Formula applied ten times?

We know that $N_{1,1} + N_{1,2} = 384 \text{ and } N_{1,1} - N_{1,2} = 128$ Then we also have $N_{2,1} + N_{2,2} = N_{1,1} \text{ and } N_{2,1} - N_{2,2} = N_{1,2}$

Notice if we sub what $N_{1,1}$ and $N_{1,2}$ into our original equation we get that $N_{1,1} + N_{1,2} = 2N_{2,1} \text{ and } N_{1,1} - N_{1,2} = 2N_{2,2}$

We observe that each time we sub back once that variables in the previous equation are equal to twice the newly introduced ones, for example when we are four layers in we have $N_{4,1} + N_{4,2} = N_{3,1} \text{ and } N_{4,1} - N_{4,2} = N_{3,2}$

And so from the following equations $N_{3,1} + N_{3,2} = N_{2,1} \text{ and } N_{3,1} - N_{3,2} = N_{2,2}$ We get that $2N_{4,1} = N_{2,1}$ and that $2N_{4,2} = N_{2,2}$ so if we were at $N_{10,x}$ then we would get 5 doublings ending with the equations $2^{5} N_{10,1} = 384 \text{ and } 2^{5} N_{10,2} = 128$

Before we proceed we notice that $128 = 2^{7}$ and $384= 2\left( 192 \right) = 2^{2} \left( 96 \right) = 2^{3} 48$ and so we determine that $N_{10,1} N_{10,2} = \frac{384 \cdot 128}{2^{10} }= \frac{48 \cdot 2^{3} \cdot 2^{7} }{2^{10} }= 48$

The product of the inputs is 49152. Each pair of operations quarters the numbers. $\frac{49152}{4^5}=48$ .

The successive pairs are : (256,128); (192,64); (128,64); (96,32); (64,32); (48,16); (32,16); (24,8); (16,8); (12,4). Therefore the product of the numbers of the 10th pair is 48.

From the introduction of the problem we can easily derive these formulas: $N_{n,1}=N_{n+1,1}+N_{n+1,2}$ $N_{n,2}=N_{n+1,1}-N_{n+1,2}$ Since this is true for any $n$ it will also be true for $n+1$ : $N_{n+1,1}=N_{n+2,1}+N_{n+2,2}$ $N_{n+1,2}=N_{n+2,1}-N_{n+2,2}$ Substituting these expressions in the first two equations and simplifying gives us: $N_{n,1}=2N_{n+2,1}$ $N_{n,2}=2N_{n+2,2}$ Which is still true for any $n$ so: $N_{0,1}=2N_{2,1} \text{|} N_{2,1}=2N_{4,1} \text{|} N_{4,1}=2N_{6,1} \text{|} ...$ $N_{0,2}=2N_{2,2}\text{|}N_{2,2}=2N_{4,2}\text{|}N_{6,2}=2N_{8,2}\text{|}...$ With $N_{0,1}=384$ and $N_{0,2}=128$ , to get $N_{10,1}$ and $N_{10,2}$ we need to substitute the expressions a total of four times into the initial equation, giving: $N_{0,1}=384=2^5 \cdot N_{10,1}$ $N_{0,2}=128=2^5 \cdot N_{10,2}$ And their product will be: $N_{10,1}\cdot N_{10,2}=\dfrac{N_{0,1}\cdot N_{0,2}}{2^{10}}$ $=\dfrac{384\cdot 128}{2^{10}}$ $=\dfrac{48\cdot 2^3\cdot 2^7}{2^{10}}$ $=\boxed{48}$ In the end, we can find any $N_{n,x}$ number with these two pairs of formulas (one for odd number indexes, and the other for even number indexes): $\text{Even:} N_{0,1}=2^{n}\cdot N_{2n,1} \text{|} N_{0,2}=2^{n}\cdot N_{2n,2}$ $\text{Odd:} N_{1,1}=2^{n}\cdot N_{2n+1,1} \text{|} N_{1,2}=2^{n}\cdot N_{2n+1,2}$