Ignore $t_0$ and $t_1$, the first terms $-1$ are: $\;(1\cdot2)\;(2\cdot2)\;(3\cdot2)\;(4\cdot2)\;(3\cdot4)\;(4\cdot4)\;(5\cdot4)\cdots$ which suggests:$$t_{n}=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+1$$
(this looks more complicated than it actually is)
Note that when $n$ is a multiple of $4$ we have $\lfloor\frac{n-1}{4}\rfloor=\lfloor\frac{n}{4}-1\rfloor$ and otherwise $\lfloor\frac{n-1}{4}\rfloor=\lfloor\frac{n}{4}\rfloor$
$\bullet\;\;n$ multiple of $4:$
$$t_{n+1}=2{\lfloor\frac{n}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+1=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+n+1$$
$$=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+2{\lfloor\frac{n}{4}\rfloor}\cdot 2+1=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n+2-2\lfloor\frac{n}{4}\rfloor\right)+1$$
$$=2{\lfloor\frac{n}{4}\rfloor}\cdot \left(n-1-2\lfloor\frac{n}{4}-1\rfloor\right)+2\lfloor\frac{n}{4}\rfloor+1$$$$=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+2\lfloor\frac{n-1}{4}+1\rfloor+1$$ $$=t_n+2\lfloor\frac{n-1}{4}+1\rfloor$$
$\bullet\;\;n$ not a multiple of $4:$
$$t_{n+1}=2{\lfloor\frac{n}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n}{4}\rfloor\right)+1=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-2\lfloor\frac{n-1}{4}\rfloor\right)+1$$
$$=2{\lfloor\frac{n-1}{4}+1\rfloor}\cdot \left(n-1-2\lfloor\frac{n-1}{4}\rfloor\right)+2\lfloor\frac{n-1}{4}+1\rfloor+1=t_n+2\lfloor\frac{n-1}{4}+1\rfloor$$
This verifies our assumption. The sum was nicely done by Patrick Li so I'll leave it there.