That’s an irrational binary number. Converting to decimal:
C_7 = 0.71671571665125114681…
Does this number show up anywhere else in the natural computational universe, or is it a one-off that’s specific to 7 under 5n+1?
It doesn’t seem to be related to well-known irrational numbers like \pi. It doesn’t seem to be the result of any typical infinite series. Is it Brun’s constant divided by two? No
Another question is whether the sequence 1011011101111010101011100110001… shows up elsewhere. Is it the center column of Wolfram’s Rule 30? No (Well, maybe it is, I didn’t even look.)
It would be interesting to tie questions about “7 under 5n+1” to other existing questions.
That’s interesting. I like the connection that any eventually periodic sequence corresponds to a rational number and any divergent sequence (which we know from Unicity of trajectory - #3 by Failix correspond to non-periodic parity sequences) corresponds to an irrational number.
In this context, a weak version of the 3n+1 problem (are there any divergent sequences) can be reformulated as: are there any irrational fractions corresponding to parity sequences? Which seems interesting (although, I have no idea if that’s useful or not …)
Also, Matt Parker talks about a similar idea on a Numberphile episode: (https://www.youtube.com/watch?v=c066hLi78B0): the “Prime Constant” which has the n’th bit set if n is prime. But AFAICT they don’t have a specific way to take advantage of that representation as a number.
And also this reminds me of Chaitin's constant - Wikipedia which encodes the halting problem for all TMs into the bits of an irrational number between 0 and 1.
@stargazer07817 just posted a great paper in Unicity of trajectory - #8 by stargazer07817 (https://math.colgate.edu/~integers/t8/t8.pdf) which considers this sequence going “the other way” as 2-adic integers (Z_2) and this representation has many interesting properties. They extend the Collatz map from the regular integers to 2-adic integers and then define the function Q: Z_2 \to Z_2 which maps a 2-adic integer to it’s parity sequence. Q is a bijection (generalizing the “Unicity of trajectories” result) and even an isometry (in 2-adic metric) which is basically equivalent to the result that the first n bits of the parity sequence are determined precisely by the lowest n bits of the starting number.
IIUC, all eventually periodic 2-adic numbers are “rational” numbers (in the sense that if you multiply them by integers you get integers). So the same observation applies here. From that paper it seems like they reference a conjecture that is broader than the Collatz conjecture that Q maps all rational numbers to rational numbers (Collatz conjecture is that it maps all integers (eventually terminating numbers) to rationals (eventually periodic numbers)).
I forgot to bring it up earlier, but there is a detail that makes the bijection between the set of infinite binary sequences and real numbers not that obvious, and that is the sequence that are ultimately periodic with a period of “1” or “0”, that can be mapped to the same number. For example 0.011111… = 0.1
There must be a way to elegantly handle those cases although it isn’t obvious to me.
Oh yeah, good point! Real numbers != bit sequences!
That said, for normal positive integers (not 2-adic) technically you cannot have repeated 1 forever at any point. I think that the numbers that have parity 1 n times are exactly the numbers \equiv -1 \pmod{2^n} since T^n(2^n - 1) = 3^n - 1 via n odd transitions.
And likewise I think the only way to get 0 repeat forever is to start from 0.
-1 has a trajectory of 11111…, and any number that falls in its cycle is going to have a trajectory that ends in 1s repeating.
If you extend to rationals with odd denominators, -1/3 → 0, so it gets a trajectory of 10000…
and that is the same as saying -1 has this trajectory in 3n+3
Agreed, I was talking about over the positive integers.
You can also extend to 2-adics and then every parity sequence is realized by a 2-adic number. This is all probably more reason to consider the parity sequences as 2-adic numbers where ...1110 != ...0001 or anything like that … I think that there is a bijection between all boolean sequences and 2-adics unlike in binary expansions of real numbers where it is not injective.
Does any start number follow a divergent trajectory under 3n+1?
You could ask:
Does any start number follow the specific divergent trajectory with a parity sequence given by \pi-3.0 = 0.1415926\ldots = 0.0010010000111111011011\ldots_2?
