Floor Function Sum

Consider the general term of summand:

$\begin{aligned} a_n & = \left \lfloor \frac {2^n}3 \right \rfloor \\ & = \left \lfloor \frac {(3-1)^n}3 \right \rfloor \\ & = \left \lfloor \frac {(3^n - (n)3^{n-1} + n(n-1)3^{n-2} - \cdots +(-1)^{n-1}\cdot 3 + (-1)^n}3 \right \rfloor \\ & = \left \lfloor 3^{n-1} - (n)3^{n-2} + n(n-1)3^{n-3} - \cdots +(-1)^{n-1} + \frac {(-1)^n}3 \right \rfloor \\ & = 3^{n-1} - (n)3^{n-2} + n(n-1)3^{n-3} - \cdots +(-1)^{n-1} \end{aligned}$

$\implies a_n = \begin{cases} \dfrac {2^n}3 - \dfrac 13 & \text{if }n \text{ is even.} \\ \dfrac {2^n}3 - \dfrac 23 & \text{if }n \text{ is odd.} \end{cases} = \dfrac {2^n}3 - \dfrac 12 + \dfrac {(-1)^n}6$

Therefore,

$\begin{aligned} S & = \sum_{n=0}^{1000} \left \lfloor \frac {2^n}3 \right \rfloor \\ & = \sum_{n=0}^{1000} \left(\frac {2^n}3 - \frac 12 + \frac {(-1)^n}6 \right) \\ & = \frac 13 \left(\frac {2^{1001}-1}{2-1} \right) - \frac {1001}2 + \color{#3D99F6} \frac 16 & \small \color{#3D99F6} \text{Note: } \sum_{n=0}^m \frac {(-1)^n}6 = \begin{cases} \frac 16 & \text{if }m \text{ is even.} \\ 0 & \text{if }m \text{ is odd.} \end{cases} \\ & = \frac {2^{1001}-1502}3 \end{aligned}$

$\implies A+B+C = 1001+1502+3 = \boxed{2506}$

Just a note: your expression is not unique (imagine increasing $A$ by $1$ and $B$ by a lot, or adding $1$ to $A$ and doubling $B$ and $C$ ). Maybe specifying the minimum value of $A+B+C$ would remove any ambiguity?

Letting $q_k=\left\lfloor\frac{2^k}3\right\rfloor$ , division by 3 yields: $2^k =3q_k+r_k$ with $r_k\in\{1,2\}$ since 2 and 3 are coprime. Now observe that doubling equation above gives $2^{k+1}=2(3q_k+r_k)=6q_k+2r_k =\begin{cases} 3\cdot2q_k+2&\text{if }r_k=1\\ 3\cdot(2q_k+1)+1&\text{if }r_k=2. \end{cases}$ Then $r_{k}+r_{k+1}=3$ and $q_k+q_{k+1}=\frac{2^k-r_r}3+\frac{2^{k+1}-r_{k+1}}3=\frac{2^k(2+1)-3}3=2^k-1$ .

Using that yields $\begin{aligned} \sum_{k=0}^{1000}\left\lfloor\frac{2^k}3\right\rfloor & = \frac12\left( \sum_{k=0}^{1000}q_k + \sum_{\ell=0}^{1000} q_\ell \right) =\frac12\left( \sum_{k=0}^{1000}q_k + \sum_{k=0}^{1000} q_{k+1} \right) - \frac12 q_{1001} \\ & =\frac12\sum_{k=0}^{1000}(q_k+q_{k+1}) - \left( \frac{ 2^{1001}-r_{1001} }6 \right) =\frac12 \sum_{k=0}^{1000}(2^k-1) - \left( \frac{ 2^{1001}-r_{1001} }6 \right) \\ & = \frac36\cdot(2^{1001}-1-1001)-\frac{2^{1001}-r_{1001}}6=\frac{2\cdot 2^{1001}-(3\cdot 1002-r_{1001})}6. \end{aligned}$ Since $r_0=1$ , we get $r_{1000} =1$ and $r_{1001}=2$ , we get $\begin{aligned} \frac{2\cdot 2^{1001}-(3\cdot 1002-r_{1001})}6 &= \frac{2\cdot 2^{1001}-(3\cdot 1002-2)}6 \\ &= \frac{ 2^{1001}-(3\cdot 501-1) }3 = \frac{ 2^{1001}-1502 }3 \end{aligned}$ Therefore $\frac{2^{1001}-1502}3=\frac{2^{\mathbf A}-{\mathbf B}}{\mathbf C}$ leads to $\mathbf A+\mathbf B+\mathbf C=1001+1502+3=2506$ , which is minimal since the numerator is even and denominator odd.