Powers of two, and adjust the last element to the value needed to have the desired average.
This jump set was proven optimal here (e.g. minimizes total number of operations):
Kangaroos, Monopoly and Discrete Logarithms, J. M. Pollard, 2000
Well, ok. But I use fixed length table for all tests, it's more practical for implementation, also I get better results for longer list than when using powers of two.
I will try your approach to see results.
Question for you: do you prefer fewer kangaroos that jump faster, or lots of kangaroos that jump slower?
I prefer faster kangs because of high DP bits that I have to use to solve high puzzles. But even so, the number of kangs is crazy because there are many GPUs and every GPU has a lot of kangs anyway.
How can you explain case #3 (the awful case with runtime 172 sqrt)? When the Tame and Wild are separated by a distance of b/2, and the average jump size is much too small, it will take a lot of jumps for them to ever meet. In the random case, it's a little better than that, but still too far from the optimal case (e.g. a correct larger average jump size).
Main question here is how many times do you solve the point to calculate average result value? In my tests I solve at least 1000 times.