Are any of the steps taken here valid? If so how could this be pushed further?

General
1

Hello, new here just a heads up I am not a mathematician and English isn’t my first language but please bare with me.

I recently learned it is possible to do this simplification from this video (I was first intrigued by a thread on this forum a couple of weeks ago) and revisited Collatz problem today:

We have M = \cfrac{3^x - 2^x}{2^k - 3^x}

Here x represents the number of odd steps and k represents the length of the loop.

So if M is an integer we have a valid circuit.

Suppose that 2^{k} - 3^{x} is a multiple of 3^{x} - 2^{x} this means that we can write 3^{x} - 2^{x} like this for some number m:

3^{x} - 2^{x} = T(m + (2^{k} - 3^{x})) - T(m) \ \dots \ (1)

Where T(n) is the nth triangular number formula.

After expending and factoring (1) we get:

3^{x} - 2^{x} = \cfrac{(2^{k} - 3^{x})(2^{k} - 3^{x} + 2m + 1)}{2}

Let us set

f(k, x) = 3^x - 2^x

g(k, x) = 2^{k} - 3^{x}

d = 2m + 1

We get the following quadratic equation:

g^2 + dg - 2f = 0

Solving for g (and skipping few steps) we get:

2^k - 3^x = \cfrac{-d \pm \sqrt{d^2 + 8(3^x - 2^x)}}{2}

But we care only about the positive solution:

2^k - 3^x= \cfrac{-d + \sqrt{d^2 + 8(3^x - 2^x)}}{2}

What I found empirically is that the RHS yields a limited number of integer solutions the greatest of which is far smaller than the difference given by the LHS.

Are any of the steps I did valid if so how could this be pushed further?

2

As an example this is the case when x=6

>>> [y for y in [(sqrt(8*f(6) + (2*m+1)**2) - (2*m+1))/2 for m in range(2000)] if y.is_integer()]
[35.0, 19.0, 14.0, 10.0, 7.0, 5.0, 2.0, 1.0]

>>> [2**k - 3**6 for k in range(20)]
[-728, -727, -725, -721, -713, -697, -665, -601, -473, -217, 295, 1319, 3367, 7463, 15655, 32039, 64807, 130343, 261415, 523559]
>>>

295 > 35

4

I’ve started reading Steiner’s paper and plan to revisit the video series. My first key takeaway was realizing that a “cycle” and a “circuit” are distinct concepts. I don’t know why I confused the two.. I will re-engage with the core problem once my review is complete. Thanks for reading.

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5

Hi! I missed this topic so feel free to tell me if I’m late and you moved away from that.
I have a few questions to help me follow your reasoning.
I’m not sure I understand how you get (1). If 2^k-3^x is a multiple of 3^x-2^x, then we can write 2^k-3^x = m(3^x-2^x). Where is the triangular number formula coming from?

6

I attempted to apply the property of expressing numbers as the difference of two triangular numbers. For example, we can write:

25 = 28 - 3 = T(5 + 2) - T(2)

where 5 is a factor of 25, and m here equals 2.

7

Sorry, I am not a specialist of triangular numbers and I don’t know this property, is this true for any number x? Unless you allow the factor to be equal to 1 and m to be equal to x, of course.

Anyway, assuming this is correct, wouldn’t then (1) be:
2^k-3^x = T(m+3^x-2^x)-T(m) ?
As it is 2^k-3^x which is a multiple of 3^x-2^x

8

I am not an expert either. According to [Polite number - Wikipedia] this is true for all odd divisors. Since both 3^x - 2^x and 2^k - 3^x are odd I think we can use this property.

If you assume 2^k - 3^x > 3^x - 2^x then you can consider the reverse. In the previous example of 25, 3^x - 2^x (which is the larger number) represents the 25 and 2^k - 3^x the odd divisor 5 and not the opposite.

9

Some polite representations from odd divisors do not allow us to use the mentioned property. However, we can fix this (i.e. include all divisors) by multiplying 3^x - 2^x by 2 (or 3). Therefore, I believe the following should always hold true for some integer m:

2(3^{x} - 2^{x}) = T(m + (2^{k} - 3^{x})) - T(m)

… provided that 2^{k} - 3^{x} is a divisor of 3^{x} - 2^{x} of course. But the goal here is to prove that there is a contradiction, which means to show it is impossible to satisfy for any fixed value of x.

