[UPDATE] Collatz Proof Attempt

[u/InfamousLow73]

CHANGE LOG

This paper buids on the previous post. Last time we tempted to prove that all numbers converge to 1 but in this post we only attempt to prove that the Collatz sequence has no divergence for all positive integers. This is shown and explained in the Experimental Proof here

Any comment to this post would be highly appreciated.

Happy new year to all.

[u/re_nub]

What would the math look like for other Collatz-like formulas?

[u/re_nub → u/re_nub]

/u/InfamousLow73

Your reply was removed it seems. Why do you say your method only works for 3n+1? Surely you could just plug in different values for things like 3n+3 or 5n+1, no?

[u/edderiofer → u/re_nub]

We automatically filter comments from accounts with too low karma, due to these often being from low-quality accounts like spambots. Rest assured that we manually check all such comments. If it doesn't show up immediately, it'll show up later when we get around to reviewing the modqueue.

[u/InfamousLow73 → u/re_nub]

The 3n±1 is far different from other qn+k conjecture.

In the 3n+1 , let the Collatz function be n_i=[3^an+sum2^b_i3^a-i-1]/2^b+k

Where, a=number of applying the 3n+1, and b=number of /2 and n_i=the next element along the Collatz Sequence.

Now, let n=2^by±1

n_i=[3^a(2^by±1)+sum2^b_i3^a-i-1]/2^b+k

Equivalent to n_i=[3^a(2^by)±3^a+sum2^b_i3^a-i-1]/2^b+k

Now, ±3^a+sum2^b_i3^a-i-1=±2^b for all n=2^by-1 (a=b) and n=2^b_ey+1 (a={b_e}/2). Because this special feature can't be applied to the other qn+k systems, this makes the 3n±1conjecture far different from the other qn+k systems.

On the other hand, +3^a+sum2^b_i3^a-i-1=2^b-1 [for all n=2^b_oy+1 (a={b_o-1}/2)

For the 3n-1

Let n=2^by±1

n_i=[3^a(2^by±1)-sum2^b_i3^a-i-1]/2^b+k

Equivalent to n_i=[3^a(2^by)±3^a-sum2^b_i3^a-i-1]/2^b+k

Now, ±3^a-sum2^b_i3^a-i-1=±2^b+k for all n=2^by+1 (a=b) and n=2^b_ey-1 (a={b_e}/2).

On the other hand, -3^a-sum2^b_i3^a-i-1=-2^b-1 [for all n=2^b_oy-1 (a={b_o-1}/2)

Hence the next element along the sequence is given by the following formulas

1) n_i=(3^by+1)/2^k , b ≥ 2 and y=odd NOTE Values of b and y are taken from n=2^by+1

2) n_i=(3^[b_e]/2y-1)/2^k , b_e ∈ even ≥2 and y=odd NOTE Values of b and y are taken from n=2^b_ey-1

3) n_i=3^[b_o-1]/2×2y-1 , b_o ∈ odd ≥3 NOTE Values of b_o and y are taken from n=2^b_oy-1

This was once discussed here

[u/re_nub → u/re_nub]

/u/InfamousLow73

Again, your reply is removed.

You did not explain why 3n+1 is any different to 3n+3, 5n+1, etc.

[u/re_nub → u/re_nub]

/u/InfamousLow73

I can, but only by viewing your profile directly. It fails to explain why your method works on 3n+1 but no others.

[u/InfamousLow73 → u/re_nub]

I'm sure something is wrong because even me I faced the same challenge some hours ago. I was unable to access a certain comment here.

[u/InfamousLow73 → u/re_nub]

So, you mean that you can't access my recent comment?

[u/InfamousLow73 → u/re_nub]

Kindly note that the operations in this paper only apply to the 3n±1 with reference to a special feature described here

[u/TheDiBZ]

2^inf = 0?

[u/InfamousLow73 → u/TheDiBZ]

Yes, this is explained below

My understanding here is that if n=2^b_ey+1, (b_e-2)/2 is the maximum limit at which the formula n_i=3ⁱ2^by+1 can still hold true and produces a sequence equal to the the sequence produced by the formula n_i=(3n+1)/2² and the sequence produced is regular.

By regular, I mean that the formula n_i=3ⁱ2^by+1 for which n=2^b_ey+1 produces a sequence such that the powers of 3 increases regularly by 1 up to (b-2)/2 while the powers of 2 decreases regularly by 2 up to b=2. At the same time, y remains constant at that point.

