I think the disconnect is that you assume that being able to do useful computation on some data implies that it must be possible to derive some insight into what the data is (side-channels or the like).
It's a fair assumption to start with. But the folks building FHE basically claim "nuh-uh", and I haven't seen anything to indicate they're wrong. Maybe some new Math grad will sort it out.
I too have been playing with the conjecture for fun. Your insight is interesting because of the appearance of 2^n, given that that always resolves to 1 for all n.
I ran some calculations looking to see if there were patterns to the next lowest number (call that number x) and could not quickly find any. So even if 7 + k*2^n follows a predicable path to its next lowest number, that number is not currently predictable.
Of course, if you can identify which n satisfies the equation x = s + k*2^n for some value of n and some "base" value s (7 is the base value in the previous example), you can predict the path of that number.
As an example, take 7 + 4*2*7 = 519. Its next lowest number is 329. 329 = 5 + 81*2^2. So for 329, s=5, k=81, n=2. So we know 329 will only take two steps to reach 247.
In my phrasing, 128k + 7 -> 81k + 5 for all positive integers k.
Pick a power of 3 n to be the coefficient for k on the right/reduced side, and then the left side will have at least one valid reducing form with coefficient power of 2 f(n) = ⌊n·log2(3)⌋+1. If there is more than one, they will have different constants. Each multiplication immediately has a division (you already got this part), and there must be a final division which is not immediately preceded by a multiplication because (3x + 1)/2 > x for all positive integers (that is, if you multiply once and then divide once, you will always be larger than just before those two things, so an "extra" division is needed to reduce). This means that there must always be at least one less multiplication than division, so the initial condition is one division and zero multiplications - the even case with n = 0. Then for n = 1 you need 2 divisions, which works because 2^2 > 3^1. Then for n = 2 you need 4 divisions, because 2^3 < 3^2 so 3 divisions is not enough. This is where f(n) comes in, to give you the next power of 2 to use/division count for a given n. When you do skip a power of 2, where f(n) jumps, you get an "extra" division, so at 16k + 3 -> 9k + 2 you are no longer "locked in" to only the one form, because there is now an "extra" division which could occur at any point in the sequence...
Except it can't, because you can't begin a reducing sequence with the complete form of a prior reducing sequence, or else it would "already reduce" before you finish operating on it, and it so happens that there's only one non-repeating option at n=2.
At n = 0, you just get D (division). At n = 1, you have an unsplittable M (multiply) D pair MD and an extra D. The extra D has to go at the end, so your only option is MDD. At n = 2, you appear to have three options for arranging your MD MD D and D: DMDMDD, MDDMDD, and MDMDDD. But DMDMDD starts with D so isn't valid, and MDDMDD starts with MDD so also isn't valid, leaving just MDMDDD.
At n = 3 there are finally 2 valid forms, 32k + 11 -> 27k + 10 and 32k + 23 -> 27k + 20, and you can trace the MD patterns yourself if you like by following from the k = 0 case.
The constants don't even actually matter to the approach. If there are enough 2^x k - > 3^y k forms when n goes off to infinity, which it sure looks like there are though I never proved my infinite sum converged, you have density 1 (which isn't enough to prove all numbers reduce) and this angle can't do any better.
PMapper author here: I can't give PMapper the love it deserves, but I fully support the work done in Fennerr's fork at https://github.com/Fennerr/PMapper .
If you listen to her voice in Her and listen to Sky back to back, they are clearly very different. Scarlett has some raspy tones to her voice that are very distinctive and come through even when she is talking like an AI.
I wonder if people just associate the "talking like an AI" thing to Scarlett and it's not actually her voice that seems familiar.
I remember hearing about this sort of thing from a grad student a few years back. She pointed out that there were interesting privacy concerns with these microbiomes, given how we leave traces of it.
