Cutting chain links until only the top orb is left

Seven orbs, numbered $1$ through $7$ from the top, hang in a vertical chain; orb $1$ is fixed to the ceiling and consecutive orbs are joined by single links. At each step one of the links still present is chosen uniformly at random and cut, and everything hanging below the cut drops away and breaks. On average, how many cuts are made before orb $1$ is the only orb remaining?

$\tfrac{49}{20}$. With $n$ links, the first cut is uniform among them and leaves a uniformly random number $0,\dots,n-1$ of links, so $a_n=1+\tfrac1n\sum_{k<n}a_k$, which simplifies to $a_n=a_{n-1}+\tfrac1n$. Hence $a_n=H_n$, and six links give $H_6=\tfrac{49}{20}$.