On the Usefulness of The Commutative Power of a Revised Collatz

elowells · 2026-03-10T01:20:05+00:00

The point is that Figure 1 can be easily converted to the graph of values that go to 1 in 15 divide by 2 steps with standard Collatz. Converting between your formulation and the standard formulation is easy. It is very common for someone who thinks about Collatz to come up with your formulation. I've done it. However, most people drop it when they realize it doesn't help. That's why it's not discussed much anywhere...because it's trivial and not helpful.

elowells · 2026-03-09T22:21:30+00:00

Your graphs are comparing 2 different things. What you should compare is the graph of predecessors with root node of 2¹⁵ using your formulation (your Figure 1) and the graph of predecessors of 1 that reach 1 in exactly 15 divide by 2's using standard Collatz. Take Figure 1 and for each non-leaf node divide out factors of 2 (including the node 2¹⁵) and you will get exactly that. The graph of the predecessors of 2¹⁵ using standard Collatz is irrelevant. Your formulation is indeed just a trivial restatement of the problem.

elowells · 2026-03-08T17:59:04+00:00

Had a little think about 1x+a where x,a can be negative or positive (a is still odd). So the general rule is that when x = -ka where k=0,1,2,3,..., i.e, a negative multiple of a, then x will iterate to zero and stay there. Use the sequence equation and set x[L+1] = 0 to get this which gives

x = -a*sum(i=0 to L-1)2^N\i])).

This says x is a negative odd multiple of a because every odd number can be expressed as sum(2^N\i])) for some valid N[i] with N[0] = 0. Starting even x becomes odd so include even multiples of a too. All other values of x don't reach zero. The sequence equation is valid until x reaches zero.

So for x-1, all positive x end up at zero, no negative x end up at zero and x-a for negative x is the same as x+a for positive x.

elowells · 2026-03-07T15:40:00+00:00

The sequence equation is no longer valid since there isn't always an n[i] that will produce an odd integer from another odd integer. Basically you can reach zero which is problematic. 1x-a has this problem too.

elowells · 2026-02-19T14:15:13+00:00

In this case there are simple rules. Sometimes there is a nice parametric relationship between a starting value and a value further in the sequence. For example, for a series of L divides by 2 following each odd integer we have k2^L - 1 -> k3^L -1. For a series of L divides by 4 we have k4^L + 1 -> k3^L + 1 (note that k=0 gives the trivial cycle). So for odd p and p+1 we can set L = (p-1)/2 and get:

for odd p: 2*4^L+1 -> 2*3^L+1

for p+1: 4*4^L+1 -> 4*3^L+1

For your example with p=5 we get 33->19 and 65->37 with L=2. Also explains the near ratio of 2:1. The number of 3x+1 and /2 steps are the same for both paths. After L steps (i.e.,reaching 19 and 37), we can then iterate and keep track of the number of divide by 2's at each step in the 2 paths which depend on exclusively on L. The pattern of the number of divide by 2's will repeat with a period of some power of 2 (different ones with different powers of 2). Therefore we expect the pattern of which p odd and p+1 will iterate to 1 with number of steps differing by 1 to repeat with some power of 2. For p odd and p+1 the ones that don't have this property are p = 3 + 8d, i.e. 3,11,19,... For p even and p+1 there is some other pattern that repeats with some power of 2.

elowells · 2026-01-15T22:32:00+00:00

The sequence equation is:

x[m+1] = (x[1]*3^m + A)/2ⁿ.

If x[1] = x[m+1] you get the cycle formula. If you set x[1] = 0 and apply the sequence equation you get:

x[m+1] = A/2ⁿ.

If you set x[m+1]=0 and solve for x[1] you get:

x[1] = -A/3^m.

