Question vaguely related to the Collatz conjecture.

Question

How would one go about proving (or disproving) the following proposition?

There is no set of positive integers $(m, n, I)$ with $n > 1$, such that

$$\frac {3^n-2^n}{2^m-3^n}=I$$

This is related to the Collatz conjecture in the following way:

If there is a non-trivial loop (i.e. one that does not contain 1) produced by the 3x + 1 iterate then that loop can be summarized as, lets call it equation 1 (I've never used MathJax before, so bear with me) $$\frac{3^n*X + b}{2^m}=X $$ To illustrate without using the Collatz iterate (since as yet no one knows of such a loop), use instead 3*X + 5. This will loop starting at 19:

19 to 31 to 49 back to 19. The number 19 is $$3^3 - 2^3$$

The loop can be summarized as $$\frac {3^3*19 + 5*19}{32} = 19$$

The 5 in the second term of the numerator arises from the number of total divisions by 2 encountered in traversing the loop, because $$2^5 - 3^3 = 5$$and

$$\frac {19*5}{2^5 - 3^3} =19$$

In general, if there are no consecutive divisions by 2, except the last that returns to the starting number, the "b" in equation 1 will be of the form $$N*(3^n-2^n)$$ where N is equal to $$2^m - 3^n$$ n is the number of elements in the loop and m is the number of divisions by 2 performed in traversing the loop. For the Collatz iterate 3X+1 N would be 1. Therefore "b" in equation 1 would be $$3^n - 2^n$$ and $$\frac {3^n - 2^n}{2^m -3^n}=I$$ where I is the lowest odd number in the loop.

as an aside, the $$3^n - 2^n$$ is equal to

$$3^{(n-1)} + (3^{(n-2)}*(2)) + (3^{(n-3)}*(2)^{2}) + (3^{(n-4)}*(2)^3)+...(3^{(n-n)}*(2^{(n-1)})$$

This sequence results from n multiplications by 3 and n divisions by 2. It is the backbone of the general numerator used in defining the elements in a Collatz loop.

Also, any number $$3^n - 2^n$$ will be the lowest number in a loop consisting of n multiplications by n, and the N used to form the iterate will $$(2^m - 3^n)$$ E.g. 3X + 13 will loop to 211 and consist of 5 elements, because $$2^8 - 3^5 = 13$$ and $$3^5 - 2^5 = 211$$ 3X + 47 will loop to 65, because $$2^7 - 3^4 = 47$$ and the loop will contain 4 elements (65, 121, 205, 331, then back to 65).

No loop constructed this way will satisfy the 3X + 1 iterate of the Collatz conjecture, because the only m,n that satisfy $$2^m - 3^n = 1$$ is (2,1), i.e. the trivial loop. Hence, the original question.

ADDENDUM

This is in response to Eric Shumard's helpful comment. It is an argument, not a proof.

The proposition is that if there is no 2-cycle, then there can be no set of positive integers (m, n, p, I) such that $$\frac {3^n - 2^n + 4*p)}{2^m - 3^n}=I$$ This is more a sketchy proposition than an inference, (and it's wrong) but here is how it is derived:

For any loop, there must be a smallest odd number in the loop. In order for this condition to hold for a candidate lowest number X₁, then $$\frac {3*X_1 + 1}{2}$$ must result in an odd number. Otherwise there will be an immediate consecutive division by two and the result will be no more than $$\frac {3*X_1 + 1}{4}$$ which will be less than X₁ for all positive X₁ > 1.

Now: In order for there to be a loop containing n odd numbers, with a total of m divisions by 2, the following equation must hold for X₁:

$$\frac {3^n*X_i + b_i}{2^m} =X_i$$ and $$\frac {b_i}{2^m - 3^n} = X_i$$Each odd element in the loop has its own b_i. The b_i are determined as follows:

Case I: 1-cycle. Rather than try to do everything in symbols, I will use an example with n=3. In this case b₁ will be $$3^3 - 2^3 = 19$$ I will do it explicitly. Starting at the smallest element in the loop, X₁ $$X_2 = \frac {3*X_1 + 1}{2}$$This cannot be an even number as stated above. The last element in the loop (again, for a 1-cycle) is

$$X_3 = \frac {3*X_2 + 1}{2} = \frac {3*(\frac{3*X_1 +1}{2}) +1}{2}$$ or $$X_3 = \frac {3^2*X_1 + 3 + 2}{4}$$ Now since this is a loop with 3 elements, $$X_1 = \frac {3*X_3 + 1}{2^{(m-n)}}$$ writing this out explicitly:

$$X_1 = \frac{3*(\frac {3^2*X_1 + 3 + 2}{4}) + 1}{2^{(m-n)}}$$ or $$X_1 = \frac {3^3*X_1 + [3^2 + (3*2) +4]}{2^{(m)}}$$

