<s>[INST] General term of a sequence. So i have the following sequence:
${1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ...}$
Where the number $i$ appears $i + 1$ times.
I would like to know the $n$-th term of this sequence. I tried to analise certain patterns within the sequence, but wasn´t able to conclude anything so far. 
 [/INST] Given $\color{red}1, 1, \color{red}2, 2, 2, \color{red}3, 3, 3, 3, \color{red}4, 4, 4, 4, 4, \color{red}5, ...$, first note that: $a_{T(k)}=k$, where $T(k)$ is a triangular number. Indeed:
$$a_{T(1)}=a_1=\color{red}1; a_{T(2)}=a_3=\color{red}2; a_{T(3)}=a_6=\color{red}3; a_{T(4)}=a_{10}=\color{red}4; a_{T(5)}=a_{15}=\color{red}5; \ ...$$
Hence:
$$T(k)\le n\le T(k+1)-1, a_n=k \iff \\
\frac{k(k+1)}{2}\le n\le \frac{(k+1)(k+2)}{2}-1 \iff \\
k^2+k-2n\le 0\le k^2+3k-2n \iff \\
\left\lceil \frac{\sqrt{8n+9}-3}{2}\right\rceil\le k\le \left\lfloor \frac{\sqrt{8n+1}-1}{2}\right\rfloor$$
Hence:
$$a_n=\left\lceil \frac{\sqrt{8n+9}-3}{2}\right\rceil \ \text{or} \ \left\lfloor \frac{\sqrt{8n+1}-1}{2}\right\rfloor.$$
</s>