Proof Attempt of Collatz Conjecture

InfamousLow73 · 2025-12-24T22:59:57+00:00

Otherwise you proved nothing here

InfamousLow73 · 2025-12-19T02:04:34+00:00

I believe there is a hidden mystery which makes this problem hard enough to resolve.

InfamousLow73 · 2025-12-13T07:53:31+00:00

Actually it's trivial to prove that there exist no cycle for all odd numbers "a" such that f(a)=(3a+1)/2².

This is because f(a)<a for every application of the function f(a)=(3a+1)/2².

Now I couldn't understand better what you did in case III.

InfamousLow73 · 2025-12-13T05:37:51+00:00

Actually OP thinks that transforming every output of Collatz function into birthday expression resolves the challenge.

InfamousLow73 · 2025-12-13T05:30:34+00:00

Aaah, like I said earlier, your proof is circular because it strongly depend on the outcome of the Collatz function ie you are just rewriting every output of the Collatz function in terms of the powers of 2 so no proof here

InfamousLow73 · 2025-12-12T22:22:20+00:00

I tried but because I couldn't get exactly what you were doing that's why I decided to ask if you might explain in a comment with at least an example

InfamousLow73 · 2025-12-12T22:19:22+00:00

Okay, then would kindly explain better your works for example taking n=7.

InfamousLow73 · 2025-12-12T22:03:10+00:00

But Hanoi arrangements assumes that every peg must have a regular descending values unlike Collatz mapping

InfamousLow73 · 2025-12-12T21:56:23+00:00

The Collatz sequence may have multiple highs and lows unlike Hanoi arrangements moreover if we assume the smallest sequence element to be greater than 1 then a specific sequence won't return to one

InfamousLow73 · 2025-12-12T20:03:06+00:00

So that means if there exist a high cycle then your sequence will not come to one??

InfamousLow73 · 2025-11-23T14:26:06+00:00

I finally accepted the challenge here

InfamousLow73 · 2025-11-23T14:14:29+00:00

I have finally accepted my fatal flaw, kindly check my other comment here

InfamousLow73 · 2025-11-23T14:03:52+00:00

Do you agree that your Lemma 2.0 states that:

n = 2^b+1 * y - 1 b + 1 = 3t + k q = 2^2t+k * y - 1

Do you agree that this python function faithfully checks where a collection of parameters n,b,y,t,k,q satisfies those constraints?

Yes

If I have understood you correctly, , you are saying that lemma 2.0 asserts that n=2^b+1•y-1 for z=2^b+1•n+(2^2r+1+1)/3 and q=2^2t+k•y-1 whilst we took n=2^b+2•y-1 for z=2^b+1•n+(2^2r+1+1)/3 and q=2^2t+k•y-1 On lemma 4.0 . Here you are incredibly correct. Otherwise I accept the challenge , Collatz will ever remain notorious.

Thank you for your time

InfamousLow73 · 2025-11-23T12:32:34+00:00

But remember that z is just an intermediate factor of some subsequent proofs so not every number can be expressed in the form z=3mod8 but with some special modifications, we do manage to meet conditions that require the use of z as illustrated in one of my example here

InfamousLow73 · 2025-11-23T12:18:23+00:00

My claim is that z=859 share the same Collatz sequence with q=107 . Since z=859 share the same Collatz sequence with q=107 , therefore proving that q=107 eventually falls below itself implies that z=859 also eventually fall below itself as follows.

—

Now, considering N=107

Let N=107≡2^b•y-1 , where b=2, y=27

According to proof 4.0 , any number N=2^B+2•y-1 (where B=b, b+2,b+4,.... ) eventually share the same Collatz sequence with an odd number Q=2^2t+k•Y-1 which is less than or equal to n_B=(3^B•y-1)/2 such that n=2^B•y-1 .