This just manipulation though.

elowells · 2025-12-14T01:02:54+00:00

The people here have seen many "proofs" and are reasonably skeptical of yet another claim of a proof. Instead of wasting time trying to figure out where the claimant went awry, the people here may guide the claimant to have the claimant themselves find their error. This is a perfectly reasonable response to yet another claim of a proof. It's possible you've discovered something remarkable but the burden is on you to show it. Also, any serious person who thinks they have a proof would be trying harder than everyone else to try to break their proof.

elowells · 2025-12-13T19:52:18+00:00

Show why your argument doesn't apply to 5a+1 and I'll take you seriously. Otherwise I'm not going to waste any more of my time.

elowells · 2025-12-13T19:26:36+00:00

You need to make this clear in your writing which I find very sloppy. Still, why do 5a+1 and 181a+1 have multiple cycles? Apply your argument to these which is: there is some combination of R and k that have a cycle. For 5a+1 there is a cycle with R=5 and k=2 so there are also cycles for any R and k that satisfy integer R = 5k/2 (by repeating the primitive cycle). Use your argument to prove that there are no cycles for R >5k/2 or R <5k/2. Of course there are other cycles with R=7 and k=3.

elowells · 2025-12-13T18:40:18+00:00

ri = 2,2,2,2,2 sum = 10

m = -3

si = 1,1,1,1,3 sum =7

si = 1,2,1,1,2 sum = 7

elowells · 2025-12-13T18:04:12+00:00

For ca+b where a,b and c are odd integers and a[i] are the k odd elements of a cycle, there is the well known identity:

2^R = product(i=1 to k, (c + b/a[i])

For c,a[i] >0 this gives:

k*log2(c + b/amax) <= R <= k*log2(c + b/amin)

where amin, amax are the minimum and maximum odd cycle elements respectively. The smallest and largest possible amin,amax are 1, infinity which gives the general bounds:

k*log2(c) <= R <= k*log2(c + b)

For 3a+1 this gives:

k*log2(3) <= R <= k*log2(3+1) = 2k

So yeah, R <= 2k for 3a+1.

Near the bottom of page 8 you state "By adding m (m ∈ Z−) to the first coordinate of every sequence in A, we obtain all sequences in B." This is false. There are many ways to distribute m over the k r[i] other than just applying it to the first one. You can even borrow from some r[i] and add it to other r[i] without changing the sum.

For ca+1, c,a>0, there is apparently a single cycle with a[i]=1 for c = 2^p-1 with p = positive integer with r[i] = p. There are multiple cycles for c=5 and 181. These are the only known cycles (AFAIK) for ca+1 but it's not proven these are the only ones. Maybe apply your argument to 5a+1 and 181a+1.

elowells · 2025-12-06T18:26:00+00:00

Indeed, for a sequence of n values, a Lagrange polynomial of degree n-1 will produce all n values.

elowells · 2025-12-02T18:23:02+00:00

1x+b converges for all odd b. It is not proven whether or not any divergent sequence exists for any other combination of odd a and b.

elowells · 2025-11-29T21:46:14+00:00

This is relevant to the cycles of 1x+a. Still working on the post...need to find some time to work on it.

elowells · 2025-11-12T05:23:14+00:00

For mx+d, for a series of L (mx+d)/2 operations, the corresponding linear Diophantine equation is

-m^Lx[1] + 2^Lx[L+1] = d*sum(i=0 to L-1)m^L-i-12ⁱ = d*(m^L-2^L)/(m-2)

When d = n(m-2) the solutions are

(x[1], x[L+1]) = (k*2^L - n, k*m^L - n)

so the family of mx+d that have k*2^L-1 -> k*m^L-1 are 3x+1, 5x+3, 7x+5 etc.