10

As an example, we can’t write 55 as a difference of triangular numbers T(m + 11) - T(m) for any m if we try to use 11 (which is a divisor for 55) here. However, we can write 110=55 \times 2 = T(11+4) - T(4) and 165 = 55 \times 3 = T(11+9) - T(9).

And since both 3^x - 2^x and 2^k - 3^x are always of the form 6k \pm 1 so multiplying by either 2 or 3 don’t affect anything and we don’t lose the generality of the property.

Thank you, your question helped me spot that mistake.

11

Thanks! I didn’t know about that. My gut reaction was to test it: “Ok, we can write 1 as T(1)-T(0), alright, 2=… T(2)-T(1), sure, but actually for any number we can write n as T((n-1)+1)-T(n-1) and that’s really nothing special.”

Now I see that I was testing powers of 2, which are the exception! The property can be used not when those numbers are odd but when they are not powers of 2, and bot 3^x-2^x and 2^k-3^x can be equal to 1, which is 2^0. Thankfully this only happens for a finite number of values, that we can deal with separately.

What do you mean by that? For example, with x=0, 3^0 - 2^0 = 0, which is not “of the form 6k \pm 1” unless I am missing something.

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12

I’m just learning in the process like you. About that my bad.. I meant when both x and k are different from 0 (because the total length of the circuit can’t be 0, of course), we would only be dealing with values of the form 6k \pm 1, as that’s what both 3^x - 2^x and 2^k - 3^x will yield for any values of x and k that we didn’t exclude. So if you multiply 3^x - 2^x by 2 or 3, that doesn’t affect the steps — we just change the 8 in the solution to 8 \times 2 or 8 \times 3, since 2^k - 3^x can’t have 2 or 3 as a factor if it’s of the form 6k \pm 1. If that makes sense.

By “solution” I mean:

Y = \cfrac{-d + \sqrt{d^2 + \mathbf{8\times2}(3^x - 2^x)}}{2}
13

For Y to take its maximum value, d needs to be 0. Suppose d = 0.

We can write:

Y = 2 \sqrt{3^x - 2^x}

And also

Y = 2^k - 3^x (since this is how we changed the variable)

The smallest value of k that makes 2^k > 3^x is approximately k \sim 1.6x.

I asked DeepSeek to check this inequality:

6. Conclusion for large x

For x \gtrsim 5.95, the inequality

2^{1.6x} - 3^x > 2\sqrt{3^x - 2^x}

holds true, and as x \to \infty, LHS \sim 2^{1.6x}, RHS \sim 2 \cdot 3^{x/2}, and since 1.6 \ln 2 \approx 1.109 > (\ln 3)/2 \approx 0.549, LHS grows faster.

Can we say that there is a contradiction when x \geq 6?

If x is the number of odd steps, can we conclude that circuits with six or more odd steps cannot form?

14

x could be 0, the cycle [0] is a valid one where x=0 and k=1, but if you want to exclude this one case that’s understandable.

Sorry, I have a hard time figuring that out.
To recap:
Let’s assume we can find x>0 and k\leq x, two positive integers such that 3^x - 2^x is a multiple of 2^k - 3^x.
Let’s note f=3^x - 2^x and g=2^k - 3^x.
f isn’t a power of two, and thus is a polite number. As g is one of its odd divisors, one of the polite representations of f is of the shape T(g+m)-T(m).
So, f=T(g+m)-T(m) = \frac{g(g+2m+1)}{2}.
By solving this equation, this is equivalent to:
2g=−d±\sqrt(d^2+8f) (where d=2m+1)
Up to this point I think I’m following you.

But then, if we were to do the same reasoning except this time instead of f we take f'=2f, there is no reason to think m' (and thus d') is going to have the same value. Moreover a polite representation can only be built from a odd divisor, and although if g divides f then 2g divides 2f, 2g isn’t odd.

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15

I am very grateful for your time and effort. Your confirmation that you are following my logic up to this point is greatly appreciated.

Just one note: in our context, k is strictly greater than x (because k = x + e, where e represents the number of downs).

Let us forget about equation (1) in my original post.

