Note that the function ~f(x) = \left\lfloor \frac{x}{2} \right\rfloor - \left\lfloor \frac{x}{7} \right\rfloor~ increases by one every multiple of 2 it hits, and decreases by 1 every multiple of 7 it hits. Since there are ~\ge 3~ multiples of 2 in between any two multiples of 7, the function is strictly increasing upon multiples of 7. We also note that for this same reason, the multiples of 7 are the local minima of this function. With this knowledge, we can binary search for the last multiple of 7 such that ~f(x) \le T~, and then just check the 6 numbers after it.

Time Complexity: ~\mathcal{O}(\log T)~