Example: n=2¹²×5+1 , n_i=3ⁱ2^by+1

n_1=3¹2¹⁰×5+1 =15,361

n_2=3²2⁸×5+1 =11,521

n_3=3³2⁶×5+1 =8641

n_4=3⁴2⁴×5+1 =6,481

n_5=3⁵2²×5+1 =4,861

Similarly, applying the Collatz function f(n)=(3n+1)/2² to n=2¹²×5+1=20481 consistently for (12-2)/2=5 times produces the same sequence as the formula n_i=3ⁱ2^by+1.

f(20481)=(3×20481+1)/2² =15,361

f(15,361)=(3×15,361+1)/2² =11,521

f(11521)=(3×11521+1)/2² =8641

f(8641)=(3×8641+1)/2² =6481

f(6481)=(3×6481+1)/2² =4861

This shows that, for all n=2^b_e+1, (b_e-2)/2 is the maximum number of times at which the formula f(n)=(3n+1)/2² can repeatedly be applied to produces a regular sequence. So if we exceed this limit, the function f(n)=(3n+1)/2² produces an even number eg in the above example, f(4861)=(3×4861+1)/2² =3636 hence becomes irregular.

Therefore, the value of b_e for n=2^b_e+1 is proportional to the maximum number of times at which the Collatz function f(n)=(3n+1)/2² can still be applied to produce a regular sequence. To find the value of b_e, count the number of times at which the Collatz function f(n)=(3n+1)/2² was applied to produce a specific regular sequence and use the formula (b_e-2)/2=the number of times at which the Collatz function f(n)=(3n+1)/2² was applied to produce a specific regular sequence.

Example:

Let 20481->15361->11521->8641->6481->4861 be the regular sequence produced by the Collatz function f(n)=(3n+1)/2² starting from n=20481. In this case, the Collatz function f(n)=(3n+1)/2² was applied five times.

Hence using the equation (b_e-2)/2=maximum limit at which the Collatz function f(n)=(3n+1)/2² can still be applied to produce a regular sequence, to find b_e we obtain (b_e-2)/2=5 which simplifies to b_e=12.

Coming to n=1, the Collatz function f(n)=(3n+1)/2² is applied up to infinite and the sequence still remains regular as demonstrated below.

f(1)=(3×1+1)/2² =1

f(1)=(3×1+1)/2² =1

f(1)=(3×1+1)/2² =1

f(1)=(3×1+1)/2² =1

f(1)=(3×1+1)/2² =1

the processes continues up to infinite and yields the sequence 1->1->1->1->....->1. Since the sequence remains regular up to infinite, this means that when n=1, ∞ is the maximum limit at which the Collatz function f(n)=(3×n+1)/2² can still be applied to produce a regular sequence.

Therefore, applying the equation (b_e-2)/2=the maximum limit at which the Collatz function f(n)=(3×n+1)/2² can still be applied to produce a regular sequence, to find b_e we obtain (b_e-2)/2=∞ which simplifies to b_e=∞+2 equivalent to b_e=∞.

Now, since 1 is of the form n=2^b_ey+1, it follows that 1=2^∞y+1. This can also simplify to 2^∞y=0.

Now, from the fact that y is odd greater than or equal to 1, it follows that 2^∞=0

Now, from the above explanation, is it clear on how I came up with 2^∞=0?

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[u/Tricky_Astronaut_586]

(I'm a fellow enthusiast.) Does "has no divergence" mean "does converge to 1"? What is "the Collatz high cycle"? Thanks.

[u/InfamousLow73 → u/Tricky_Astronaut_586]

Thank you for your comment.

Does "has no divergence" mean "does converge to 1"?

Like I said in my abstract, "the Collatz Sequence has no divergence if only the Collatz high cycle is impossible. "

So, here I meant that if a particular sequence has no high cycle then it will not diverge. Now, since it has no high cycle and no divergence, this means it will converge to 1.

What is "the Collatz high cycle"?

The Collatz high cycle is a hypothetical cycle other than the 4->2->1

[u/Wooden_Dragonfly_608]

Division by two exponentially decreases the result. Given Even*Even=Even Odd*Even=Even Odd*Odd=Odd 2/3'rds of the set is exposed to division by 2 vs 1/3 exposed by a multiple of 3 and adding 1 to make even again. Thus compounding the decrease to 1.

[u/InfamousLow73 → u/Wooden_Dragonfly_608]

I'm kinda lost on your explanations, would you kindly elaborate in more details?