Japan Mathematical Olympiad

Assume that $a_{n+1}\ge a_{n+2}$ holds for some positive integers $n$. Since any positive integer $i$ with $a_i\le a_{n+2}+c$ must also satisfy $a_i\le a_{n+1}+c$, it follows that $a_n\ge a_{n+1}$. Hence, if $a_n<a_{n+1}$ holds for some positive integer $n$, then $a_{n+1}<a_{n+2}$ follows, and by induction, we have $a_n<a_{n+1}<a_{n+2}<a_{n+3}<\dots$.

Similarly, when $a_n>a_{n+1}$ holds for some positive integer $n$, we have $a_n>a_{n+1}>a_{n+2}>\dots$, especially $a_{n+d}\le a_n-d$ for any non-negative integer $d$. This leads to a contradiction since $a_{n+a_n}\le 0$.

Suppose that $a_n=a_{n+1}$ for all positive integers $n$. Then, as all integers $i\ge 2$ satisfy $a_i=a_2\le a_2+c$, this contradicts the fact that there are exactly $a_1$ positive integers $i$ satisfying it. Therefore, there exists a positive integer $k$ such that $a_k<a_{k+1}$, and from the above discussion, we have $a_1\le a_2\le \cdots \le a_k<a_{k+1}<a_{k+2}<\dots$.

For integers $n\ge k$, since any integer $i>n+c+1$ satisfies $a_i>a_{n+c+1}\ge a_{n+1}+c$, we have $a_n\le n+c+1$. Therefore, if we set $b_n=a_n-n$ ($n\ge k$), then we have $b_n\le c+1$ and $b_k\le b_{k+1}\le b_{k+2}\le \dots$. Thus, there exist an integer $d$ and an integer $M\ge k$ such that for $n\ge M$, $b_n=d$, i.e., $a_n=n+d$. Since $a_1\le a_2\le \cdots \le a_{M+c+1}<a_{M+c+2}<\dots$, for any positive integer $i$, $a_i\le a_{M+1}+c=a_{M+c+1}$ and $i\le M+c+1$ are equivalent. Therefore, $a_M=M+c+1$, which is also equal to $M+d$, and for integers $n\ge M$, we have $a_n=n+c+1$.

Suppose that for an integer $N\ge 2$, we have $a_n=n+c+1$ for all $n\ge N$. Since $a_1\le a_2\le \cdots \le a_{N+c}<a_{N+c+1}<\dots$, for any positive integer $i$, $a_i\le a_N+c=a_{N+c}$ and $i\le N+c$ are equivalent. Therefore, $a_{N-1}=N+c$.

Thus, by induction, we have $a_n=n+c+1$ for any positive integer $n$. This indeed satisfies the problem condition because $a_i\le a_{n+1}+c$ and $i\le a_n$ are equivalent for any positive integer $i$.