Puzzle Library

Let $ω (n)$ denote the number of distinct prime divisors of a positive integer $n$ .
So $ω (1) = 0$ and $ω (360) = ω (2^{3} \times 3^{2} \times 5) = 3$ .

Let $S (n)$ be $\sum_{d ∣ n} 2^{ω (d)}$ .
E.g. $S (6) = 2^{ω (1)} + 2^{ω (2)} + 2^{ω (3)} + 2^{ω (6)} = 2^{0} + 2^{1} + 2^{1} + 2^{2} = 9$ .

Let $F (n) = \sum_{i = 2}^{n} S (i!)$ . $F (10) = 4821.$

Find $F (10 000 000)$ . Give your answer modulo $1 000 000 087$ .

Write-up coming later

The complete problem is available here. An approach, code, and answer will be added later.