A polite representation of \mathbf{2f} can be written in the form T(g + m') - T(m'), provided that g is an odd divisor of f.

To illustrate, suppose f = 55 and g = 11. We can write 110 = 2 \times 55 = 120 - 10 = T(11+4) - T(4).

We have the following derivation:

2f = \cfrac{g(g + 2m' + 1)}{2}

4f = g(g + 2m' + 1)

g(g + d') - 4f = 0

g^2 + d'g - 4f = 0

Solving for g:

g = \cfrac{-d' + \sqrt{d'^2 + \mathbf{16f}}}{2}

For \cfrac{-d' + \sqrt{d'^2 + 16f}}{2} to take its maximum value, we simply set d' = 0, which gives:

g = 2\sqrt{f}

Back-substituting f and g:

2^k - 3^x = 2 \sqrt{3^x - 2^x}

Please let me know if you are still following my reasoning up to this point.

16

We use odd g to build the polite representation of 2f. We can even do the same for 3f — it doesn’t matter. If g is an odd divisor of f, this approach works for both 2f and 3f, but not always for f itself. For example, we can build a polite representation for 55 using 5 (since 55 = T(5 + 8) - T(8)), but we can’t do that using 11. So the trick to generalizing is to multiply f by 2 or 3 to ensure that we can build polite representations for 2f or 3f using all the odd divisors of f and the condition that g must be an odd divisor of f remains unchanged if that makes sense.

17

So, if I follow correctly you are now telling me that actually the property you used sooner (that there exists m such that f=T(g+m)-T(m) when g divides f) isn’t true…
From what I read on wikipedia, this property isn’t stated, there is just a correspondance between each odd divisor and each polite representation but for 55 and 11 the corresponding one is 55 = T(10)-T(0).
Is there something I missed that makes you think that by taking a bigger f it is going to become true?

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18

Thank you for asking the right questions, you’ve helped me identify the weak points. I realize I shouldn’t have called it a property just because it worked in a few examples I tested, as it obviously isn’t one.

To check, I just multiplied two primes (one small and one relatively large) by 2 and 3, but that didn’t work. It always seems to only work for the smallest prime. And when it does work, we can find the other prime in the sum right in the middle. For example, 55 = 9 + 10 + 11 + 12 + 13. But since we are dealing with numbers of the form 6k \pm 1, so f perhaps would have more than one prime factor. The test still shows that multiplying by either 2 or 3 didn’t do the trick.

I’m not sure, but if we multiply by bigger numbers that are co-prime with f and g , maybe that would work. However, it would need to be proven, and even then I think it would introduce another unknown variable to deal with. So the proof might become more complicated, or perhaps this is a dead end.

Once again, thank you so much for your time and insight. I really appreciate you pushing my thinking on this. I learned few things.

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19

Don’t worry, even if I may sound dry I think you have interesting ideas and that’s great!

What you describe in this last post is true for any number that is the product of two odds:
11*9 = 99, so you can write 99 = 11+11+11+11+11+11+11+11+11
= 10+11+11+11+11+11+11+11+12 (transfer 1 from 1st term to the last)
= 9 + 10+11+11+11+11+11+12+13 (transfer 1 from 1st and 2nd term to the 2 last ones)
= 8+9 + 10+11+11+11+12+13+14
= 7+8+9 + 10+11+12+13+14+15
= T(15)-T(6)
But also:
99 = 9+9+9+9+9+9+9+9+9+9+9
= 4+5+6+7+8+9+10+11+12+13+14
=T(14)-T(3)

But when one of the factors is relatively big compared to the whole number, a problem ensues:
33=11*3=3+...+3
=(-2)+(-1)+0+1+2+3+4+5+6+7+8
=T(8)-T(2)
Here you don’t have 8=11+m

That said your fix could work: Let’s assume you were looking for a polite representation for pq, where p is big enough that we have the previous situation.
Using pq*2^n instead:
33*2^n = 3*2^n + 3*2^n+ ... 3*2^n
= (3*2^n-5) + (3*2^n-4)+ ... (3*2^n+5)
=T(3*2^n+5) -T(3*2^n-6) (as long as we pick n \geq 1 so that (3*2^n-6) is positive)
=T(11+m)-T(m) with m=(3*2^n-6)

Not sure where you go from there, though.

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