Because c gets quite large there is a small advantage to altering your short cut test to avoid a multiplication inside the loop on a (cost for this is a single divide outside). There are a lot of cutoffs on rad[a] is too large. (and there might be even more if you apply this heuristic on b instead).

for c in range(limit):
     for a in range(1, c // 2):
         if rad[a] * rad[c] >= c:
             continue

Can be made marginally faster by changing it to the equivalent form (we know that c, rad[c] > 0)

for c in range(limit):
     alim = c // rad[c];
     for a in range(1, (c + 1) // 2):
         if rad[a]  >= alim:
             continue

Because c gets quite large, there is a small advantage to altering your short cut test to avoid a multiplication inside the loop on a (cost for this is a single divide outside). There are a lot of cutoffs on rad[a] is too large (and there might be even more if you apply this heuristic on b instead).

for c in range(limit):
     for a in range(1, c // 2):
         if rad[a] * rad[c] >= c:
             continue

Can be made marginally faster by changing it to the equivalent form (we know that c, rad[c] > 0)

for c in range(limit):
     alim = c // rad[c];
     for a in range(1, (c + 1) // 2):
         if rad[a]  >= alim:
             continue