Simplifying the 3n+1 problem where any starting odd number will always
terminate too 16,8,4,2,1,4,2,1,4,2,1...
I know it is still classed as a conjecture but I believe the above statement
puts it clearly as a theorem as stated in the reasoning below
..
All numbers that cross over are not repeated in the same sequence

Giving all the odd numbers in the first column where all evens are not shown.
Odds will appear as a start of a sequence.
Anything represented in a previous sequence will appear just once.
Note sequence 3,10,5.. where 5 appears so later 5 is stand alone.
The second number in each sequence is just the multiple of 6 too the next
second number of the next sequence.
If an appears breaks up that multiple such as sequence 9 has 28 as the second
number then we know sequence 7 will have 28+6 = 34 somewhere in its sequence.
Also for the third number in each sequence the multiple is 3 between instead
of 6
The same holds true for each appears between, where as an example sequence
9 term 3 = 14 where two (appears) in the next two sequences = 3+3=6 +3 for
sequence 15 = 9 then 9+14= 23 which is the 3rd term of seq. 15
So the second term of each sequence has a difference of 6 too the next
sequence and the third term in each sequence has a difference of 3 too the
next sequence.
1 3 10 5 16 8 4 2 1
5 appears in sequence 3
7 22 11 34 17 52 26 13 40 20 10 appears in sequence 3
9 28 14 7 appears in sequence 7
11 appears in sequence 7
13 appears in sequence 7
15 46 23 70 35 106 53 160 80 40 appears in sequence 7
17 appears in sequence 7
19 58 29 88 44 22 appears in sequence 7
21 64 32 16 8 4 2 1
23 appears in sequencw 15
25 76 38 19 appears in sequence 19
27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 appears in sequence 15
29 88 44 22 appears in sequence 7
31 appears in sequence 27
33 100 50 25 appears in sequence 25
35 appears in sequence 15
37 112 56 28 14 7 appears in sequence 7
39 118 59 178 89 268 134 67 202 101 304 152 76 38 19 appears in sequence 19
41 appears in sequence 27
43 130 65 196 98 49 148 74 37 appears in sequence 37
45 136 68 34 appears in sequence 7
47 appears in sequence 27
49 appears in sequence 43
51 154 77 232 116 58 29 appears in sequence 29
53 appears in sequence 15
55 166 83 250 125 376 188 94 47 appears in sequence 27
57 172 86 43 appears in sequence 43
59 178 89 268 134 67 202 101 304 152 76 appears in sequence 25
61 appears in sequence 27
63 190 95 286 143 430 215 646 323 970 485 1456 728 364 appears in sequence 27
65 196 98 49 appears in sequence 43
67 appears in sequence 39
69 208 104 52 appears in sequence 7
71 appears in sequence 27
73 220 110 55 appears in sequence 55
75 226 113 340 170 85 256
77 232 116 58 29 appears in sequence 29
79 238 119 358 179 538 269 808 404 202 appears in sequence 59
81 244 122 61 appears in sequence 61
83 250 125 376 188 94 appears in sequence 27
85 appears in sequence 75
87 262 131 394 197 592 296 148 74 37 appears in sequence 37
89 268 134 67 appears in sequence 39
91 appears in sequence 27
93 280 140 70 35 appears in sequence 15
95 286 143 430 215 646 323 970 485 1456 728 364 appears in sequence 27
97 292 146 73 appears in sequence 73
99 298 149 448 224 112 56 23 appears in sequence 15
101 304 152 76 38 19 appears in sequence 19
103 appears in sequence 27
105 316 158 79 appears in sequence 79
107 322 161 and so on --->oo

This pattern of the second term +6 holds for each next sequence.
Sequence 101 second term is 304 +6 =310 so 304+6 = 310 which appears in
sequence 27
This pattern of the third term +3 holds for each next sequence.
Sequence 101 third term is 152 +3 =155 so 152+3 = 155 which appears in
sequence 27

Please point out any errors.

So second term +6 and third term +3 defines the next odd numbered starting sequence.

