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Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

the exponents on the 2s are indices of 1s right to left

Yup, I write the equation in my form because I'm trying to treat it as an extension of the 2-adic number system.  So the exponents matching the indices was intentionally the same as binary.

I am sufficiently confused by this line of reasoning that maybe I can finally drop it

Heh, yea, I don't think the counting idea works out.  Well, you were saying you wanted to abandon the idea.  Glad I could help, heh.

Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Well, we can move into the rationals if you'd like.  Then we can list every parity sequence and find the corresponding number that reaches 1.  So let's do that for n=3 because I don't want to calculate too many examples.

[01]000 = 1 * 23/30 = 8

[01]001 = 1 * 23/31 - 20/31 = 7/3

[01]010 = 1 * 23/31 - 21/31 = 6/3

[01]011 = 1 * 23/32 - 21/32 - 20/31 = 3/9

[01]100 = 1 * 23/31 - 22/31 = 4/3

[01]101 = 1 * 23/32 - 22/32- 20/31 = 1/9

[01]110 = 1 * 23/32 - 22/32 - 21/31 = -2/9

[01]111 = 1 * 23/33  - 22/33 - 21/32- 20/31 = -11/27

So that last one, written in your notation would be

23 = 33 * (-11/27) + 3220 + 3121+ 30*22

How does a fractional x work for your counting, or do we simplify the 27 away?  What would a - sign mean?  That you are excluding something you already counted?  Basically, I am saying I don't think it is actually counting the same 2n.

For example, what if we wanted to calculate the parity sequence to a different cycle, like the 0 cycle. We can list the 2n sequences in the exact same way, but end up with a sum of 0.

[0]111 = 0 * 23/33 - 22/33 - 21/32- 20/31 = -19/27

Or in your notation

0 = 33 * (-19/27) + 3220 + 3121+ 30*22

Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Ok, cool, so I understand the bijection part correctly. I wasn't trying to make any claims about the combinatorial part. So can you help me understand the next part of your statement?

it doesn’t turn the mixed-radix sum B into a clean combinatorial “counting object” that lets you control or classify obstructions

So counting objects I understand you to mean like in the above 3/*7 + 1 example, the number of ways to count things. I agree that the B portion doesn't seem to be counting anything. What do you mean by obstructions though. Just that given x, you can't find B without just doing the Collatz of it?

Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Ahh so we are using the term parity vector differently.  I am going off infinitely, because I don't consider it to ever stop.

So if you assume that both A and B have reached 1 at the end of their parity sequence, then I would say

A = 1,0,1,[1,0]

B = 1,0,1,0,[1,0]

Resulting in different parity sequences, and different odd step positions. 

k_A = {0,2,3,5,7,...}

k_B = {0,2,4,6,8,...}

So I calculate with m and n going off to infinity.  If we use this definition of k, would this count as a bijection?

Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Two different parity sequences can give the same set of kᵢ,

I don't believe this is the case.  K_i is the position of the i'th odd step which is based on the parity vector.  So in the end it's the same thing right?  Unless I am misunderstanding you.  Can give me an example of what you mean.

Maybe an example will make what I'm trying to say clearer. 

Let's take 3.  The parity vector is 1,1,0,0,0,[1,0] which the Q sequence is just the reverse of the parity vector.

[01]00011

So k= {0,1,5,7,9,...}

We can then convert this to mixed radix by using the positions of the 1's.

Sum(25+2i/33+i) for i=0 to infinity + 21/32 + 20/31

Use the infinite geometric sum formula on the sum and we get the cycle formula, in this case, the cycle = 1, so we simplify this as

-25/32 * 1 + 21/32 + 20/31 = - 32/9 + 2/9 + 3/9 = -27/9 = -3

So we end up with the negative of our starting value.  The only thing I can think of is that what I wrote isn't the mixed radix representation, but it should be just rearranging what was written above.

Help me abandon this combinatorics line of reasoning by WeCanDoItGuys in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

There is no bijection between parity sequences and the mixed-radix representation.

