value of $k_n$ is at least $2n$ as every term before has already been "paired up" and we need to "pair up" the next term as well or else it would be unique. This means $$a_{k_{n+1}} > a_1 \cdot \frac{k_1}{1} \cdot \frac{k_2}{k_1 + 1} \cdot ... \cdot \frac{k_{n+1}}{k_{n} + 1} \geq a_1 \cdot \frac{2}{1} \cdot \frac{4}{3} \cdot ... \cdot \frac{2n+2}{2n+1} = a_1 \bigg(1 + \frac{1}{1} \bigg)\bigg(1 + \frac{1}{3} \bigg)\bigg(1 + \frac{1}{5} \bigg)\cdots\bigg(1 + \frac{1}{2n+1}\bigg) > a_1 \cdot \bigg(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n+1}\bigg)$$ But since the original sequence is infinite and it can be derived that $\sum_{n=0}^{\infty }\frac{1}{2n+1}$ does not converge from the diverging summand $\sum_{n=1}^{\infty }\frac{1}{n}$, then the final expression in our inequality chain can be arbitrarily large for arbitrarily large $n$, contradicting the fact that it is less than $a_{k_{n+1}}$ and thus less than one, as desired $\blacksquare$

Suppose otherwise. Then each $\frac{a_n}{n}$ is equal to some other $\frac{a_m}{m}$. Let $i_1 = 1$. Define the sequence $i_1,i_2,\cdots$ by letting, for any $i_k$, $i_{k+1}$ be $1$ more than the maximal $i\ge i_k$ with $\frac{a_i}{i} \ge \frac{a_{i_k}}{i_k}$. Such maximal $i$ exists because $\frac{a_i}{i} \ge \frac{a_{i_k}}{i_k}$ implies $1>a_i \ge i \cdot \frac{a_{i_k}}{i_k}$. This way, we get the bound \[a_{i_{k+1}} > \frac{i_{k+1}-1}{i_k} \cdot a_{i_k}.\]Now, write \[a_{i_{k+1}} > a_1\prod_{j=1}^k \frac{i_{j+1}-1}{i_j} = a_1\prod_{j=1}^k \prod_{i_j < t < i_{j+1}} \frac{t}{t-1}.\]We now find a lower bound for this product. Suppose there was a particular minimum $j$ for which $i_{j+1} = 1+i_j$. Observe $j>1$, since $\frac{a_1}{1}$ is equal to another $\frac{a_n}{n}$ with $n>1$. The condition $i_{j+1} = 1+i_j$ implies that $\frac{a_{i_j}}{i_j}$ is greater than $\frac{a_i}{i}$ for all $i>i_j$, so there is some $i'<i_j$ with $\frac{a_{i_j}}{i_j} = \frac{a_{i'}}{i'}$. This $i'$ has $a_{i'}/i'$ not equal to any $a_{i_{j'}}/i_{j'}$ for $j'\ne j$. Thus for any $j\ge 1$, \[i_{j+1}-i_j-1 \ge 1+\#\{j'>j: i_{j'+1} = i_{j'}+1\}.\]This means that the following iterated algorithm will eventually yield $i_{j+1}-i_j = 2$ for all $j<k$ and $i_{k+1}-i_k \ge 2$: If $j<k$ and $i_{j+1}-i_j-1 \ge 2$, replace $i_{j+1}$ with $i_{j+1}-1$, which has the effect of replacing $\frac{i_{j+1}-1}{i_{j+1}-2}$ with $\frac{i_{j+1}}{i_{j+1}-1}$ and thus reducing the overall product value. Thus the product is at least \[\frac 21\cdot \frac 43\cdot \cdots \cdot \frac{2k}{2k-1} = \prod_{s=1}^k \left(1+\frac{1}{2s-1}\right) > \sum_{s=1}^k \frac{1}{2s-1}.\]Note that this sum is unbounded as $k\to \infty$. Thus for any $N>0$, there exists $k$ for which $1>a_{i_{k+1}} > Na_1$. This is absurd because we cannot have $\frac 1N > a_1$ for all $N>0$.