Dan

On 5/30/2022 5:02 PM, Dan joyce wrote:

> This pattern of the second term +6 holds for each next sequence.
> Sequence 101 second term is 304 +6 =310 so 304+6 = 310 which appears in
> sequence 27
> This pattern of the third term +3 holds for each next sequence.
> Sequence 101 third term is 152 +3 =155 so 152+3 = 155 which appears in
> sequence 27
>
> Please point out any errors.
>
> So second term +6 and third term +3 defines the next odd numbered starting sequence.
>

1) In the 3n+1 sequence, if n is odd, the next term is 3n+1. The next
odd number will be n+2. The next term in its sequence would be
3(n+2)+1, or 3n+6+1. This does differ from the next term of n (3n+1) by
6. (3n+6+1) - (3n+1) = 6.

2) If n is odd, 3n+1 must be even (3n is odd, odd+1 is even) So the
second following term is 1/2 the first following term. For the two odd
numbers n and n+2, the first following terms are 3n+1 and 3n+6+1. The
next terms will be (3n+1)/2 and (3n+6+1)/2. The differences between
these two are 6/2 or 3.

3) Some famous mathematician (I who) told his students not to waste time
with the 3n+1 problem, it's a trap which will suck all your attention.

Your conjectures about the next term of n+2 being 6 greater than the
next term of n (n odd) and the second term of n+2 being 3 greater than
the second term of n both appear perfectly correct.

Michael Moroney wrote :
> On 5/30/2022 5:02 PM, Dan joyce wrote:
>
>> This pattern of the second term +6 holds for each next sequence.
>> Sequence 101 second term is 304 +6 =310 so 304+6 = 310 which appears in
>> sequence 27
>> This pattern of the third term +3 holds for each next sequence.
>> Sequence 101 third term is 152 +3 =155 so 152+3 = 155 which appears in
>> sequence 27
>>
>> Please point out any errors.
>>
>> So second term +6 and third term +3 defines the next odd numbered starting
>> sequence.
>>
>
> 1) In the 3n+1 sequence, if n is odd, the next term is 3n+1. The next odd
> number will be n+2. The next term in its sequence would be 3(n+2)+1, or
> 3n+6+1. This does differ from the next term of n (3n+1) by 6. (3n+6+1) -
> (3n+1) = 6.
>
> 2) If n is odd, 3n+1 must be even (3n is odd, odd+1 is even) So the second
> following term is 1/2 the first following term. For the two odd numbers n
> and n+2, the first following terms are 3n+1 and 3n+6+1. The next terms will
> be (3n+1)/2 and (3n+6+1)/2. The differences between these two are 6/2 or 3.
>
> 3) Some famous mathematician (I who) told his students not to waste time with
> the 3n+1 problem, it's a trap which will suck all your attention.

You forgot to write forget. You elided who you alluded to.

> Your conjectures about the next term of n+2 being 6 greater than the next
> term of n (n odd) and the second term of n+2 being 3 greater than the second
> term of n both appear perfectly correct.

Dan Joyce being opaque about the Collatz conjecture. Probably because Dan does not want to give credit to AP for proving Collatz in 2016.
On Monday, May 30, 2022 at 4:02:28 PM UTC-5, Dan joyce wrote:
> Simplifying the 3n+1 problem where any starting odd number will always

Here is AP's proof of Collatz--
World's First Proof of Collatz Conjecture// Math proof series, book 6 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 20Aug2021. This is AP's 19th published book.

Preface: Old Math's Collatz conjecture, 1937, was this: If you land on an even number, you divide by 2 until you come to an odd number. If you come to or land on an odd number, you do a 3N+1 then proceed further. The conjecture then says that no matter what number you start with, it ends up being 1..

What the Collatz proof of math tells us, is that so very often mathematicians pose a conjecture in which their initial formulation of the conjecture is murky, obfuscation and poorly designed statement. Such poorly designed statements can never be proven true or false. An example that comes to mind of another poorly designed conjecture is the No Odd Perfect Conjecture, in which the statement is obfuscation of factors. So for the odd number 9, is it 1+3, or is it 1+ 3 + 3. So when a mathematics conjecture is full of obfuscation and error in the statement, then these type of conjectures never have a proof. And takes a person with a logical mind to fix and straighten out the conjecture statement and then provide a proof, thereof.

Cover picture: when I think of Collatz, I think of a slide, a slide down and so my French curve is the best slide I can think of, other than a slide-ruler, but a slide ruler is slide across.

