Find the value of $ [1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$ Assume that [x] is the floor function. I am not able to find any patterns in the numbers obtained. Any suggestions?
$$[1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$$

I think I have an answer. Look at $\frac{1}{3}$ in binary format and you will notice that it is $0.\overline{01}$.
This tells us something, namely that $[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to:
$$\sum \limits_{i=0}^{50} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]=\sum \limits_{i=0}^{49} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]+[2^{100}/3]$$
$$\sum \limits_{i=0}^{49} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]+[2^{100}/3]=\sum(4^i-1)+[2^{100}/3]=\frac{4^{50}-151}{3}+[2^{100}/3]$$
Edit 1: Added after I knew what to shoot for:
I suppose we can simplify this by noting that $4 \bmod 3 = 1$ thus $[2^{2i}/3] = \frac{2^{2i}-1}{3}$. This would give us:
$$\frac{4^{50}-151}{3}+2^{100}/3 - 1/3 = \frac{2^{101}-152}{3}$$
Edit 2: Going for a more general solution (boring but useful I guess)
Consider $m=2n+1$ odd: $$\sum \limits_{i=0} ^{2n+1} \lfloor\frac{2^i}{3}\rfloor = \sum \limits_{i=0} ^n (4^i-1) = \frac{4^{n+1}-3n-4}{3}=\frac{2^{m+1}-3\frac{m-1}{2}-4}{3} =\frac{2^{m+2}-3 m-5}{6}  $$
For $m=2n$ even:
$$\sum \limits_{i=0} ^{2n} \lfloor\frac{2^i}{3}\rfloor = \sum \limits_{i=0} ^{n-1} (4^i-1) + \lfloor \frac{4^n}{3} \rfloor = \frac{2^{2n+1}-3n-2}{3}=\frac{2^{m+1}-\frac{3m}{2}-2}{3}=\frac{2^{m+2}-3m-4}{6} $$
We can then rewrite this as:
$$\frac{2^{m+3}-6m-9+(-1)^{m}}{12}$$