Most probably, the same core idea as in the above posts, but I still would put down the following solution hoping it's a bit different. Let $A:=\{a_n/n: n\in\mathbb N\}.$ Define $f:\mathbb{N}\to A$ as $f(n)=a_n/n$. We must prove the set $f^{-1}(a)$ consists of only one element for some $a\in A$ . For the sake of contradiction, assume $f^{-1}(a)$ is a set of at least two elements for every $a\in A$ Note that the family of sets $\{f^{-1}(a): a\in A\}$ is a disjoint cover of $\mathbb{N}$. Let $I_a$ be the closed interval with left and right ends respectively the minimal and maximal element of $f^{-1}(a), a\in A.$ It is straightforward to see that $f^{-1}(a)$ consists of only finitely many elements, so the definition is correct. Further, $\mathcal{I}:= \{I_a : a\in A\}$ is also a cover of $\mathbb{N}$, though it may not be disjoint. It's not difficult to be seen that the family $\mathcal{I}$ contains a subfamily (to avoid clumsy notations we denote it again by $\mathcal{I}$) which also is a cover of $\mathbb{N}$, and can be further split into two families $\mathcal{I}_1, \mathcal{I}_2,$ each one of disjoint sets. For any $I\in \mathcal{I}, I=[a,b]$ we define $g(I):= b/a$ (the fact that $b>a$ is essential for the following arguments). It follows $$\prod_{I\in \mathcal{I}_i} g(I)$$is finite for $i=1,2$ because otherwise the sequence $\{a_i\}_{i\in\mathbb N}$ would be unbounded.It means $$\prod_{I\in \mathcal{I}} g(I)<\infty\qquad(1)$$Let $I\in \mathcal{I}, I=[a,b].$ \begin{align*} g(I)=\frac{b}{a}&=1+\frac{b-a}{a}\\ &\ge 1+\frac{1}{a}+\frac{1}{a+1}+\dots +\frac{1}{b-1}\\ &\ge 1+\frac{1}{2}\left(\frac{1}{a}+\frac{1}{a+1}+\dots +\frac{1}{b-1}+\frac{1}{b} \right)\\ &=1+\frac{1}{2}\sum_{n\in I\cap\mathbb{N}}\frac{1}{n} \end{align*}Denote $\displaystyle g_1(I):=\frac{1}{2}\sum_{n\in I\cap\mathbb{N}}\frac{1}{n}.$ Further, we get \begin{align*} \prod_{I\in \mathcal{I}} g(I) &= \prod_{I\in \mathcal{I}} \left(1+g_1(I)\right)\\ &>\sum_{I\in\mathcal{I}} g_1(I)\\ &\ge \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}=\infty \end{align*}which contradicts $(1)$. (the fact that $\mathcal{I}$ is a family of (closed) intervals that cover $\mathbb{N}$ was used in the last inequality)

Let a season be a set of $a_i$ with the same value of $\frac{a_i}{i}$, and the finale be the last $a_i$ with this common value. AFTSOC that all seasons have size $\geq 2$. Now, we generate the sequence $f_i$ where $f_0=1$, and $f_{k+1}$ is the finale of $f_k+1$. (This is fine if $f_{k+1} = f_k+1$). Additionally, $f_k$ is well-defined because an infinitely long season clearly gives us $a_i>1$. Then, we have that \[a_{f_{n+1}} = \frac{f_{n+1}}{f_n+1} a_{f_n+1} >\frac{f_{n+1}}{f_n+1} a_{f_n} \]Thus, we have that \[a_{f_n} \geq \prod_{i=2}^{n} \frac{f_{n}}{f_{n-1}+1} a_{f_1}\] $\textbf{Remark: }$ Now, note that if any point we have $f_{k+1}\geq f_k+4$, then we can weaken the bound by instead going from $f_k\to f_k+2\to f_{k+1}$ instead of $f_k\to f_{k+1}$. This is because $\frac{f_{k+1}}{f_{k}+1} \cdot \frac{f_k}{f_{k-1}+1}>\frac{f_{k+1}}{f_k+3}\cdot \frac{f_k+2}{f_k+1}\cdot \frac{f_k}{f_{k-1}+1}$. Thus, we may assume that for all $k$ we have $f_{k+1}\leq f_k+3$, and thus, $f_k\leq 1+3k$. Back to the solution. The old bound becomes \[a_{f_n} \geq \prod_{i=2}^{n} \frac{f_{n}}{f_{n-1}+1} a_{f_1}\]Note that we have $f_n\geq 2n$ because if $f_n\leq 2n-1$, then there are $n$ seasons that end in the first $2n-1$ terms, which is absurd because each season has at least two episodes. Thus, we now have \[a_{f_n} \geq f_n \cdot \prod_{i=1}^{n-1} \frac{f_i}{f_i+1} a_1\geq 2n \cdot \prod_{i=1}^{n-1} \frac{2i}{2i+1} a_1\]Now, note that \[(\prod_{i=1}^{n-1} \frac{2i+1}{2i})^2\leq \prod_{i=1}^{n-1} \frac{2i+1}{2i}\cdot \prod_{i=1}^{n-1} \frac{2i}{2i-1} = 2n-1\]Thus, \[a_{f_n} \geq 2n\cdot \frac{1}{\sqrt{2n-1}} a_1\]so no matter how small $a_1$, we can still find some $n$ such that $\frac{2n}{\sqrt{2n-1}}> \frac{1}{a_1}$ so $a_{f_n} > 1$, a contradiction. Thus, the original hypothesis was incorrect, and there exists a season with size $<2$ and we're done. $\blacksquare$.

