User talk:Mugbuff

March 2009
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Alan Tyte's "proof"
May I provoke an even more extreme response? Rather than Collatz's conjecture, here is Tyte's conjecture: Take any non-negative integer, multiply it by 3 and add 1. If the result is even divide by 2 until an odd integer is reached. Repeat and eventually the sequence will converge to 1. If this conjecture is true, then Collatz is true, as it is just a sub-set of Tyte's conjecture. Whereas Collatz fails for one even number, i.e. 0, Tyte applies to all non-negative even and odd integers including zero. To avoid the more extreme anger at such presumption, I am no longer claiming a proof but only a structure that is worth very serious consideration as strongly suggesting the validity of Collatz due to the probability of showing that all integers belong to the same graph. You will shoot me down completely if you give me an integer for which it is not possible to determine its n-chain position. I do not understand what 3n+7 has got to do with 3n+1. 3n+1 is unique in that it has a spine of 0's providing the pivots for the one n-chain at each Ln. 'My' conclusion that Collatz is true was not based on the existence of an n-chain structure but on the formats required to generate pivots. For more than one graph to exist it must be possible to construct a spine of pivots which allow looping. There are only six equations used when creating a spine and as the length of the spine increases the possibility of a loop appears to become more and more remote because of the general trend for values to increase as you step up. In support of this, where there are multiple graphs, loops are soon found (as for 3n+65 or 3n-1). Mugbuff (talk) 19:56, 25 March 2009 (UTC)

No-one has commented on my paper for a year,


 * Which means what, exactly? That your proof is valid? Non Sequitur.

so I have removed the previous discussion.


 * I did not give you permission to delete my comments.

Forget the word 'proof' which seems to drive everyone mad. I am still utterly convinced that I have described a valid structure which allows the allocation of every positive integer which converges to 1 to a unique position on the graph.


 * You have not shown that those unique positions are connected. All the integers in 3n+7 occupy unique positions on the 3n+7 graph. But 3n+7 fails the conjecture, so the truth of the conjecture does not follow from unique positions.

Whether this 'proves' Collatz is open to debate


 * Not really. As I show in the part you deleted (a portion of which I will be restoring), it's a Non Sequitur Fallacy, the debate is settled.

but I have seen no other approach that delivers an equivalent comprehensive structure.


 * Check out Ken Conrow's site. His doesn't work either, but his Left Descent Assemblies are such a structure.

For example, 27 is a small value that takes a comparatively large number of steps, namely 41, to converge to 1. It is possible to show that, using my definition of n-chains, 27 is the second term in a 29-chain whose pivot, 59339, is the second term of a 33-chain whose pivot, 3911993580, is the second term of a 40-chain whose pivot, 3569835538891, is the second term of the 41-chain at L41 whose pivot is 0. Note the presence of two even pivots. Since 27 is a link in a 29-chain, it is also the pivot of a 28-chain, 27-chain, and so on down to a 1-chain. For example the 2-chain would start 27 220 1763, 220 would be the pivot of a 1-chain starting 220 881. 1763 and 881 both converge to 1 in 41 steps. I would welcome more debate,


 * Why? You have not refuted my objections.

Because I do not consider tham valid. Mugbuff (talk) 20:38, 25 March 2009 (UTC)

but I cannot accept any argument that starts by critising the integration of even integers into the structure. (a
 * Show me someone who accepts such nonsense.

Until the significance of incorporating even numbers into the graph is appreciated, I am afraid that it is pointless to discuss the matter with you, we must just agree to differ. I can only continue to reiterate that it is their inclusion which allows the construction and comprehensiveness of a valid structure. I hope by now that it is clear that I think the significant parts of a single step is its start value (odd or even) and its final value (always odd), the intermediate even values do not appear on the graph. Successive values on an n-chain result from the required division by increasing powers of 2. The step up path from any even integer or any multiple of 3 in the graph, follows the same path as the odd integer one or two terms along the 1-chain containing such an integer. However such integers are also the pivots for higher chains which would be missed if such integers were not incorporated into the graph. The ultimate example is that 0 is the pivot for one n-chain in the infinite number of n-chains that exist for every value of n. Mugbuff (talk) 08:57, 27 March 2009 (UTC)

I repeat that no odd integer is treated in any way other than by using the standard Collatz algorithm.


 * Uh, that's not the issue, is it? Can you say the same for even numbers? No, you can't.