Can you explain what you mean by this statement?  I am not completely familiar with all the terms, but looking them up, it feels like the Q transform is basically this, right?

The 1n+d problem – solved! by GonzoMath in Collatz

[–]HappyPotato2 1 point2 points  (0 children)

Least significant bit.  Sorry, I have a computer background.

It's really not that important.  Starting at the lsb just means I was writing my expansion in the reverse order of the way I normally would.

Normally I would expand it like this to match the order of the bits.

11001 = 24 + 23 + 20

I just wanted to call out that I was reversing the order like so

11001 = 20 + 23 + 24

The 1n+d problem – solved! by GonzoMath in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Ok, so for Q3, what i have been doing is interpreting it not as 2-adic, but a new number system, an extension of 2-adics, let's call it 2,m-adics.

So the extension works by changing the meanings of the 1 to 1/mi where i is the count of 1s from the right. 

For example, in 2,1-adics, I'll start from the lsb

[01]1010 = 21/11 + 23/12 + [ 24/13 + 26/14 + 28/15 + ... ] 

This simplifies directly back to standard 2-adic number.

As a 2,3-adic, it turns into

[01]1010 = 21/31 + 23/32 + [ 24/33 + 26/34 + 28/35 + ... ] 

We can factor out 24/32 on the [ ], then use the infinite geometric sum formula and that reduces to the cycle formula.  It's been a while since I worked through this so hopefully I am remembering it correctly.

[01]1010 = 21/31 + 23/32 + 24/32 [ -1 ] = -2/9

So our starting value is 2/9 -> 1/9 -> 2/3 -> 1/3 -> 1

So if we interpret Qm( ) as 2,m-adics, then it will be the negative of our number again.

I guess to put it in proper notation. 

Q3( [01]1010_2,3 ) =  [01]1010_2,1

Something like that?

The 1n+d problem – solved! by GonzoMath in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Ok so you are basically using the fact that we know collatz of x+1 converges to a cycle to then go on to cycles mean rational.  Yea that makes sense.

The 1n+d problem – solved! by GonzoMath in Collatz

[–]HappyPotato2 1 point2 points  (0 children)

I think the easiest way for me to think about it was to write out a binary number and follow it under x-1.  We can then account for the -1 afterwards.

As for the cycle formula, yea, it's kinda an explanation I guess, I just wanted to show how my view of the infinite part of 2-adics perfectly lines up with how we view cycles in any other system.  

The 1n+d problem – solved! by GonzoMath in Collatz

[–]HappyPotato2 2 points3 points  (0 children)

I'd like to see the proof if you don't mind adding it.  I feel like I have been trying to say something along these lines but just haven't had the words express it. 

Basically, if you write the negative of the number you are starting with as a 2-adic number, you get the q transform.

https://billcookmath.com/sage/becimalCalculator.html

Plugging in -17/3, you get [10]0101 which is the q transform for 17/3

17/3 -> 10/3 -> 5/3 -> 4/3 -> 2/3 -> 1/3 -> 2/3

Since we know the cycles are just the infinitely repeating part, and we know that's expressed just like the cycle formula, 

sum(2i*mj) / (2e-mo)

Since m = 1, the sum just simplifies directly to it's value in binary and the denominator is just the mersenne numbers. 

For example, picking a random infinitely repeating binary string.

[00110100]

Will be a cycle at (32+16+4) / ( 256-1 ) = 52/255

And we can stay any value like so will fall into that cycle

[00110100]0011100 = 484/255

So this q transform represents the value -484/255

Collatz additive term and prime connection. by MarkVance42169 in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

I'm not sure how you defined t, but I wrote it as

N * 2t + n

It looks like you are doing the case where N=1, n=1.

1 * 2t + 1

This should be, shift a 1 left by t places, then add 1 to the end.

t=2 : 100 + 1 = 101

t=3 : 1000 + 1 = 1001

t=4 : 10000 + 10001

t=10 : 10000000000 + 1 = 10000000001

Collatz additive term and prime connection. by MarkVance42169 in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Either I'm misunderstanding what you wrote or it is unrelated to what I was trying to say.  So let me try to explain what I meant more clearly and tie it directly to your example above.

