Packing an infinite sequence of disks
Solution 1:
Part I - Exact result for small $a$
Let $\rho_c \approx 0.75487766624669$ be the unique real root of the cubic equation $\rho^3 + \rho^2 - 1 = 0$. When $a \le a_c = (1 + \rho_c)^2 \approx 3.079595623491439$, $$Q(a) = (\sqrt{a}-1)^2$$
Let $\rho = \sqrt{q}$. Given any two circles with area $1$ and $q$, their diameter will be $\frac{2}{\sqrt{\pi}}$ and $\frac{2\rho}{\sqrt{\pi}}$. Any geometric figure containing both of them must have a diameter $\ge \frac{2(1+\rho)}{\sqrt{\pi}}$. In particular, if a circle of area $a$ contains both of them, we will have $$2\sqrt{\frac{a}{\pi}} \ge \frac{2(1+\rho)}{\sqrt{\pi}}\quad\iff\quad a \ge (1+\rho)^2 \quad\iff\quad q \le (\sqrt{a}-1)^2$$
This gives us an upper bound for all $a$:
$$Q(a) \le (\sqrt{a}-1)^2$$
When $a \le a_c = (1 + \rho_c)^2$, we can actually pack a sequence of circles of area $\rho^{2k}, k = 0, 1, \ldots$ where $\rho = \sqrt{a}-1$ inside a circle of area $a$. This leads to a lower bound
$$Q(a) \ge (\sqrt{a}-1)^2$$
and above assertion will follow. To see how this is done.
We start with a circle of radius $\sqrt{\frac{a}{\pi}} = \frac{1+\rho}{\sqrt{\pi}}$ at the origin. Let us call it $C_a$.
We then place two new circles, $C_b$ and $C_0$, with radius $\frac{1}{\sqrt{\pi}}$ and $\frac{\rho}{\sqrt{\pi}}$ inside $C_a$. We position them in such a way so that their centers lying on the $x$-axis and $C_a, C_b$ and $C_0$ are kissing each other.
There are now two slots inside $C_a$, one above and one below the $x$-axis where we can place a $4^{th}$ circle kissing $C_a, C_b$ and $C_0$. We place a circle kissing $C_a, C_b, C_0$ in the upper slot and call it $C_1$. We place another circle kissing $C_a, C_b, C_0$ in the lower slot and call it $C_{-1}$.
Now above $C_1$, there is a slot inside $C_a$ we can place a $4^{th}$ circle kissing $C_a, C_b$ and $C_1$. We place such a circle there and call it $C_{2}$.
- Now above $C_2$, there is a slot inside $C_a$ we can place a $4^{th}$ circle kissing $C_a, C_b$ and $C_2$. We place such a circle there and call it $C_{3}$.
We can repeat these steps repeatedly. If we do the same thing to those slots below $x$-axis, we will ultimately obtain a sequences of circles $\ldots, C_{-2}, C_{-1}, C_0, C_1, C_2, \ldots$. such that for every integer $k$, the two quadruples of circles $( C_a, C_b, C_{k}, C_{k+1})$ and $( C_a, C_b, C_{k}, C_{k-1})$ are kissing among themselves.