I don't want to. I think Collatz's handling of even integers is trivial. In effect his handling sometimes starts part way through a step down from some odd integer as in the sequence 16 8 4 2 1 which is just part of the step down from 5, or the sequence 56 28 14 7 which is part of the step down from 37, and similarly for other even numbers; for numbers of format k.3^m.2^n (k prime to 2 and 3, n>0) Collatz just reduces them to k.3^m. Mugbuff (talk) 20:38, 25 March 2009 (UTC)

It is never required to step down to an even integer or to an oddd integer that is a multiple of 3. Omitting even integers precludes the production of a definable structure.


 * Which means said structure is invalid. --Mensanator (talk) 05:15, 25 March 2009 (UTC)

Why ? because I do nothing wrong ? Mugbuff (talk) 20:38, 25 March 2009 (UTC) Mugbuff (talk) 22:03, 24 March 2009 (UTC)




 * First of all, of the pivots you claim comprise these n-chains, some of them do not exist on the Collatz graph as you created them by an inconsistant application of the 3n+1 rule. Secondly, you assume that every 6-level step down converges to the same number (which is only true if all the graph componts are connected). Thus, you are making the truth of the Collatz Conjecture an axiom of your proof of the Collatz Conjecture.

Every 6-chain consists of integers which step down to the same integer after 6 steps, that is what a 6-chain is. Above every node is a 6-chain that steps down to that node. There are an infinite number of 6-chains each one stepping down to a unique node.Mugbuff (talk) 20:38, 25 March 2009 (UTC) 1/4 of the positive integers have format 4i+1 (where i    is a positive integer) and are therefore at the same level as i.


 * For example, for i in {0,1,2,3,4,5,6,7} we can see how this holds locally on the graph...

i=0 5__16   i=2 37__112    i=3 53__160    i=4 64__208 =1    8             56             80            104       1___4         9___28        13___40        17___52           2             14             20             26       0___1         2____7         3___10         4___13   i=5 85__256  i=6 101__304   i=7 117__352 128          152            176       21___64       25___76        29___88            32            38             44        5___16        6___19         7___22


 * ...and how it holds globally (* elements of loop cycle).

101_304            152          24__76   37_112     38     117_352       56   6_19_58      176    9__28        29_______88       14                 44    2___7_________________22    69_208                          11_34    104                             17_____52 53_160                                    26     80 85_256                                    13_____40    128                                           20 21__64                                        3__10     32                                            5_____16                                                   8                                                1*_4*                                                   2*                                                 0_1*


 * All well and good. But a question remains to be answered: is 3n+1 true because this works or does it work because 3n+1 is true?

At this point we are not talking about 3n+1 but the fact that, on whatever graph, all 4i+1 step down to the same level as i using 3n+1 Mugbuff (talk) 20:59, 25 March 2009 (UTC)
 * We can answer that by repeating the example with 3n+C. Keeping in mind that in 3n+C, the "4p+1" rule becomes "4p+C" and that the trivial loop cycle is C-4C-2C-C instead of 1-4-2-1.


 * Using 3n+7, it is still ture locally...

i=0 35_112   i=1 51__160    i=2 67__208    i=3 83__256 =7    56             80            104            128        7__28        11___40        15___52        19___64           14             20             26             32        0___7         1___10         2___13         3___16   i=4 99__304  i=5 115__352   i=6 131__400 152          176            200       23___76       27___88        31__100            38            44             50        4___19        5___22         6___25


 * ...BUT NOT GLOBALLY!!

131_400      200    31_100        50     6__25_82           41_130               65_202                  101_310                      155_472           67_208         236              104         118           15__52          59_184               26              92 99_304            2__13______________46    152                               23_____76 83_256                                      38    128                                   4__19_____64   147_448                                   32       224                                3__16       112                                    8     115_352        56                                    4         176    7*__28*                                   2      27__88        14*                                   1_10*      44  51_160    0____7*                                      5*______22*     80                                                         11*_____40*                                                                 20*                                                                 10*


 * Because the graph components are not connected (there are two distinct loop cycles, the trivial loop 7-28-14-7 and the counterexample 10-5-22-11-40-20-10).


