Ellip-tastic Packing!

Let the center of the ellipse be $O$ at the origin, let its semi-major axis be $a$ , let its semi-minor axis be $1$ , let the radius of each circle be $r$ , and let the following circles have centers $A$ and $B$ :

Then the ellipse has an equation of $\cfrac{x^2}{a^2} + y^2 = 1$ , and the right-most circle has an equation of $(x - a + r)^2 + y^2 = r^2$ .

Combining these two equations and solving for $x$ gives $x = a$ and $x = \cfrac{a(a^2 - 2ar + 1)}{a^2 - 1}$ .

For them to share one point of tangency, $x = a = \cfrac{a(a^2 - 2ar + 1)}{a^2 - 1}$ , which solves to $r = \cfrac{1}{a}$ .

Also, $AB = 2r$ , $OA = a - 3r$ , and $OB = 1 - r$ , so that by the Pythagorean Theorem on $\triangle AOB$ , $(a - 3r)^2 + (1 - r)^2 = 4r^2$ .

Substituting $r = \cfrac{1}{a}$ into $(a - 3r)^2 + (1 - r)^2 = 4r^2$ solves to $a = \frac{1}{3}(-1 + \sqrt[3]{62 - 3\sqrt{183}} + \sqrt[3]{62 + 3\sqrt{183}})$ .

The eccentricy is then $E = \cfrac{\sqrt{a^2 - 1}}{a} \approx 0.8855623$ , which makes $\lfloor 10^6 E \rfloor = \boxed{885562}$ .