
Largest Prime Factors of Consecutive Numbers
Let $f(n)$ be the largest prime factor of $n$.
Let $g(n) = f(n) + f(n + 1) + f(n + 2) + f(n + 3) + f(n + 4) + f(n + 5) + f(n + 6) + f(n + 7) + f(n + 8)$, the sum of the largest prime factor of each of nine consecutive numbers starting with $n$.
Let $h(n)$ be the maximum value of $g(k)$ for $2 \le k \le n$.
You are given:
- $f(100) = 5$
- $f(101) = 101$
- $g(100) = 409$
- $h(100) = 417$
- $h(10^9) = 4896292593$
Find $h(10^{16})$.