--------------------------
Table of Contents
--------------------------

Part I: AP researches the Collatz conjecture in 2016 and proves it with 4 Consecutive Evens.

Alright tonight I dozed off for about an hour and once awoken I had Collatz proof of mine on my mind. Trouble was I could not remember the structure of my proof so had to review what my 2016 proof was.

It is very well and nicely contained. It is a very strong proof.

What it says is this-- quoting from my book.

Proof at last Re: 3/8 increments of fall in Collatz Re: Mechanism shown for 75, as to why Collatz works; preliminary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1
Mechanism shown for 75, as to why Collatz works; preliminary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1

Alright, what that mechanism means is given any odd number, say we are given 75, then the even numbers surrounding 75 from 3N-3, 3N-1, 3N+1, 3N+3 will always have a even number that has three divisions by 2. So for 75 the 3N is 225. And thus we have a choice 3N-3=222, or 3N-1=224 or 3N+1=226, or 3N+3 =228. One of those choices has at least three divisions by 2 and perhaps one has more than two. So, let us see. 222=2*111, 224=2*2*2*2*2*7, 226=2*113, 228=2*2*57.

So here we have the even number 224 which has a 2 as factor for 5 times.

So Old Math can prove that in any given odd number larger than 1, that we always have a three factor of 2 even number produced. So that Collatz mechanism is that we multiply by 3 any odd number, but we end up by dividing by at least 8 any even number produced, because one of those even numbers is always divisible by 8.

--- end quote ---

So what the proof basically is-- is given any 4 consecutive even numbers such as the example of 222, 224, 226, 228 that one of them is factorable --at minimum-- by 8 and thus has a reduction of at least 8 times. In my case example 224 is factorable by 2 for 5 times, mind you not 10 but 2*2*2*2*2 = 32.

And what solves the proof is that we alter the statement of Collatz to mean that we Have A Choice of Even Number. We have a choice of picking 3N-3, 3N-1, 3N+1, or 3N+3 which 222, 224, 226, 228 are a result of the odd number 75.

So the Proving Mechanism is this choice of 4 even numbers and we must pick, yes, must select the highest factorable even number of those four numbers. That means in each Collatz up tick from a odd number is no more than 3 times but every even number will be a down slide by a factor of at minimum of 8 times. So, in the proof of Collatz, every number will head for 1, because upward tick is 3 times, but every downward slide is 8 times downward.

This is a feature of math proofs, we tend to forget them as the years pass and need to constantly review how the proof worked.

Now many insane babbling critics will say I altered the statement of Collatz. Yes, I altered a fog statement by illogical degenerates in mathematics to a statement that bears the mechanism of Collatz. There can never exist a proof of Old Original Collatz, for it is a cloudy mess of why the slide occurs. The slide occurs from the structure of even numbers. The smallest set of even numbers is 2, 4, 6, 8 and it is the 8 that is divisible by 2 in four times. It is the mandatory picking of the largest even number in the 4 consecutive evens that proves Collatz.

For the fools that say AP proved something different from Collatz. No, AP only showed that Old Collatz was not a intelligent statement. And I can dream up millions of statements that may look like some math statement but is just a garbled mess, never provable.

AP

On 6/1/2022 5:43 AM, Archimedes Plutonium wrote:
> Alright tonight I dozed off for about an hour and once awoken I had Collatz proof of mine on my mind. Trouble was I could not remember the structure of my proof so had to review what my 2016 proof was.

> What it says is this-- quoting from my book.
>
> Proof at last Re: 3/8 increments of fall in Collatz Re: Mechanism shown for 75, as to why Collatz works; preliminary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1
> Mechanism shown for 75, as to why Collatz works; preliminary Full Collatz 3N+-1 simultaneously 3N+-1 proof; disproof of half-baked Collatz 3N+1
>
> Alright, what that mechanism means is given any odd number, say we are given 75, then the even numbers surrounding 75 from 3N-3, 3N-1, 3N+1, 3N+3 will always have a even number that has three divisions by 2. So for 75 the 3N is 225. And thus we have a choice 3N-3=222, or 3N-1=224 or 3N+1=226, or 3N+3 =228. One of those choices has at least three divisions by 2 and perhaps one has more than two. So, let us see. 222=2*111, 224=2*2*2*2*2*7, 226=2*113, 228=2*2*57.
>
> So here we have the even number 224 which has a 2 as factor for 5 times.
>
> So Old Math can prove that in any given odd number larger than 1, that we always have a three factor of 2 even number produced. So that Collatz mechanism is that we multiply by 3 any odd number, but we end up by dividing by at least 8 any even number produced, because one of those even numbers is always divisible by 8.
>
> --- end quote ---
>
It is true for any 4 consecutive even numbers, one will be divisible by
8. How is that helpful? Your claim is that one of {3N-3, 3N-1, 3N+1,
3N+3} is divisible by 8, while correct, isn't helpful since the next
term is 3n+1, not pick one from the set {3N-3, 3N-1, 3N+1, 3N+3}.
Perhaps you didn't understand the problem?

