No, the loop would have been slower. That is why they found the starting point with that bizarre two-instruction sequence. For any table size between (2^^n)+1

No, the loop would have been slower. That is why they found the starting point with that bizarre two-instruction sequence. For any table size between (2^^n)+1 and 2^^(n+1), inclusive, it returned 2^^n as the starting point. Then of course it had to loop after that.