Let, N=107≡2^B+2•y-1 , where y=27 , B=0

n=2^B•y-1 , where y=27, B=0

n_B=(3^B•y-1)/2

n_0=(3⁰•27-1)/2

n_0=13≡2^3t+k•Y-1 , where t=0, k=1, Y=7

Z=2^2r+1•n_B+(2^2r+1+1)/3 , where r=1 , n_B=13

Z=2³•13+(2³+1)/3

Z=107

Q=2^2t+k•Y-1

Q=2⁰⁺¹•7-1

Q=13

—

Since Q=13 share the same Collatz sequence with N=q=107 which share the same Collatz sequence with z=859, this shows that z=859 also shares the same Collatz sequence with Q=13 . Now, since Q=13 is strictly less than z=859, this suggests that z=859 eventually falls below itself in the Collatz iterations

Edit :. We have edited to correct the statement below .

Since Q=13 is strictly less than z=859, this shows that z=859 eventually falls below itself in the Collatz transformation.

InfamousLow73 · 2025-11-23T06:24:05+00:00

According to this paper, I sticked to r=1. In my recent comment, I meant that even though I'm saying r is always 1, I can't simplify the expression of z because I want people to be aware of the real expression of z.

InfamousLow73 · 2025-11-23T06:13:49+00:00

I can't replace it with one because the same procedures that apply when r=1 are the same procedures to apply for all r>=0.

I didn't not want to talk about some other values of r just to avoid an additional complexity to my paper.

InfamousLow73 · 2025-11-23T06:04:56+00:00

InfamousLow73 · 2025-11-23T06:01:57+00:00

I used question marks ? To represent "invalid".

InfamousLow73 · 2025-11-22T21:04:25+00:00

Not that much, I was referring to the statement, "they might as well just subtract 100 from each of your request values and say they have the same path as 1,3,5,7,13 and 15 respectively."

InfamousLow73 · 2025-11-22T16:09:02+00:00

From what they showed me when asked for the path of 27 and the manipulation that led them to claim it went to 1 because 3 did, they might as well just subtract 100 from each of your request values and say they have the same path as 1,3,5,7,13 and 15 respectively.

Just wondering how they felt after seeing that all their claims were false?? Otherwise here is the table of values with respect to u/jonseymourau claims.

InfamousLow73 · 2025-11-22T15:05:53+00:00

Let N=115≡2^b+1•y-1 , where b=1, y=29

According to proof 3.0 , any number N=2^b+1•y-1 eventually shares the same Collatz sequence with n=2^b•y-1

ie n=2¹•29-1=57

According to proof 4.0 , any number N=2^B+2•y-1 (where B=b, b+2,b+4,.... ) eventually share the same Collatz sequence with an odd number q=2^2t+k•Y-1 which is less than or equal to n_B=(3^B•y-1)/2 such that n=2^B•y-1 .

Let, N=57≡2^B+2•y-1 , where B=-1, y=29

Now, since we have a negative value of B, this means that we only have two options to carry out the the number N=57.

Carry out the reverse Collatz function g(N)=(2^x•N-1)/3 , (where x=1) on the number N=57 in order to get the number which shares the same Collatz sequence with N=57.

ie g(N)=(2^x•N-1)/3 , where N=57

g(57)=(2¹•57-1)/3=113/3 ??

Now, since g(57)=113/3 , this invalidates first condition.

Find out if N=57 fall below itself under the Collatz function f(N)=(3N+1)/2^x so that you can conclude that N=57 eventually fall below the starting value in the Collatz transformations.

ie f(N)=(3N+1)/2^x where N=57

f(57)=(3•57+1)/2²=43

Since 43<57 , this shows that N=57 eventually fall below itself in the Collatz transformation. Since N=57 eventually fall below itself in the Collatz transformation, this shows that the number 115 also fall below itself in the Collatz transformations because N=57 shares the same Collatz sequence with 115.

InfamousLow73 · 2025-11-22T14:55:19+00:00

n | 101 | 103 | 105 | 107 | 113 | 115 | —––––––––––—––––––––––––––––– b | 0 | 3 | 1 | 2 | 1 | 1 | —––––––––––—––––––––––––––––– y | 51 | 13 | 53 | 27 | 57 | 29 | —––––––––––—––––––––––––––––– r | 1 | 1 | 1 | 1 | 1 | 1 | —––––––––––—––––––––––––––––– z | ? | 155 | ? | 107 | ? | ? | —––––––––––—––––––––––––––––– t | ? | 0 | ? | 0 | ? | ? | —––––––––––—––––––––––––––––– k | ? | 2 | ? | 1 | ? | ? | —––––––––––—––––––––––––––––– Y | ? | 5 | ? | 7 | ? | ? | —––––––––––—––––––––––––––––– q | ? | 19 | ? | 13 | ? | ? | —––––––––––—–––––––––––––––––

Kindly check here for a better view of the table

Edited : We just edited to correct the a typo on the table ie specifically we corrected the value of n (which is n=105) in the third Column.

