Something looks familiar about this recurrence...

If we write $P_n(ix) = i^nQ_n(x)$ then we have $Q_0(x) = 1$ , $Q_1(x) = x$ and $Q_n(x) \; = \; 2xQ_{n-1}(x) + Q_{n-2}(x)$ Thus $Q_0(\tfrac12) = 1$ , $Q_1(\tfrac12) = \tfrac12$ and $Q_n(\tfrac12) \; = \; Q_{n-1}(\tfrac12) + Q_{n-2}(\tfrac12)$ so that $Q_n(\tfrac12) = \tfrac12L_n$ for all $n \ge 0$ . Moreover $Q_0'(x) = 0$ , $Q_1'(x) = 1$ and $Q_n'(x) \; = \; 2xQ_{n-1}'(x) + Q_{n-2}'(x) + 2Q_{n-1}(x)$ so that $Q_0'(\tfrac12) = 0$ , $Q_1'(\tfrac12) = 1$ and $Q_n'(\tfrac12) \; = \; Q_{n-1}'(\tfrac12) + Q_{n-2}'(\tfrac12) + 2Q_{n-1}(\tfrac12) \; = \; Q_{n-1}'(\tfrac12) + Q_{n-2}'(\tfrac12) + L_{n-1}$ Since $nF_n \; = \; (n-1)F_{n-1} + (n-2)F_{n-2} + L_{n-1}$ we deduce that $\big[Q_n'(\tfrac12) - nF_n\big] \; = \; \big[Q_{n-1}'(\tfrac12) - (n-1)F_{n-1}\big] + \big[Q_{n-2}'(\tfrac12) - (n-2)F_{n-2}\big]$ and since $Q_n'(\tfrac12) - nF_n = 0$ when $n = 0$ and $1$ , we deduce that $Q_n'(\tfrac12) \; = \; nF_n \hspace{2cm} n \ge 0 \;.$ Thus $S = 2017F_{2017}$ and $T = 17F_{17}$ , and hence the fractional part of $\tfrac{S}{T}$ is $\tfrac{4142}{27149}$ , making the answer $\boxed{4142}$ .

This approach seemed simpler than considering the Chebyshev polynomials $P_n(x)$ trigonometrically.