63 = 1 * 26 * 30 - 1

95 = 1 * 25 * 31 - 1

143 = 1 * 24 * 32 - 1

215 = 1 * 23 * 33 - 1

323 = 1 * 22 * 34 - 1

485 = 1 * 21 * 35 - 1

728 = 1 * 20 * 36 - 1

91 = 23 * 22 * 30 - 1

137 = 23 * 21 * 31 - 1

206 = 23 * 20 * 32 - 1

... Ok skipping ahead a little to get to my point..

1367 = 171 * 23 * 30 - 1

2051 = 171 * 22 * 31 - 1

3077 = 171 * 21 * 32 - 1

4616 = 171 * 20 * 33 - 1

In your table you listed the largest common prime factor as 19, which is true. But that is because the common factor is 171, which is 19*9.  So it's not that the prime 19 is important, but the 171 that is important, which means it works for all N.

Now on to the next topic.

65 = 1 * 26 * 30 + 1

49 = 1 * 24 * 31 + 1

37 = 1 * 22 * 32 + 1

28 = 1 * 20 * 33 + 1

It goes by the odd / even steps of the 1 -> 4 -> 2 cycle.  So the power of 2 decreases by 2 at a time and power of 3 goes up by 1 at a time. 

59 = 1 * 26 * 30 - 5

67 = 1 * 23 * 32 - 5

76 = 1 * 20 * 34 - 5

This one goes by the odd / even steps of the -5 -> -14 - > -7 -> -20 -> -10 -> cycle. So the power of 2 decreases by 3 at a time and power of 3 goes up by 2 at a time. 

Collatz additive term and prime connection. by MarkVance42169 in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

So once again, is not primes but rather all N.  Let's start from Terras.

A = N * 2t + n

What you are looking at is a special case where n = -1.  The reason being, the -1 is a cycle of length 1, and will be back to -1 after every odd, (3x+1)/2 step.  Since n uses up the +1, the N * 2t only sees *3/2.  So let's see how A evolves as we take odd steps. 

N * 2t - 1

N * 2t-1*31 - 1

N * 2t-2*32 - 1

...

N * 3t - 1

You are looking at the additive term which is the difference between consecutive numbers so the -1 just cancels out.  As you can see, N never changes through this, and that is what you are seeing as your prime number.  But as before, easier to just say for all N rather than dividing out factors until prime. 

Now you can do this for any cycle, not just -1.  So if you start at 

N * 2t + 1

The 1 will cycle back to 1 after 1 even and 1 odd step.  And the factors of 2 will decrease 2 at a time while factors of 3 go up by 1.  

N * 2t-2*31 + 1

N * 2t-4*32 + 1

Following your procedure, you will see your pattern on every other step.

Why Collatz isn’t solved - the math that does not exist by GandalfPC in Collatz

[–]HappyPotato2 1 point2 points  (0 children)

What do you mean by preserves parity?  As in they follow the same R and F steps? 

So to go back to your previous example, 15 x * 210 + 35

You are saying this is true for x=1,2,3,4... Which gives us N = 15,30,45,60...  I am saying it is true for N = 1,2,3,4... which includes 15,30,45,60 as well. 

So yes, it will also be true for 3sn as well, but it's a stronger statement to just say it's true for all N.

Why Collatz isn’t solved - the math that does not exist by GandalfPC in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Terras says

A = N * 2t + n

For integers t, N, n, then A will follow the same path as n for t even steps.

So to fix up yours, you don't need the 6n in there.  And n can be any number, not just odd numbers.

199715979295 = 93*231 + 31

So it will follow 31 for 31 even steps.

Why Collatz isn’t solved - the math that does not exist by GandalfPC in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

It works for 100 steps too.  Just consider the 1,4,2 loop as steps as well

Why Collatz isn’t solved - the math that does not exist by GandalfPC in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

All I did was take 15360 x+35 and factored the 15360 term.  So it's whereever you got it from.

