No Fibbing 4

Since $\prod_{x=1}^X \frac{G_xG_{x+2}}{G_{x+1}^2} \; = \; \frac{\prod_{x=1}^X G_x \times \prod_{x=3}^{X+2}G_x}{\prod_{x=2}^{X+1}G_x^2} \; = \; \frac{G_1G_2\left(\prod_{x=3}^X G_x^2\right)G_{X+1}G_{X+2}}{G_2^2\left(\prod_{x=3}^XG_x^2\right)G_{X+1}^2} \; = \; \frac{G_1G_{X+2}}{G_2G_{X+1}}$ for any $X \ge 1$ we see that $\prod_{x=1}^\infty \frac{G_xG_{x+2}}{G_{x+1}^2} \; = \; \lim_{X \to \infty} \prod_{x=1}^X\frac{G_xG_{x+2}}{G_{x+1}^2} \; = \; \tfrac14\lim_{X \to \infty} \frac{G_{X+2}}{G_{X+1}} \; = \; \tfrac14(2+\sqrt{5})$ using problem 2 in this series. Thus the answer is $2+5+4=\boxed{11}$ .