Group size maximising expected unshared birthdays

For a group of $n$ people with independent birthdays uniform over $365$ days, let $b(n)$ be the expected number of days on which exactly one person in the group has their birthday. Two integers $n$ tie for the largest value of $b(n)$. What is their sum?

$729$. By linearity, $b(n)=n\left(\tfrac{364}{365}\right)^{n-1}$, since each person is alone on their day with that probability. The ratio $b(n+1)/b(n)=\tfrac{n+1}{n}\cdot\tfrac{364}{365}$ equals $1$ exactly at $n=364$, so $b(364)=b(365)$ is the maximum.