This sequence of circles form part of an Apollonian gasket (see figure at end of Part I). Let $\frac{\rho_k}{\sqrt{\pi}}$ be the radius of $C_k$. If we apply Descartes' four circle theorem to the pair of quadruples of circles for each $k$, we find $\lambda = \frac{1}{\rho_{k+1}}$ and $\frac{1}{\rho_{k-1}}$ are the two roots of a quadratic equation
$$\left( \lambda + 1 + \frac{1}{\rho_k} - \frac{1}{1+\rho}\right)^2 = 2 \left(\lambda^2 + 1 + \frac{1}{\rho_k^2} + \frac{1}{(1+\rho)^2}\right)$$
This leads to $$\frac{1}{\rho_{k+1}} + \frac{1}{\rho_{k-1}} = 2 \left(1 + \frac{1}{\rho_k} - \frac{1}{1+\rho}\right) \quad\iff\quad \frac{1}{\rho_{k+1}} - \frac{2}{\rho_k} + \frac{1}{\rho_{k-1}} = \frac{2\rho}{1+\rho} $$
The RHS is an inhomogeneous recurrence relation in $\frac{1}{\rho_k}$ with constant terms. Since the associated polynomial has a double root at $1$, the general solution of it has the form
$$\frac{1}{\rho_k} = A + B k + C k^2$$
for suitable chosen constants $A, B, C$. Since $\rho_0 = \rho$ and by construction, $\rho_1 = \rho_{-1}$, we find $$\frac{1}{\rho_k} = \frac{1}{\rho} + \frac{\rho}{1+\rho}k^2 \quad\iff\quad \rho_k = \frac{\rho}{1 + \frac{\rho^2}{1+\rho} k^2} $$ If $\rho$ is chosen such that
$$\rho^{2k} \le \rho_k \quad\text{ for all } k \ge 1\tag{*1}$$
then the circles $C_k$ will be large enough to hold a sequences of circles of area $1, q, q^2, q^3, q^4, \ldots$ in following order $C_b, C_0, C_{1}, C_{-1}, C_{2}, C_{-2}, \ldots$.
A few plots of $\rho_k - \rho^{2k}$ for various $k$ shows that $(*1)$ is true for small $\rho$. The smallest $\rho$ that cause trouble fails at $k = 1$. This leads to a determination of the critical value $\rho_c$ as a root of
$$\frac{\rho}{1 + \frac{\rho^2}{1+\rho} } - \rho^2 = 0 \quad\implies\quad \rho^3+\rho^2-1 = 0$$
appeared in the assertion at the beginning of this answer.
Following is a picture showing the circle arrangement at $\rho = 0.75$.
- The pink circle is circle $C_b$.
- The orange circles with label $C_k, k = 0, \pm 1, \pm 2$ isn't circle $C_k$.
Circle $C_k$ is the blue circle behind it. - Together with $C_b$, the orange circles are those circles with area $q^k$
which we want to fit inside the circle with area $a$.
The $\rho = 0.75$ used in this figure is very close to $\rho_c$. As one can see, circles $C_0, C_1, C_2, \ldots$ get blocked by the corresponding orange circles and nearly invisible. When we increase $\rho$ beyond $\rho_c$, the orange circle on top of $C_1$ grows too large and $C_1$ no longer able to contain it. Since $C_1$ is one of the largest circle one can place inside $C_a \setminus ( C_b \cup C_0 )$, $Q(a)$ no longer equal to $(\sqrt{a}-1)^2$.
Part II - Lower bound estimation for large $a$
Let us switch to case of large $a$ and establish a non-trivial lower bound
for $Q(a)$.
Let $\rho = \sqrt{q}$ once again.
For any $r \ge 3$, let $m = \lfloor \frac{r-1}{2}\rfloor \ge 1$. Start with a regular hexagon with side $2m$, we subdivide it into a bunch of equilateral triangles of side $2$. This give us a hexagonal patch on a triangular lattice with spacing $2$. This patch will contains $3m(m+1)$ faces and $3m(m+1) + 1$ vertices. If we place a circle of unit radius on each of the vertices, the resulting collection of circles can be placed inside a big circle of radius $r$. Since $r \ge 2m+1 > r-2$, we have
$$3m(m+1) + 1 = \frac{3(2m+1)^2+1}{4} > \frac34(r-2)^2$$
This means for any $r \ge 3$, we can pack at least $\left\lceil \frac34(r-2)^2 \right\rceil$ unit circles inside a circle of radius $r$.
As a consequence, for any $a > 9a_c$, we can pack at least
$$N = \left\lceil \frac{3a}{4a_c}\left(1 - 2\sqrt{\frac{a_c}{a}}\right)^2 \right\rceil$$
circles of area $a_c$ inside a circle of area $a$.
