Are you smarter than a 8th grader?

Think of the zeroes as the "first" generation, and of the ones as the "later" generation, and we have the typical procedure for generating the Fibonacci sequence: every zero turns into a one, and every one remains while generating a new zero.

If we label the Fibonacci sequence as $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3,$ etc. then we get $A(n) = F_{n+1}.$ Next, the number of zeroes in pattern $X(n)$ is equal to the number of ones in pattern $X(n-1)$ , which in turn is equal to the number of digits in pattern $X(n-2)$ . Also, every zero will be followed by a one, except if the zero is placed at the end. It is easy to see that $X(n)$ ends in zero precisely if $n$ is even. Therefore $B(n) = \left\{\begin{array}{ll} A(n-2)-1 & \text{if}\ n\ \text{is even,} \\ A(n-2) & \text{if}\ n\ \text{is odd.} \end{array}\right.$ Thus, $A(23) + B(23) = F_{24} + F_{22},$ and my favorite way to calculate this is as $\frac{(\tfrac12+\tfrac12\sqrt5)^{24} - (\tfrac12-\tfrac12\sqrt5)^{24} + (\tfrac12+\tfrac12\sqrt5)^{22} - (\tfrac12-\tfrac12\sqrt5)^{22}}{\sqrt5}$ $=\boxed{64079}.$

Lol this was cool :)

Define two more recursions.

$\begin{aligned} M(t) & = \text{ Number of times 1 occurs in X(t) } \\ N(t) & = \text{ Number of times 0 occurs in X(t) } \end{aligned}$

For the sake of convenience, a 0 implies one 1 in the next term and a 1 implies a 10 in the next term. $0 \implies 1 \\ 1 \implies 10$ Now observe that $\begin{aligned} M(x) & = N(x-1) + M(x-1) \\ N(x) & = M(x-1) \end{aligned}$

We can simplify this into $N(x) = N(x-1) + N(x-2)$ with $N(1) = 0, N(2) = 1$ .

Now observe that $A(x+1) = 2M(x) + N(x) = N(x+1) + N(x+2) = N(x+3) \implies A(23) = N(25)$ Thus we can recursively count $N(25)$ and get that $A(23) = 46368$ .

Next observe that for $n$ odd, $B(n) = N(x) \implies B(23) = N(23)$ .

Again add recursively (or just use your old work) to get that $B(23) = 17711$ .

Therefore, $A(23) + B(23) = \boxed{64079}$ .

Wow. Madoka Magica the anime. I've never expected to find such things on Brilliant :)

Assume that $X(0)=0$ , which is perfectly fine because that leads to $X(1)=1$ , which is true.

Take a look at the sequence : $0, 1, 10, 101, 10110, ...$ , you may notice that $X(n)$ is made from 2 parts, the first part is $X(n-1)$ and the second part is $X(n-2)$ . We will prove this for every positive $n$ larger or equal to $2$ by mathematical induction.

Clearly $X(2)=10$ is made from $X(1)=1$ and $X(0)=0$ .

Suppose that $X(k)$ is comprised of $X(k-1)$ and $X(k-2)$ . We will prove that $X(k+1)$ is comprised of $X(k)$ and $X(k-1)$ .

When changing the digits of $X(k)$ to form $X(k+1)$ , we're basically doing three tasks : Changing the digits of the $X(k-1)$ part, changing the digits of the $X(k-2)$ part and combining both parts together. By changing the digits of $X(k-1)$ , we obtain $X(k)$ , and by changing the digits of $X(k-2)$ , we obtain $X(k-1)$ . And when combining two parts together, we are perfectly sure that $X(k+1)$ is made of $X(k)$ and $X(k-1)$ .

Hence, it is clearly that $A(n) = A(n-1) + A(n-2)$ . Also, $A(0)=1=F(1)$ and $A(1)=1=F(2)$ . Hence, the general formula of $A(n)$ is $A(n)=F(n+1)$ . Thus, $A(23)=F(24)=\boxed{46368}$ .

Observe that $X(0)$ ends with $0$ , $X(1)$ ends with $1$ , $X(2)$ ends with $0$ and etc. Also, apart from $X(0)$ , every number in the sequence always begins with $1$ . It really easy to prove that $0$ and $1$ appear alternatively in the sequence, also to prove that every number begins with $1$ apart from $X(0)$ . This should be left as an exercise to the reader :)

Because $X(n)$ is made of $X(n-1)$ and $X(n-2)$ , hence $B(n)=B(n-1)+B(n+2)$ for $n$ being an even number and $B(n)=B(n-1)+B(n-2)+1$ for $n$ being an odd number (we have to add $1$ because when combining $B(n-1)$ and $B(n-2)$ , the last number of $B(n-1)$ , which is $0$ , combined with the first number of $B(n-2)$ which is $1$ to form another $01$ ).

Observe that $B(1)=0=F(0)$ , $B(2)=0=1-1=F(1)-1$ , $B(3)=1=F(2)$ , we can use the above formula to find the general formula of $B(n)$ , which is $B(n)=F(n-1)-1$ for even number n and $B(n)=F(n-1)$ for odd number n. Hence, $B(23)=F(22)=\boxed{17711}$ .

Finally, $A(23)+B(23)=46368+17711=\boxed{64079}$ .

This is really long, so just read Alan Yan's solution and enjoy :)