This sounds fun, until you realize your life is at stake

It is easy to see that the minimum score is attained if we only have a card of $2048$ when the game ends and the rest of the cards are $2$ since we do not perform unnecessary operations or gain excess points by fusing extra cards together, ie. if we play optimally, the number of operations and points "awarded" are both minimized.

Define the function $f(n)$ and $g(n)$ where $f(n)$ denotes the least possible points "awarded" when a card of $2^{n}$ is produced while $g(n)$ denotes the least possible number of operations performed when a card of $2^{n}$ is produced. In other words, since $2^{11}=2048$ , we want to find $f(11)+g(11)$ .

It is trivial to note that $f(1)=0$ . To compute $f(2)$ , note that to obtain a card of $4$ , we fuse together two cards of $2$ , so $f(2)=2.2^{1}=4$ . To compute $f(3)$ , since we are fusing together two cards of $4$ , we end up with $2.2^{2}+2f(2)$ points. Note that $2.2^{2}$ are the points obtained upon fusing while $2f(2)$ points are cumulative from our previous operations leading to the production of two cards of $4$ .

This can be easily generalized. Therefore, we obtain the following recurrence formula:

$f(n)=2f(n-1)+2^{n}$ .

This recurrence formula is nice and all, but let’s try and find a closed form for $f(n)$ . Let’s use $n=3$ and $n=4$ as motivating examples.

For $2^{3}=8$ , note that the points awarded come from $2^{3}$ when the card is produced and $2^{2}.2$ as the cumulative points. So we can write $f(3)=2^{3}.1+2^{2}.2$ .

Similarly, for $2^{4}=16$ , the points awarded come from $2^{4}$ and $2f(3)=2(2^{3}.1+2^{2}.2)$ . Therefore, we can write $f(4)=2^{4}.1+2^{3}.2+2^{2}.2^{2}$ .

These observations lead naturally to the conjecture:

$f(n)=2^{n}.1+2^{n-1}.2+ \cdots + 2^{2}.2^{n-2}$

$=(n-1)2^{n}$ .

Substituting this into our recurrence formula above, we find that this formula is indeed true.

Now, to calculate $g(n)$ , first notice that $g(n+1)=2g(n)+1$ . To get the card $2^{n+1}$ form fusing two cards of $2^{n}$ , we need $1$ operation while $2g(n)$ denotes the cumulative operations leading to the production of the two $2^{n}$ cards.

Let’s try to find a closed form for $g(n)$ . After writing out a few values, we find that $g(1)=0, g(2)=1, g(3)=3, g(4)=7, g(5)=15$ . It does not take too much ingenuity for us to conjecture that $g(n)=2^{n-1}-1$ . Substituting this into our recurrence formula, we find that this is indeed true.

Therefore,

$g(n)+f(n)= 2^{n-1} -1 + (n-1)2^{n}$

$g(11)+f(11)=2^{10}-1+(10)2^{11}=\boxed{21503}$

It is easy to see that this minimum score is indeed attainable. In fact, it is suggested by the recurrence formula. To get the card $2^{n+1}$ , we focus on assembling two cards of $2^{n}$ separately before fusing them together, starting from $n=1$ . This way, we can ensure that no extra cards is produced.

When you make the first $4$ , you receive the minimum of $2^2+1$ , where $2^2$ are the points awarded and $1$ is the number of operation.

When you make the first $8$ , you receive the minimum of $2(2^2+1)+2^3+1=2^3+2^3+2+1=(3-1) \times 2^3+2+1$ , where $2(2^2+1)$ is because you've made 2 number $4$ 's.

When you make the first $16$ , you receive the minimum of $2(2^3+2^3+2+1)+2^4+1=2^4+2^4+2^4+2^2+2+1=(4-1) \times 2^4+2^2+2+1$ , where $2(2^3+2^3+2+1)$ is because you've made 2 number $8$ 's.

So, when you make the first $2^n$ , you receive the minimum of $(n-1) \times 2^n +2^{n-2}+2^{n-3}+ ...+2+1$ .

With $n=11$ , when you make the first $2048=2^{11}$ , you receive the minimum of $10 \times 2^{11}+2^9+2^8+...+2+1=20480+2^{10}-1=\boxed{21503}$ .