An Odd-Only Generalization

First notice, $\angle CA_iA_j$ contributes to $F$ if and only if $i-j = \pm 1$ and the minimum of $i$ and $j$ is an odd integer.

Now, we have, in degree measure, $\angle A_1CA_{n+1} = 180$

[As $A_1, C, A_{n+1}$ are collinear]

$\implies \angle A_1CA_2 + \angle A_2CA_3+ \angle A_3CA_4+ \angle A_{4}CA_{5}+ \ldots \ldots \ldots +\angle A_nCA_{n+1} = 180$

$\implies \angle A_1CA_2 + \angle A_{2+n}CA_{3+n}+ \angle A_3CA_4+ \angle A_{4+n}CA_{5+n}+ \ldots \ldots \ldots +\angle A_nCA_{n+1} = 180$

[As $\angle A_iCA_j$ and $\angle A_{i+n}CA_{j+n}$ are vertically opposite angles when $i=j-1$ and $j\leq n$ ]

$\implies 180-( \angle A_1+ \angle A_2) + 180-( \angle A_{2+n}+ \angle A_{3+n}) + 180-( \angle A_3+ \angle A_4)+ 180-(\angle A_{4+n}+A_{5+n})+ \ldots \ldots \ldots +180-(\angle A_n+A_{n+1}) = 180$

$\implies n\times 180-( \angle A_1+ \angle A_2+ \angle A_{2+n}+ \angle A_{3+n} + \angle A_3+ \angle A_4+\angle A_{4+n}+A_{5+n}+ \ldots \ldots \ldots +\angle A_n+A_{n+1} )= 180$

$\implies 180n - F(n) = 180$ [As, when $i \neq j$ , $A_{i+n}$ and $A_{j+n}$ are different points]

$\implies F(n) = 180(n-1).$

So, $F(2017)= 180\times 2016 = \boxed{362880}$ .