Paying to roll a d20 more times for its maximum

You have a fair $20$-sided die. Before seeing any outcome, you must commit to how many extra rolls to buy beyond the first, at $\text{\char36}1$ per extra roll (zero is allowed). After all the rolls you are paid the largest value that came up. If you choose the number of extra rolls to maximise expected profit (payout minus cost), what is that maximal expected profit?

$\tfrac{1079}{80}=13.4875$. With $n$ rolls the maximum has mean $\sum_{k=1}^{20}\left(1-\left(\tfrac{k-1}{20}\right)^n\right)$, and profit is that minus $n-1$. The values are $10.5$, $12.825$, $13.4875$, $13.4833$ for $n=1,2,3,4$, and the marginal gain keeps shrinking, so three rolls (two extra) is optimal.