dgrozev wrote: . It's not difficult to be seen the family $\mathcal{I}$ can be split into two subfamilies $\mathcal{I}_1, \mathcal{I}_2$, each one of disjoint sets. Why that is true?

@shalomrav: Sorry, my bad, it's not written as it should be. I meant that the family $\mathcal{I}$ contains a subfamily (to avoid clumsy notations we denote it again by $\mathcal{I}$) which is also a cover of $\mathbb{N}$, and can be split into two families, each one of disjoint sets. We start with an interval $I_1\in \mathcal{I}$ which left end is $1$, that is $I_1=[1,i_1].$ Further, we choose $I_2$ to be the interval with the greatest right end which covers $i_1+1$. There is only one like that, since no two intervals in $\mathcal{I}$ can share a common end (the family of sets $\{f^{-1}(a): a\in A\}$ is a disjoint cover of $\mathbb{N}$). Let the right end of $I_2$ be $i_2$. We proceed by choosing $I_3$ be the interval with the greatest right end that covers $i_2+1$ and so on. Obviously $I_j,j=1,2,\dots$ is a cover of $\mathbb{N}$ and each of the families $\mathcal{I}_1:=\{I_j : j \text{ is odd}\} $ and $\mathcal{I}_2 :=\{I_j : j\text{ is even}\}$ consists of disjoint intervals.

I really like this problem. This feels like the converse of problems which are mainly combinatorics but the main step is a casual AM-GM or some algebraic manipulation, which just feel like some sort of toxic relationship lol.

Let $b_i = \frac{a_i}{i}$. Suppose for the sake of contradiction that every value in $b$ appears at least twice. Call an index $i$ terminal if there is no $j > i$ such that $b_j = b_i$, and nonterminal otherwise. Lemma 1: Let $s_1 < s_2 < \dots < s_k$ be nonterminal indices. Then there exists $m$ such that \[a_m \ge a_1 \prod_{i=1}^k \frac{s_i + 1}{s_i}.\]Proof: We perform the following algorithm that generates a set $S$ of disjoint intervals. There is a counter $c$ initially set to 1. Do the following: 1. Advance $c$ to the next nonterminal index in $s$, possibly doing nothing if $c$ is already on such an index. 2. Now there exists $d > c$ such that $b_d = b_c$ since $c$ is a nonterminal index. Insert $[c, d-1]$ into $S$. 3. Advance $c$ to $d$. 4. If $c > s_k$, then stop. Otherwise return to step 1. Let $S = \{[l_1, r_1], [l_2, r_2], \dots, [l_n, r_n]\}$ be the intervals of $S$ listed in increasing order. Then for all $i$, $b_{l_i} = b_{r_i + 1}$ by looking at step 2 of the algorithm. Therefore $a_{r_i + 1} = \frac{r_i + 1}{l_i} a_{l_i}$. Also, $a_{r_i + 1} \le a_{l_{i+1}}$ since $a$ is increasing and $r_i + 1 \le l_{i+1}$. Chaining these relations, between $a_1 \le a_{l_1}, a_{r_1 + 1}, a_{l_2}, a_{r_2 + 1}, \dots, a_{l_n}, a_{r_n + 1}$, together gives that \[a_{r_n + 1} \ge a_1 \prod_{i=1}^n \frac{r_i + 1}{l_i}.\]Let $T$ be the set of integers that lie in the union of the intervals in $S$. Now this relation can be rewritten as \[a_{r_n + 1} \ge a_1 \prod_{x \in T} \frac{x + 1}{x}.\]But notice that $T \subseteq \{s_1, s_2, \dots, s_k\}$ since the only time indices can be skipped in the algorithm is in step 1, which cannot skip over anything in $s$. Therefore we have \[a_{r_n + 1} \ge a_1 \prod_{x \in T} \frac{x + 1}{x} \ge a_1 \prod_{i=1}^k \frac{s_i + 1}{s_i},\]as desired. $\square$ Lemma 2: For any positive integer $n$, at least half of $\{1, 2, \dots, n\}$ are nonterminal. Proof: Since we assume each value in $b$ appears at least twice, there is a map sending any terminal index $x \le n$ to the greatest nonterminal index $y < x$ such that $b_y = b_x$. This map is injective because $b_x$ is pairwise distinct as $x$ varies over terminal indices. Therefore there are at least as many nonterminal indices at most $n$ as terminal indices, as desired. $\square$ To finish, consider the following blocks of indices: \[[1, 1], [2, 4], [5, 16], \dots, [4^n + 1, 4^{n+1}], \dots.\]For any block $[4^n + 1, 4^{n+1}]$, by Lemma 2 on the prefix up to $4^{n+1}$, at least $4^n$ indices in the block are nonterminal (there are at least $2 \cdot 4^n$ in the prefix, subtracting at most $4^n$ that should be excluded because they are before $4^n + 1$). Furthermore, in each block, the product of $\frac{x+1}{x}$ with $x$ ranging over all nonterminal indices in the block is at least \[\frac{4^{n+1} - 4^n + 2}{4^{n+1} - 4^n + 1} \frac{4^{n+1} - 4^n + 3}{4^{n+1} - 4^n + 2} \frac{4^{n+1} - 4^n + 4}{4^{n+1} - 4^n + 3} \dots \frac{4^{n+1} + 1}{4^{n+1}} = \frac{4^{n+1} + 1}{4^{n+1} - 4^n + 1},\]which is approximately $\frac{4}{3}$ for large $n$. Putting the nonterminal indices of sufficiently many blocks together and applying Lemma 1 lets us force $a_m$ to be arbitrarily large, contradicting the fact that $a_i < 1$ for all $i$.