The quantity in brackets is where then 19 comes from, i.e. is $$3^3 - 2^3$$

Case II: Now, to get a 2-cycle, we interpose a division by 2 prior to the final cascade of divisions by 2; i.e. $$X_3 = \frac {3*X_2 +1}{4 * 2}$$

So $$X_3 = \frac {3^2*X_1 + 3 + 2}{8}$$

and

$$X_1 = \frac {3^3*X_1 + [3^2 + (3*2) + 8]}{2^{(m)}}$$ Note that the last term in the bracket is now 8 instead of 4 and the sum is 23, or $$3^3 - 2^3 + 4*1$$Any time that we do an additional division by 2 we increase the power of 2 that corresponds to that iteration of the loop, as well as all higher powers of two. For example, if we had n = 5, the b for a 1-cycle would be $$3^5 - 2^5$$ or

$$3^4 + (3^3*2) + (3^2*2^2) + (3*2^3) + (2^4) = 211$$

If we were to do a single consecutive division by 2 after X₄ then b would be

$$3^4 + (3^3*2) + (3^2*2^2) + (3*2^3) + 2^5 = 227$$

If instead we were to do the consecutive division after X₃ then b would be

$$3^4 + (3^3*2) + (3^2*2^2) + (3*2^4) + 2^5 = 251$$

We increase the exponent on the 4th term by 1, and this carries through the remaining terms, in this case the final term. Note that the second b is $$3^5 - 2^5 + 4*4$$ and the third is $$3^5 - 2^5 + 4*10$$ Note also, that if were to make a 3-cycle by doing an additional consecutive division by 2 after X₄, then we would add an additional $2^5$ to the last term and get $$3^4 + (3^3*2) + (3^2*2^2) + (3*2^4) + 2^6 = 283$$ which is $$3^5 - 2^5 + 4*18$$

Notice that every time we add a division by 2 anywhere in the loop prior to the final cascade of divisions of 2 that lead back to the beginning, p increases; it cannot be negative. Also, and I could be wrong about this, but if 3-cycles are also of the form $$3^n -2^n + 4*p$$ it would seem to be the case for any other m-cycle. This is because, the powers of 2 will always increase from left to right, so if we multiply any term by 2 we have to multiply all subsequent terms by the same factor.

The reason p is multiplied by 4 and not 2 is that there cannot be consecutive divisions by 2 in the first iteration, so we would never multiply the second term of b by 2 in going from one m-cycle to another. All increases in the value of b will be in increments of 4*p_i.

I suspect I may be missing some restriction on the values that p can assume, but I don't know where, i.e. some value of p that is inadmissible in a 2-cycle that would be admissible in a 3-cycle. I don't know enough about the way Steiner constructed his proof to tell. For a 3-cycle, you would alter the terms in b twice, and this might produce some ps that cannot occur in 2-cycles. What would be needed to show that there are no loops in the Collatz conjecture using this reasoning is that there is no set of positive integers (m, n, p, I) such that $$\frac {3^n - 2^n + 4*p}{2^m - 3^n} = I$$ that holds for all p. I suspect this cannot be inferred from the non-existence of 2-cycles, even though 2-cycles would produce the same form.

In fact, for n = 5, and m = 8, p = 9 satisfies the equation $$\frac {3^5 - 2^5 + 4*9}{2^8 - 3^5} = 19$$ However, the only permissible ps for n= 5 and a 2-cycle are 4, 10 and 19.

FINAL ADDENDUM

Thank you to Eric Shumard and Collag3n for their responses. My original question could have been phrased "What insights are necessary to prove..." and the responses provided insights. This exercise also illustrated for me the fallacy of assuming that if condition a is necessary to the existence of condition b, the non-existence of condition b implies the non-existence of condition a. Sometimes it does and sometimes it does not.

To expand on the assertions before this final addendum, I should explain how I arrived at them, and what the constraints are. First, to make sure that I am not using ambiguous terminology, as I understand it, a k-cycle means a loop in the Collatz conjecture such that there are exactly k-1 $X_i$ in the loop such that $X_i$ > $X_{i+1}$. For a 2-cycle, there is only one $X_i$ that satisfies this constraint. That is my understanding, and the understanding that is used in this argument.

The assertion that the only permissible ps for a loop with n = 5 elements are 4, 10, and 19 is probably unnecessarily opaque, or at least complicated. A cleaner way of making the assertion is to multiply those numbers by four and state, for n = 5, that a s-cycle consisting of 5 elements has the form $$\frac {3^5 - 2^5 + S_i}{2^8 - 3^5}$$ where S_i = 16, 40, or 76. This is only valid if the consecutive divisions by two that occur before reaching the last element of the loop is limited to one additional division; i.e. $$X_{i+1}=\frac {3*X_i +1}{4}$$ If there are two additional divisions, $S_i$ = 32, 80, or 152, or twice that obtained for a single additional division by 2.