For any $\rho$ such that $\rho^N \le \rho_c$, let $D_0, D_1, D_2, \ldots$ be a sequence of circles with $\text{Area}(D_k) = q^k = \rho^{2k}$. We can split them into $N$ groups:
$$ \{\;D_0, D_N, D_{2N}, \ldots\;\}, \{\;D_1, D_N+1, D_{2N+1}, \ldots\;\}, \ldots\\ \{\;D_{N-1}, D_{N+(N-1)}, D_{2N+(N-1)}, \ldots\;\}. $$ Since $\rho^N \le \rho_c$, we can fit each group inside a circle of area $a_c$. As a result, we can pack all $D_k$ inside a single circle of area $a$. This establishes a lower bound for $Q(a)$ as
$$ Q(a) \ge \rho_c^{\frac{2}{N}} \ge \rho_c^{ \frac{8a_c}{3a}\left(1-2\sqrt{\frac{a_c}{a}}\right)^{-2} } \sim 1 - \frac{\lambda_\ell}{a} + O\left(a^{-3/2}\right) \tag{*2} $$ where $\displaystyle\;\lambda_\ell = \frac{8|\log\rho_c|a_c}{3} \approx 2.30928260910042$.
Since the total area of the circles $D_k$ cannot exceed $a$, we find a trivial upper bound for $Q(a)$
$$\frac{1}{1-Q(a)} \le a \quad\implies\quad Q(a) \le 1 - \frac{1}{a}$$
Compare this with $(*2)$, we find $(*2)$ implies for large $a$, we can cover at least $\sim 40\%$ of area $a$ by the circles $D_k$. Intuitively, this looks a little bit too low to me. I believe there is still a huge room of improvement for the lower bound estimation.
Update - a more satisfactory lower bound estimate
After more fooling around, I find above lower bound estimate is indeed far from optimal.
For any integer $n \ge 24$, let
$$\alpha_n = \frac{\sqrt{3}}{2(n-\frac12)},\quad \quad\theta_n = \frac{2\pi}{n-\frac12}, \quad r_n = \frac{\pi}{n} \quad\text{ and }\quad z_n = e^{ -(\alpha_n+ i)\theta_n }.$$
Identify the plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$. For each $k \in \mathbb{N}$, place a a circle of radius $r_n |z_n|^k$ at the point $z_n^k$. We find all these circles lying inside a circle centered at origin with radius $1+r_n$. When $n \ge 24$, these small circles won't overlap. The ratio of the area between the circle at $z_n^k$ and $z_n^{k+1}$ are all equal to $e^{-2\alpha_n\theta_n}$.
For $a \ge 75$, if we associate a integer $n$ to it by the formula $n = \left\lfloor\pi (\sqrt{a} - 1)\right\rfloor$, $n$ will be $\ge 24$. Scaling the circle configuration in previous paragraph appropriately, we get $$\begin{align} Q(a) & \ge Q\left(\left(\frac{n}{\pi} + 1\right)^2\right) = Q\left(\left(\frac{1+r_n}{r_n}\right)^2\right)\\ & \ge e^{-2\alpha_n\theta_n} = \exp\left[-\frac{2\sqrt{3}\pi}{(n-1/2)^2}\right] \ge \exp\left[-\frac{1}{\delta_c(\sqrt{a}- k)^2}\right] \end{align} \tag{*3} $$ where $\delta_c = \frac{\pi}{2\sqrt{3}} \approx 0.90689968211711$ is the optimal packing density for tiling the plane with equal circles and $k = 1 + \frac{3}{2\pi}$. One consequence of $(*3)$ is
$$\liminf_{a\to\infty} \frac{1}{a(1-Q(a))} \ge \delta_c$$
What this means is in the limit of very large $a$, at least $\sim 90\%$ of the area $a$ can be covered by smaller circles whose radii are in geometric progression.