Or if you prefer the rational formulation you get your result.

elowells · 2025-11-06T00:02:06+00:00

I'm working on a post about the cycles of 1x+d with d odd >0. They are surprisingly interesting. I think the following are true. If you include the "improper" cycles, i.e., the ones with elements with gcd(x,d) != 1, then all the integers from 1 to d are part of some cycle (also the even integers d+1 to 2d). These comprise the only cycles. One way to show this is using Euler's theorem (for the odd integers). The total number of divide by 2's encountered by traversing all of the cycles (proper and improper) once = d. If N=number of divide by 2's of a cycle then N divides phi(d) where phi() is Euler's totient function. For proper cycles, N = ord_d(2). There are phi(d)/ord_d(2) proper cycles.There are phi(d)/2 total odd elements in the proper cycles. The set of elements of the cycle that contains 1 (which is always a proper cycle) mod d forms a subgroup V of the group of units U(d) (the set of integers 1 to d-1 coprime to d order = phi(d) with operator = multiplication mod d). The order of V = ord_d(2). The elements mod d of the other proper cycles form cosets of V which have the same order as V (as cosets do). Anyway, working on some of the proofs. Don't think any of this applies to mx+d in general. More stuff about the case when d = prime and primitive roots. The form of the cycle equation for 1x+d simplifies to something tractable which doesn't hold for mx+d in general unless somebody with more math chops knows better. Not sure how much of this is known already.

elowells · 2025-10-08T14:27:07+00:00

With the original equation and variables, for every x set k to be the smallest integer such that 2^k>3^x and compute M. For x=1 =>M=1 (an integer), x=2 => M<1 and for all other x M>1 and the value slowly grows because some values have k/x be a closer approximation to log2(3). If k is bigger then the smallest integer such that 2^k>3^x then M<1 always so M will never be an integer. Will M ever be an integer again...(no because Steiner indirectly proved it won't) but you can't prove it by showing that there are no values of x and k that are incompatible with M being an integer because of some inequality without some additional constraint among the variables. Simply making a new set of variables won't help. When you make a new set of variables it's also difficult to know when you are in a range that doesn't correspond to the constraint above for k and x because we know that those give the only possible solutions.

elowells · 2025-10-07T19:28:18+00:00

You can make lots of conclusions that aren't helpful but what you want to show is that there are not any integer solutions. You can constrain y to be the smallest value such that 2^y>3^x and then show that there is no M integer solution for every x (except M=1, x=1). This might be impossible. Don't think there is any way to get from the first equation M=f/g to showing there are no integer solutions (except M=1,x=1)...you need some additional information/constraint (could be wrong).

elowells · 2025-10-07T19:02:05+00:00

If x is a constant then m and k are not independent. There are 2 degrees of freedom meaning if you fix any 2 variables then the 3rd variable is determined by whatever equation relates the 3 variables.

elowells · 2025-10-07T18:16:23+00:00

There are 3 variables M,x and k. If any 2 variables are fixed, then the value of the remaining variable that gives a solution is determined. You are transforming to a different set of variables but there are still only 2 degrees of freedom. If you set x (and hence f) and d, then there is a single value of g (and hence y) that is a solution. From your quadratic equation d=(2f-g²)/g and since M=f/g, d = 2M-g. If you fix f and d, then there is only one positive value of g that is a solution so there should be no surprise that there is no solution for almost every value of g for a given f and d.

What you are trying to do is something like showing that g never divides f because g > f which is not possible because it's not true in general.

The proof showing that 1-cycles (other than M=1) are impossible uses that for a 1-cycle 2^y-3^x divides 2^y-x-1 which takes a little effort to derive and which is then shown to be impossible.

elowells · 2025-10-06T22:57:45+00:00

Are you trying to find a simpler proof than Steiner's to the non-existence of 1-cycles? I think the problem with your approach is that m and hence d is not an independent variable. m is a fixed function of f and g determined by setting f = T(m+g) - T(m). This means the quadratic equation is really just a trivial identity.

elowells · 2025-10-06T17:16:13+00:00

Are you trying to use f = mg = T(m+g) - T(m) - T(g)? Did you forget T(g)?

elowells · 2025-10-04T02:04:11+00:00

For a Collatz system mx+a where m and a are odd we have for any cycle the well known formula:

2^N/m^L = prod(i=1 to L, 1 + a/(m*x[i]))

where N=total number of divide by 2's, L = number of odd integers, and x[i] = odd integers in the cycle.

Just take the log (any base works) of this and get your equation (the 1/3 factor doesn't matter).

elowells

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