Polygon

It is a standard result that the constructible regular $n$ -gons are such that $n \ge 3$ can be written in the form $n \; = \; 2^a p_1 p_2 \cdots p_t$ where $a \ge 0$ and $p_1,p_2,\ldots,p_t$ are distinct Fermat primes (primes of the form $F_m = 2^{2^m} + 1$ for some integer $m \ge 0$ ) when $t > 0$ . The sum of all such reciprocals is $A \; = \; \left(\sum_{a \ge 0} 2^{-a}\right) \prod_{p} \left(1 + \tfrac{1}{p}\right) - 1 - \tfrac12 \; = \; 2\prod_{p} \left(1 + \tfrac{1}{p}\right) - \tfrac32$ where the product is over all Fermat primes. Now $F_0=3$ , $F_1=5$ , $F_2 = 17$ , $F_3=257$ and $F_4 = 65537$ are all prime, but $F_m$ is not prime for $5 \le m \le 32$ . Thus $A \; \approx \; 2 \times \tfrac43 \times \tfrac65 \times \tfrac{18}{17} \times \tfrac{258}{257} \times \tfrac{65538}{65537} - \tfrac32 \; = \; \tfrac{4869735552}{1431655765}$ and hence the answer is $\lfloor 10000A \rfloor = \boxed{19014}$ . This approximation to $A$ is certainly good enough to estimate $A$ to the required $4$ decimal places, since the next Fermat prime, if it exists and whatever it may be, will be seriously huge, and will therefore not change the answer obtained by this approximation. Of course, if there are no other Fermat primes beyond $F_4$ , then the formula given above is not an approximation.