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?
$$[1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$$

Just to better asses the solution given by Kitter Catter, premised that
$$
\eqalign{
  & x = \left\lfloor x \right\rfloor  + \left\{ x \right\}  \cr 
  & \left\lfloor { - x} \right\rfloor  =  - \left\lceil x \right\rceil   \cr 
  & \left\lceil x \right\rceil  = \left\lfloor x \right\rfloor  + \left\lceil {\left\{ x \right\}} \right\rceil  \cr} 
$$
with $\left\lfloor x \right\rfloor $ being the floor, $\left\lceil x \right\rceil $ the ceil, 
and $\left\{ x \right\}$ the fractional remaining
then:
$$
\eqalign{
  & \left\lfloor {{{2^{\,n + 1} } \over 3}} \right\rfloor \quad \left| {\;0 \le n} \right.\quad  = \left\lfloor {2{{2^{\,n} } \over 3}} \right\rfloor  = \left\lfloor {2^{\,n}  - {1 \over 3}2^{\,n} } \right\rfloor  =   \cr 
  &  = 2^{\,n}  + \left\lfloor { - {{2^{\,n} } \over 3}} \right\rfloor  = 2^{\,n}  - \left\lceil {{{2^{\,n} } \over 3}} \right\rceil  = 2^{\,n}  - \left\lfloor {{{2^{\,n} } \over 3}} \right\rfloor  - \left\lceil {\left\{ {{{2^{\,n} } \over 3}} \right\}} \right\rceil  =   \cr 
  &  = 2^{\,n}  - \left\lfloor {{{2^{\,n} } \over 3}} \right\rfloor  - 1 \cr} 
$$