Does this work? Strange feeling solution that looks at the beginning of chains (more or less) instead of the ending Suppose otherwise. Separate the positive integers into families based on the value of $\frac{a_i}{i}$. Call an integer $i$ sweet if $\frac{a_i}{i}<\frac{a_1}{1},\ldots,\frac{a_{i-1}}{i-1}$. Obviously each sweet integer is the smallest element of a family. Claim: There are infinitely many sweet integers (therefore families). Proof: $\frac{i}{a_i}$ is unbounded above. $\blacksquare$ Claim: Let $C=\tfrac{1}{a_1}>1$. For each family $\mathcal{F}$, we have $C\min_{\mathcal{F}} i \geq \max_{\mathcal{F}} i$. Proof: Otherwise, if $x=\min_{\mathcal{F}} i$ and $y=\max_{\mathcal{F}} i$, then $$\frac{a_x}{x}=\frac{a_y}{y} \implies a_y=\frac{ya_x}{x}>\frac{Cxa_x}{x}>Ca_1=1,$$contradiction. $\blacksquare$ Now let $1=i_1<i_2<\cdots$ be the sweet integers. We prove a key estimate. Claim: We have $a_{i_j} \leq \prod_{k=j}^\infty \frac{i_k}{i_{k+1}-1}$. Proof: We will prove by induction on $a \geq 0$ that $a_{i_j} \leq \prod_{k=j}^{j+a-1} \frac{i_k}{i_{k+1}-1}$ for all $j$, with the base case being trivial (empty product equals $1$). For the inductive step, if $a_{i_{j+1}} \leq \prod_{k=j+1}^{j+a}\frac{i_k}{i_{k+1}-1}$, then $$\prod_{k=j+1}^{j+a} \frac{i_k}{i_{k+1}-1}\geq a_{i_{j+1}}>a_{i_{j+1}-1}\geq \frac{i_{j+1}-1}{i_j}a_{i_j},$$with the last inequality from the fact that $i_j$ is sweet and no sweet integers exist in $(i_j,i_{j+1})$. Rearranging implies the desired conclusion. $\blacksquare$ We will also need a crucial "independent" lemma. Lemma: Let $S \subseteq \mathbb{Z}^+$ have positive upper density. Then $\prod_{x \in S} \frac{x+1}{x}$ diverges. Proof: Suppose it converged. Then, $$\prod_{x \in S} \left(1+\frac{1}{x}\right)>1+\sum_{x \in S} \frac{1}{x},$$but the sum is well-known to diverge for any $S$ with nonzero density. $\blacksquare$ This sets up the main claim of the problem. Claim: The set of sweet $i$ has density $1$ (equivalently, the set of non-sweet $i$ has density $0$). Proof: By our previous claim, we should have $$a_1 \leq \prod_{j=1}^\infty \frac{i_j}{i_{j+1}-1} \leq \prod_{j=1}^k \frac{i_j}{i_{j+1}-1}$$for all $k$. Since we are assuming convergence issues don't exist (otherwise we would get issues already), we can perform algebraic manipulations, writing $$a_1 \leq \frac{i_1}{i_{k+1}-1}\prod_{j=2}^k \frac{i_j}{i_j-1} \leq \frac{1}{i_k}\prod_{j=2}^k \frac{i_j}{i_j-1}.$$On the other hand, we have $$\frac{1}{i_k}\prod_{j=i_2}^{i_k} \frac{j}{j-1}=\frac{1}{i_2-1},$$which upon division means that $\prod_{\substack{i_2 \leq j \leq i_k\\j\text{ non-sweet}}} \frac{j}{j-1}$ is bounded above by $\frac{1}{a_1(i_2-1)}$, i.e. $\prod_{j\text{ non-sweet}} \frac{j}{j-1}$ converges, by sending $k \to \infty$ (the reason for proving the first claim is to ensure we can do this). Using the lemma on $S=\{a-1 \mid a\text{ non-sweet}\}$ implies the desired result. $\blacksquare$ Now, this result implies that for any (fixed) $0<r<1$, there exists some constant $N_r$ such that for all $N>N_r$, at least $rN$ of the integers in $[1,N]$ are sweet, else the lower density of sweet integers is at most $r$. But by the problem condition, there is an injection sending each sweet integer $i$ to a non-sweet integer (in the same family) which is at most $Ci$ according to our second claim, hence at least $rN$ of the integers in $[1,CN]$ are non-sweet, so we should have $\frac{C-r}{C}\geq r$, which is false as $r \to 1^-$, implying the desired contradiction. $\blacksquare$

