Problem Archive

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

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

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

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