Comment by InfoSecErik
5 days ago
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.
5 days ago
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.