As I mentioned, 3n+1 is guaranteed to be even, so the conjecture could
be modified so for odd n, the next in the sequence is (3n+1)/2 rather
than 3n+1 without really changing the problem other than skipping many
always-even numbers.

> So what the proof basically is-- is given any 4 consecutive even numbers such as the example of 222, 224, 226, 228 that one of them is factorable --at minimum-- by 8 and thus has a reduction of at least 8 times. In my case example 224 is factorable by 2 for 5 times, mind you not 10 but 2*2*2*2*2 = 32.

But in your specific case of 75, the next in sequence is 226, not 224.
>
> And what solves the proof is that we alter the statement of Collatz to mean that we Have A Choice of Even Number. We have a choice of picking 3N-3, 3N-1, 3N+1, or 3N+3 which 222, 224, 226, 228 are a result of the odd number 75.

That is a completely different problem. The real Collatz next term is
3n+1, not "pick and choose".

On Tuesday, May 31, 2022 at 2:05:50 PM UTC-4, Michael Moroney wrote:
> On 5/30/2022 5:02 PM, Dan joyce wrote:
>
> > This pattern of the second term +6 holds for each next sequence.
> > Sequence 101 second term is 304 +6 =310 so 304+6 = 310 which appears in
> > sequence 27
> > This pattern of the third term +3 holds for each next sequence.
> > Sequence 101 third term is 152 +3 =155 so 152+3 = 155 which appears in
> > sequence 27
> >
> > Please point out any errors.
> >
> > So second term +6 and third term +3 defines the next odd numbered starting sequence.
> >
> 1) In the 3n+1 sequence, if n is odd, the next term is 3n+1. The next
> odd number will be n+2. The next term in its sequence would be
> 3(n+2)+1, or 3n+6+1. This does differ from the next term of n (3n+1) by
> 6. (3n+6+1) - (3n+1) = 6.
>
> 2) If n is odd, 3n+1 must be even (3n is odd, odd+1 is even) So the
> second following term is 1/2 the first following term. For the two odd
> numbers n and n+2, the first following terms are 3n+1 and 3n+6+1. The
> next terms will be (3n+1)/2 and (3n+6+1)/2. The differences between
> these two are 6/2 or 3.
>
> 3) Some famous mathematician (I who) told his students not to waste time
> with the 3n+1 problem, it's a trap which will suck all your attention.
>
> Your conjectures about the next term of n+2 being 6 greater than the
> next term of n (n odd) and the second term of n+2 being 3 greater than
> the second term of n both appear perfectly correct.

If this is a true pattern of 3n+1 --->oo what other pattern(s) are there?

Dan

Dan joyce wrote:

> On Tuesday, May 31, 2022 at 2:05:50 PM UTC-4, Michael Moroney wrote:
>> On 5/30/2022 5:02 PM, Dan joyce wrote:
>> > This pattern of the second term +6 holds for each next sequence.
>> > Sequence 101 second term is 304 +6 =310 so 304+6 = 310 which appears
>> > in sequence 27 This pattern of the third term +3 holds for each next
>> > sequence.
>> > Sequence 101 third term is 152 +3 =155 so 152+3 = 155 which appears
>> > in sequence 27
>>
>> Your conjectures about the next term of n+2 being 6 greater than the
>> next term of n (n odd) and the second term of n+2 being 3 greater than
>> the second term of n both appear perfectly correct.
>
> If this is a true pattern of 3n+1 --->oo what other pattern(s) are
> there?

morone is angry being banned from posting in sci.physics.relativity.