Here is an example for how those numbers were obtained, using an n = 5 loop as an example:

For a 1-cycle, b would be $3^n - 2^n$ or $$3^4 + (3^3*2) + (3^2*2^2) + (3*2^3) + (2^4) = 211$$ If the additional division by 2 were to occur after $X_4$, as we did above, we would multiply the last term of b by 2, or we would add the last term to b: $2^4*2 = 2^5$ or $2^4 + 2^4 =2^5$. So, in this case, $S_1 = 16$ If, instead, the additional division by 2 occurs after $X_3$, then we would add the last two terms to the b for the 1 cycle, or $$3*2^3 + 2^4 = 40$$ If the additional division by 2 occurs after $X_2$ then we would add the last three terms of the b for the 1-cycle: $$3^2*2^2 + 3*2^3 + 2^4 = 76$$

Let j be the number of additional consecutive divisions before reaching $X_n$. For the example above, j = 1. The $S_i$ for a two cycle is given by $$S_i = (3^i - 2^i) * 2^{n-i-1}*2^j$$ where i ranges from 1 to n-2. Therefore the proposition for 2-cycles becomes:

There is no set of positive integers (m, n, i, j, I) with n > 1, m > j, and $1 \le i \le n-2$ such that $$\frac {3^n - 2^n + (3^i - 2^i) * 2^{n-i-1}*2^j}{2^m - 3^n} = I$$

This does not seem to be a very promising approach to the Collatz conjecture. However, we may note for completeness that none of the permissible numerators of the preceding equation can be a multiple of 3, nor obviously a prime number. We may just note for instance that for j = 1 and n=5 the 3 permissible numerators, 211, 227 and 251 are all prime numbers. We can also note that $X_1$ can never be of the form 4*N + 1, where N is an integer > 0.

Answer 1

As Eric stated, this is the Steiner proof. It is very difficult to find online, so I put a summary here:

It goes like this (I replaced the original Baker's bound by the tighter Rhin's bound, and assumed we work in the positive numbers):

To have a 1-cycle you start at an odd number $a=b2^n - 1$, you climb up to $b3^n - 1$ and then divide by $2^{m-n}$ to reach $a$ again:

$$b2^n - 1=\frac{b3^n - 1}{2^{m-n}}$$ In other words $$a=\frac{(a+1)\frac{3^n}{2^n}-1}{2^{m-n}}$$ or $$a(2^m-3^n)=3^n-2^n$$ $$a=\frac{3^n-2^n}{2^m-3^n}$$

In a cycle, $\frac{3^n}{2^m}<1$ (since $a=\frac{3^n}{2^m}\cdot a + \delta$ with $\delta>0$) and with that we can write

$$0 < \frac{2^m}{3^n} - 1 = \frac{2^{m-n} -1}{b3^n} = \frac{2^{m-n} - 1}{2^{m-n}(b2^n - 1) + 1} < \frac {1}{2^n}$$

For an $x>1$ you have $0<\log x<x-1$ so you can write

$$0 < m \log 2 - n \log 3 < \frac{2^m}{3^n} - 1 $$ which leads to $$\begin{array}{|c|}\hline m \log 2 - n \log 3 < \frac {1}{2^n}\\\hline\end{array}$$

But, the smallest $m$ possible for this inequality $\frac{3^n}{2^m}<1$ is $m=\lceil n\cdot log_2(3)\rceil$ so we have $$|m\log 2 - n\log 3|\ge |\lceil n\cdot log_2(3)\rceil\log 2 - n\log 3|$$

and the Rhin bound says: $|\lceil n\cdot log_2(3)\rceil\log 2 - n\log 3| > 0.00218n^{-13.3}\approx (n\cdot log_2(3))^{-13.3}$

Now, we know that (growth of $x^a$ vs $a^x$):

$$\frac{1}{(n\cdot log_2(3))^{13.3}}>\frac {1}{2^n}$$ for $n>\frac{-13.3W_{-1}(\frac{-\log(2)}{13.3 \log_2(3)})}{log(2)}\approx96.52...$ ($W$ is the productlog)

and with some manual check, we can see it is still true for $1<n\le96$ (the case $n=1$ being straightforward anyway)

leading to a contradiction $$\begin{array}{|c|}\hline m \log 2 - n \log 3 > \frac {1}{2^n}\\\hline\end{array}$$ So, there is no $a$ within the 1-cycle Collatz context ($m$ is not free) such that $$a=\frac{3^n-2^n}{2^m-3^n}$$

(More on the Rhin bound here)

Answer 2

What you described is known as a 1-cycle and Steiner proved that they can't exist for 3x+1. Steiner, R. P. (1977). "A theorem on the Syracuse problem". Proceedings of the 7th Manitoba Conference on Numerical Mathematics. pp. 553–9. MR 0535032.