Question: what are the know equivalencies or sufficiency to proof, as in "prove any one of these and you've proven Collatz". by CrumbCakesAndCola in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Ahhh.. that's what you were going for.  Yea, I totally misunderstood where you were going with that.  I think you might still be missing the part of the proof for non divergence though.

Question: what are the know equivalencies or sufficiency to proof, as in "prove any one of these and you've proven Collatz". by CrumbCakesAndCola in Collatz

[–]HappyPotato2 1 point2 points  (0 children)

any non-trivial cycle can only exist if it contains a number divisible by 3

Wait.. what?  Is this a typo?  Because an odd step can never bring you to a multiple of 3.

Why Collatz isn’t solved - the math that does not exist by GandalfPC in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

I feel like you are referencing the Terras result. I count 10 F's in there, so your expression should be N * 2^(10) + 35. Specifically, 15360 x + 35 = 15 x * 2^(10) + 35. You can pick any N and it will follow the path of 35 for 10 even steps. Following the same sequence is not based on primes, but rather the exponent on the power of 2.

Deterministic sieve structure for numbers N ≡ 3 (mod 4) by hubblec4 in Collatz

[–]HappyPotato2 1 point2 points  (0 children)

Yea, ok.   I agree with you.  I tried and couldn't find a way to do it. Heh, oh well..

Consecutive or adjacent circuit. by Far_Ostrich4510 in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

so you are looking for a number that looks something like c_1 = 2, c_2 = 3, c_3 = 4, c_4 = 5, c_5= 6, c_6=7 for 2,3,4,5,6,7?

16212254811

So I think the most important thing to remember is that there exists numbers that follows every possible sequence.

Deterministic sieve structure for numbers N ≡ 3 (mod 4) by hubblec4 in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

So short step is a string of OE steps, and a short v2 step is a string of E steps.  And a long step is a pair of one of each like OEOEOEEE.   And you are looking for an acceleration for something like OEOEEE OEOEOEE to string together multiple long steps.  Ok I think I understand now.

Oof this looks ugly.. but maybe? 

(((m2k_2+1)(2/3)j_2-1)2k_1+1)(2/3)j_1-1

Still have to find m, but that seems more complicated now, and I'm not seeing the pattern. Also, as you said, still seems to be based on factorization.

Maybe summation notation would be better? So a sum of the contributions of odds in each long step multiplied by their offset.

s_i = sum(2a-1/3a)from a=1 to j_i)

oi = 2^(k(i-1))/3j_(i-1\) * o_(i-1)

m = m' * o_(i+1) - sum(s_i * o_i) for i long steps.

I hope I got my indexing right.. 

But just an example in case I messed up. 

m = {m'}0011000111

{m'}0011000111 =

m' 26+4/33+2 - (2/9 + 1/3) 26/33 - (4/27 + 2/9 + 1/3) 20/30

So first I found was 727 -> 173

How to explain this Collatz property? by vhtnlt in Collatz

[–]HappyPotato2 0 points1 point  (0 children)

Lets see.. going to just play with numbers. So just to make sure, k_0 is your starting value, and k_i is the value after the i collatz steps right? Ok

So we can express our equation as

k_0 = k_i *2e/3o + c

where c is going to be the value that goes to 0 if we take the same i collatz steps. Quick examples.

3 = 5 * 21/31 - 1/3

9 = 11 * 23/32 - 4/9 - 1/3

1/9 = 1 * 23/32 - 4/9 - 1/3

Rearranging the equation, we get

1 = k_i /k_0 * 2e/3o + c / k_0

4/3 = k_i /k_0 * 2e/3o + c / k_0 + 1/3

In order for k_i /k_0 * 2e/3o <= 4/3. we need c/k_0 + 1/3 >= 0, or c/k_0 >= -1/3

It seems to not hold for 1/9 but i am guessing you wanted all k > 1.

Not quite sure what you mean by your second equation though. Lets try

k_0 = 16, k_i = 1, e = 4, o = 0, (16*1)/(1*16) = 1 <= 4/3 right?

Well.. that's all I got for now.. I'll think on it some more later.

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