 * So we see that although the premise "all 4i+1 are on the same level as i" is true, the 3n+7 example demonstrates that "connected" 'connected'does not appear in my paper, I do not follow your argument. As a matter of interest a value for c in your commenet that gives a lot of separate graphs is 65, I found 16 separate graphs for 3x+65; the largest least value in a graph was 2223. I am quite confident that my nchain struture is true within each graph. Mugbuff (talk) 19:53, 5 April 2008 (UTC)


 * You agree that whenever you have separate graphs the conjecture is false? So, obviously 3n+65 fails the conjecture? Yet, as you say, the n-chains are still true. So, the truth of the conjecture doesn't follow from the truth of the n-chains which are true in 3n+65 even though the conjecture is false for 3n+65. So, n-chains being true in 3n+1 doesn't prove that 3n+1 is true. Do you follow that argument? --Mensanator (talk) 00:08, 8 April 2008 (UTC)

No.


 * Ok, what part of "3n+65 fails the conjecture" don't you understand? Do you not understand that multiple graphs means the conjecture is false? Do you not understand that 3n+1 must be proved to have exactly one graph in the positive domain? Do you not understand the concept "necessary but not sufficient"? That means it must be true but by itself doesn't prove the conjecture. If all you've done is shown this n-chain structure is true in 3n+1, then your proof is wanting because, as you say above, the n-chain structure is also exhibited by 3n+65 which we know fails the conjecture because there are at least 16 unconnected graphs. There must be something else that distinguishes the 3n+1 case from the 3n+65 case. That's why I say it is a Non Sequitur, the truth of the conjecture doesn't follow from n-chains, otherwise 3n+65 would be true also. I have no idea what's missing, but it suffices to show that n-chains still work despite the conjecture being false to know that your "proof" is incomplete.

Consider the pivot spines for 3n+65 and compare them to the one known simple pivot spine for 3n+1 Mugbuff (talk) 20:59, 25 March 2009 (UTC)




 * Collatz strings
 * Any number must have one of the 6 formats 3e, 3e+1, 3e+2, 3o, 3o+1or 3o+2
 * where e is an even number and o is an odd number.
 * Above every number is an infinite string of numbers that converge to that number,
 * or to a number at the same level, after one 3x+1 cycle.
 * The string is calculated from 4p+1 where p is the previous term.
 * The first number in such a string is calculated :
 * for 3e      it is  4e
 * for 3e+1  it is  e
 * for 3e+2 it is  16e+12
 * for 3o     it is  4o
 * for 3o+1 it is  8o+3
 * for 3o+2 it is  2o+1
 * Examples
 * 102 = 3 x 34          the string is 136  545  2181 etc, each term converges to  409
 * 409 is second term of the string starting with 102
 * 102 409 1637 etc,  converge to  307
 * 103 = 3 x 34 + 1   the string is   34  137   549  etc,  each term converges to  103
 * 103 413  1653 etc,   converge to  155
 * 104 = 3 x 34 + 2   the string is  556 2225  8901 etc,  each term converges to  1669
 * 1669 is third term of string starting with 104
 * 104 417  1669 etc,  converge to  313
 * 105 = 3 x 35          the string is  140  561  2245 etc, each term converges to  421
 * 421 is third term of string starting  26  105
 * 26 105 421  etc,  converge to  79
 * 106 = 3 x 35 +1    the string is  283  1133  4533 etc, each term converges to  425
 * 425 is second term of string starting with 106
 * 106 425 1701  etc,  converge to 319
 * 107 = 3 x 35 +2    the string is  71  285  1141 etc, each term converges to  107
 * 107 is start of a string
 * 107 429 1717  etc,  converge to 161
 * String facts
 * 1.     Every number is part of a string
 * 2.     Every even number, 2n, is the start of a string
 * 3.     Every odd number of format 4n+3 is the start of a string
 * 4.     After the start value every term in the string has format 4n+1
 * 5.     Above every 6n+1  is a string starting with  2n
 * 6.     Above every 6n+5  is a string starting with  4n+3
 * 7.     Every string cycles through formats  6n+1, 6n+5, 6n+3, 6n+1, etc
 * 8.     Above 2/3 of the terms in a string is another string
 * Conclusion
 * The above applies to every positive integer from 0 to infinity, hence no number exists that
 * Is not part of an infinite structure of strings.
 * It is a fact that the start values of strings form a definite pattern.
 * Pattern available but needs a lot of space to fully explain and demonstrate.
 * The pattern includes all positive integers.
 * If all integers converge to 1 then so do all odd.
 * (Above every 6n+1 and 6n+5 in the numbers tested up to 268 is an infinite string) Mugbuff (talk) 12:27, 3 August 2022 (UTC)