{"app":{"locale":"en-GB"},"problem":{"content":"

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 $ \\Sigma_{d | 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 $F(n)=\\Sigma_{i=2}^n S(i!)$.$F(10)=4821.$

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

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$2^{\omega(n)}$