Yeah, it seems extremely unlikely (but as with many Collatz things) hard to prove that any finite number will have the exact parity sequence of \pi - 3. Specifically, we can calculate the binary digits of such starting value from bottom up, at each bit, it seems equally plausible that that bit is 0 or 1, in order to be a finite number, it must be eventually all 0s. Some early values for parity sequence 0010010000…:
For 3n+1, a 50/50 split of 0s and 1s will crash any start number to 1. (Assuming \pi is normal, which, nobody knows.)
Speaking of 5n+1, @Failixposted the idea that a divergent 5n+1 trajectory’s parity sequence has to have at least some minimal amount of 0’s. I wonder if a 50/50 split is okay (e.g., doesn’t involve too much “loitering”).
I have the same feeling about 5n+1. As @Failix pointed out, there are only countably many parity sequences starting from integers, but there are infinitely many parity sequences / real numbers, so the chance that any specific one is realized by an integer seems improbable.
What do you mean by a “is a 50/50 split ok”? Do you mean, is there any way to prove that the ratio can’t be too close to 50/50? (In analogy to how the ratio can’t be too close to 0-100 as shown by @Failix ) I would guess that it can (and will) get very close to 50-50.
A 5n+1 trajectory needs at least 43% odds to diverge. That’s a lower bound … and @Failix gave an upper bound.
So 50% should be well within that range, and empirically that’s what you see with start number 7.
(I was thinking that you might need more than 43% to avoid “diverging too slowly” and running out of numbers to hit. For example, (5/2)^.431n (1/2)^.569n = 10000 is solved by n = 17693 steps, and there might not be 17k numbers to hit on the way to 10k. But this doesn’t really go anywhere – with 44% odds you avoid any such problems.)
In the 5N+1 system the tree of all possible paths to 1 has connection nodes that are integers ending in a 6. Branches in the tree consist of an odd root of values 1, 3, 5 … followed by doubling even numbers. Any branch root with a units digit of 5 is a bare branch and has no connection nodes. Branches with units digit of 1, 3, 7 and 9 have a first node at depths 5, 2, 4, and 3 respectively. Following nodes are located at every fourth following even value. The tree is sparsely connected. Node values on a tree have a separation value of 10 (6, 16, 26, 36 …) and each branch except those with a units digit of 5, has nodes with a separation value of 16 within the branch.
The majority of starting values diverge to unbounded values in disconnect tree fragments and only a small proportion of branches connect to the tree with a root at 1. The odd branch root values of 1 3 15 19 25 51 63 97 do reach 1. The majority of other branches below 100 most probably diverge to infinite values. This would include the number 7. Node connections in the lower part of the tree are not dense enough to connect all low valued branches. There must be some disconnected subtrees that remain disconnected from the tree rooted at 1.
When I said the that majority of values probably diverge, I should have said diverge or form cycles. The value 7 is not guaranteed to diverge, but if it doesn’t then it must form a cycle. The important point is that the density of connections is too low for all branches to have access to the tree rooted at 1. Nodes of 6, 16, 26, 36 … have an arithmetic separation of +10. Branches have spacing of x16 between nodes. Density rules out a single connected tree. There is not room to connect all low valued branches before the node connection values on a branch grow too large. Pieces of the tree that don’t connect can only be a cycle or a divergent growth. This proves existence of either a cycle or of divergent growth without providing an example of either.
There is a similar proof for the Collatz conjecture. I would love to present a proof of the Collatz system by very simple methods but I am still building my trust level and I am not sure how to get access to start a new topic. I have two different methods one using induction and the other using a sorting method. Both methods avoid any mention of Baker related hard issues or of Proposition B. Would a proof of connectivity of a single Collatz tree that rules out integer cycles qualify as a back door method for proving integer cycles don’t exist without the need for Baker hard methods?
Yeah, my understanding of the situation for 5n+1 is that we can prove that most numbers must diverge because of some sort of bulk rules (like the density rules you mention), but we cannot prove that any specific value diverges.
Specifically for 7, it’s still unknown if it does eventually go to 1 (is part of the 1 tree), right? It seems extremely unlikely, but IIUC, there is not proof that it is separate.
The situation for 5n+1 is even worse, in that we cannot prove anything at all about the existence of a diverging sequence, despite it being highly suspected.
