, 2023

Solution:
Let $p_{k} = \frac{128 - k}{k}$, so
$$\binom{127}{k} = p_{1} p_{2} \cdots p_{k}$$
Now, for $k \leq 63$, unless $32 \mid \gcd(k, 128 - k) = \gcd(k, 128)$, $p_{k} \equiv -1 \pmod{8}$. We have $p_{32} = \frac{96}{32} = 3$. Thus, we have the following characterization:
$$r_{k} = \begin{cases} 1 & \text{ if } k \text{ is even and } k \leq 31 \\ 7 & \text{ if } k \text{ is odd and } k \leq 31 \\ 5 & \text{ if } k \text{ is even and } k \geq 32 \\ 3 & \text{ if } k \text{ is odd and } k \geq 32 \end{cases}$$
We can evaluate this sum as
$$\begin{aligned} 4 \cdot (0 + 1 + 2 + 3 + \cdots + 63) & + 3 \cdot (-0 + 1 - 2 + 3 - \cdots - 30 + 31) \\ & + (32 - 33 + 34 - 35 + \cdots + 62 - 63) \\ = 4 \cdot 2016 + 3 \cdot 16 + (-16) = 8064 + 32 = 8096. \end{aligned}$$