By analogy with what is known on Collatz, we may reasonably expect the following statement to be provable for the 5n+1 variant: For some (all?) constant a>1, almost all positive integers n reach a value higher than n^a. At the present time, this is pure speculation…
1 Like
That’s interesting… a possible version of Tao’s result flipped on its head for 5n+1. Along those lines, it would be interesting to know if Tao’s method generalizes from 3n+1 to 3n+k or better, to any “contracting” Collatz-like rule of the form
(a_0 n + b_0)/d … if n \equiv 0 \mod d
(a_1 n + b_1)/d … if n \equiv 1 \mod d
…
(a_{d-1} n + b_{d-1})/d … if n \equiv d-1 \mod d
where a_0 a_1 ... a_{d-1} < d^d (with all a and d coprime).
This 1978 paper demonstrates specific start numbers that never reach 1 for the 1093n+1 problem. Basically, there’s a “wall” that prevents start number 3 from ever reaching 1. If you track it, 3 appears to diverge to infinity, but I suppose it might hit a loop.
This 1995 paper extends that to the 21n+1 problem, the 39n+1 problem, and so on. They crazily show that for almost all q, the associated qn+1 problem definitely has trajectories that don’t hit 1. But if I understand right, those trajectories, too, could all hit loops rather than diverging to infinity. (Please correct me if this is wrong!)
The 5n+1 problem also has numbers that definitely don’t reach 1, such as loopers 26 and 27. But in the case of 5n+1, there’s no “wall” preventing an apparent diverger like 7 from one day plummeting to 1 (unlikely as it seems).
At least with cycles, if you find one, you can easily convince someone – like, “Here’s the cycle.”
While if you found a divergent number, you’ve still got a heck of job to convince someone that it’s divergent. (Same goes if you found a set of numbers for which some unknown member is divergent.)
1 Like
Yes, we know very little on divergent numbers, whatever the qn+1 variant.
In the case of 21n+1, the inverse tree has a regular structure that can be entirely described by a simple regexp (@cosmo, this is for you). We know exactly which numbers get to 1, and virtually nothing on the divergent sequences.
2 Likes
Cool! (I believe that my work on regexes is generalisable to all qn+1 such that 2 is a primitive root of the multiplicative group of Z/q^kZ for all k)
1 Like
Hi! (love the youtube videos.)
As a non-mathematician a perhaps trivial question occurs to me: how irrational is this number? Ie. how approximatable is it? Is that even interesting or important?
Great, @unmode, I get this series of approximations
\frac{2}{3}, \frac{3}{4}, \frac{5}{7}, \frac{38}{53}, \frac{43}{60}, \frac{210}{293}, \frac{969}{1352}, \frac{1222}{1705}, \frac{2191}{3057}, …
the latter of which is within 2 \cdot 10^{-8} of the value.
Not sure how irrational it is, and I guess it could turn into a repeating pattern … if start number 7 happens to get caught in a cycle.