Any number can only occur finitely many times in the sequence evidently. Claim: For integers $n_1 < n_2 < n_2 + 1 < n_3 < \dots$, it follows that \[ \frac{n_2}{n_1} \cdot \frac{n_3}{n_2 + 1} \cdot \frac{n_4}{n_3 + 1} \cdots \frac{n_{2k+2}}{n_{2k+1} + 1} \]diverges. Proof. Note that $\frac{b}{a} \cdot \frac{c}{b+1}$ is minimized when $b = a + 1$ for integers $b$. As such, the product is minimized when $n_3 - n_2 = \dots = 2$ and $n_2 - n_1 = 1$. The product then becomes \[ \frac{n_1 + 1}{n_1} \cdot \frac{n_1 + 3}{n_1 + 2} \cdot \frac{n_1 + 5}{n_1 + 4} \cdots \]This then becomes at least \[ \frac{p_1 + 1}{p_1} \cdot \frac{p_2 + 1}{p_2} \cdot \dots \]for primes $p$ which diverges by the harmonic series. $\blacksquare$ Then, if for each $a_i$ we have $\frac{a_j}{a_i} = \frac{j}{i}$, letting $\frac{a_{n_{i+1}}}{a_{n_i + 1}} = \frac{n_{i+1}}{n_i + 1}$ finishes.

REDACTED

We will show that if there is no isolated value, then the sequence is unbounded. We say that an index $i$ is a portal if there exists an index $j>i$ for which $\frac{a_i}{i}=\frac{a_j}{j}$. Claim: Portals have positive density. Partition the indices into "groups" based on the value of $\frac{a_i}{i}$. Each group consists of at least two indices, and in each group, all indices except for the last (if finite) are portals, which shows the claim (in fact, this argument shows $\rho \geq \frac12$). Now, we will employ the following algorithm to generate unbounded $a_i$: Start with $i=1$. Then, for each step in the algorithm, -if $i$ is not a portal, increase $i$ by $1$, which at least does not decrease $a_i$. -if $i$ is a portal, jump to the smallest $j>i$ such that $\frac{a_i}{i}=\frac{a_j}{j}$. This multiplies $a_i$ by $\frac{j}{i}$. Note that in order to show that the product of such $\frac{j}{i}$ diverges, it suffices to show that the numbers traversed via portals (as opposed to traversed through incrementing) have positive density, since the $a_i$ multiplies by $\frac{j}{i}$ when we use a portal. Due to the "greedy" nature of our algorithm, each portal is either used or skipped when travelling through another portal (they cannot be "stepped on but unused"). Thus, since all of the portals have positive density, this means that either used portals or skipped portals have positive density. In the former case, we are done, since each time we use a portal we traverse at least one index. In the second case, we are also done as skipped portals is a subset of numbers traversed via portal.