If you look at $\lfloor \sqrt {a_n} \rfloor$ there are two in a row of the same almost all the time. This is not surprising, because if $a_n$ is a little larger than $m^2$, and it takes two additions of $m$ to get greater than $(m+1)^2=m^2+2m+1$. The curious pattern is that it takes three when $m$ is a power of $2$. This is because you hit $m^2=2^{2p}$ exactly and two additions of $m$ just get you to $m^2+2m \lt (m+1)^2$. I haven't proven this yet, but it will allow a closed form for $a_n$ in terms of the sum of two triangular numbers plus $2^{p}-1$ for the extra powers of $2$.
A recursion that almost shows that if we ever have $a_n=2^{2p}$ the pattern is as claimed is as follows: Let $a_n=2^{2p}$. We will add three copies of $2^p$, then two copies of each integer from $2^p+1$ through $2^{(p+1)}-1$. Having done so, we will have $a_{n+2^{(p+1)}+1}=2^{2p}+3\cdot 2^p+2\sum_{i=2^p+1}^{2^{P+1}-1} i \\=2^{2p}+3 \cdot 2^p+2\frac{(2^{p+1}-1)2^{p+1}}2-2\frac{2^p(2^p+1)}2 \\ =2^{2p+2}$ I haven't proved we don't hit an exact square before that.
Since $a_1=1$, we can say that $a_{2^{m+1}+m-1}=2^{2m}$ The places you hit $2^{2p}$ are at $n=1,4,9,18,35,68 \ldots = 2^{p+1}+p-1$, which is OEIS A083706.