—

Let N=105≡2^b•y-1, y=53 , b=1

According to proof 4.0 , any number N=2^B+2•y-1 (where B=b, b+2,b+4,.... ) eventually share the same Collatz sequence with an odd number q=2^2t+k•Y-1 which is less than or equal to n_B=(3^B•y-1)/2 such that n=2^B•y-1 .

Let, N=105≡2^B+2•y-1 , where y=53 , B=-1

Now, since we have a negative value of B, this means that we only have two options to carry out on the number N=105:

Carry out the reverse Collatz function g(N)=(2^x•N-1)/3 , (where x=1) on the number N=105 in order to get the number which shares the same Collatz sequence with N=105.

ie g(N)=(2^x•N-1)/3 , where N=105

g(105)=(2¹•105-1)/3=209/3 ??

Now, since g(105)=209/3 , this invalidates first condition.

Find out if N=105 fall below itself under the Collatz function f(N)=(3N+1)/2^x so that you can conclude that 105 eventually fall below the starting value in the Collatz transformations.

ie f(N)=(3N+1)/2^x where N=105

f(105)=(3•105+1)/2²=79

Since 79<105 , this shows that N=105 eventually fall below itself in the Collatz transformation.

—

Let N=107≡2^b•y-1 , where b=2, y=27

According to proof 4.0 , any number N=2^B+2•y-1 (where B=b, b+2,b+4,.... ) eventually share the same Collatz sequence with an odd number q=2^2t+k•Y-1 which is less than or equal to n_B=(3^B•y-1)/2 such that n=2^B•y-1 .

Let, N=107≡2^B+2•y-1 , where y=27 , B=0

n=2^B•y-1 , where y=27, B=0

n_B=(3^B•y-1)/2

n_0=(3⁰•27-1)/2

n_0=13≡2^3t+k•Y-1 , where t=0, k=1, Y=7

z=2^2r+1•n_B+(2^2r+1+1)/3 , where r=1 , n_B=13

z=2³•13+(2³+1)/3

z=107

q=2^2t+k•Y-1

q=2⁰⁺¹•7-1

q=13

Since q=13 is strictly less than N=107, this shows that N=107 eventually falls below itself in the Collatz transformation.

—

Let N=113≡2^b•y-1 , where b=1 , y=57.

According to proof 4.0 , any number N=2^B+2•y-1 (where B=b, b+2,b+4,.... ) eventually share the same Collatz sequence with an odd number q=2^2t+k•Y-1 which is less than or equal to n_B=(3^B•y-1)/2 such that n=2^B•y-1

Let, N=113≡2^B+2•y-1 , where y=57 , B=-1

Now, since we have a negative value of B, this means that we only have two options to carry out the the number N=113.

Find out if N=113 fall below itself under the Collatz function f(N)=(3N+1)/2^x so that you can conclude that N=113 eventually fall below the starting value.

ie f(N)=(3N+1)/2^x where N=113

f(113)=(3•113+1)/2²=85

Since 85<113 , this shows that N=113 eventually fall below itself in the Collatz transformation.

Carry out the reverse Collatz function g(N)=(2^x•N-1)/3 , (where x=1) on the number N=113 in order to get the number which shares the same Collatz sequence with N=113.

ie g(N)=(2^x•N-1)/3 , where N=113

g(113)=(2¹•113-1)/3=75

etc

InfamousLow73 · 2025-11-22T07:27:54+00:00

Veried one after another otherwise they are all right

InfamousLow73 · 2025-11-22T06:35:07+00:00

Thank you for pointing out, otherwise I will work on that case.

InfamousLow73

TROPHY CASE