I'm not sure if it helps, but for 130 bits, we only need to divide the point by 2, 101 times. The issue is figuring out where to subtract 1 to avoid floating-point errors.
the problem of dividing by 2 is that you need 2**101 pubkeys (according to your approach).
update:
dividing by 3 you need 3**14 pubkeys, to reduce puzzle130 down to the equivalent of puzzle 105.
So reducing 25 bits, if dividing by 2 we need 2^25 public keys with 1 one of them to be the correct result, but dividing by 3 we need 3^14 keys, one of them would be correct, how did you calculate it?