We first note that $a_1 = 1$, $a_2 = 3$ and $a_3 = 6$ hold. It is quite straight-forward to note that a block of length $a_{n-1} + 1$ starts with the number $a_n + 1$, and that this block therefore ends on the number $a_n + (a_{n-1} + 1)$, which yields $a_{n+1} = a_n + a_{n-1} + 2$.

This recursion has the constant solution $a_n \equiv -2$, and the homogeneous recursion $a_{n+1} = a_n + a_{n-1}$ is obviously of Fibonacci type. Writing the Fibonacci sequence in the form $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, $F_3 = 2$ and so on, we can easily check that $a_n = F_{n+3} - 2$ holds for $n = 1, 2, 3$, and this therefore yields an explicit formula for $a_n$. $\square$