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 2018-03-02, 13:34 #1 Steve One   Feb 2018 1001002 Posts Collatz Conjecture Proof This does not make any mistake or omission that you think it will! 1. The solution lies in correct 'tumbler' allignment, as in cracking a safe. 2. The correct allignment is (4 + 10n). 3. As can be seen in the truncated section below, within 4 operations of 3n + 1 rules, every number, odd and even, (1 + n5) (2 + n5) (3 + n5) (4 + n5) and (5 + n10), is 'hit' and passed through on the way hopefully, to 1. (5 + 10n) occurs down 3rd column from (16 + n30). (16 - 1)/3 = 5. (45 - 1)/3 = 15 etc. Remember, it is only necessary to KNOW 5 15 25 35 45 go down to 1. (5 15 25 35) etc occur elsewhere but l wanted them within the truncated space so as to have 'proof' right up front. No rule is broken. 4 2 1 14 7 24 12 6 3 34 17 44 22 11 54 27 64 32 16 8 [[[[[[[[[4]]]]] 74 37 84 42 21 94 47 104 52 26 13 114 57 124 62 31 134 67 144 72 36 18 [[[[[[[9] 154 77 164 82 41 174 87 184 92 46 23 194 97 204 102 51 214 107 224 112 56 28 [[[[[14]]]]] 4. Therefore, if we prove every member of (4 + 10n) goes down to one, then ALL numbers go down to 1 under 3n + 1 rules. Period! 5. If we saw; A B A C B A D C B A , nobody would question that if we KNOW 'A' 'works', then we KNOW all work. There is no room figuratively or visually to doubt. 6. If we saw; A B C A D E B A F G C B A, after a little hesitation, acceptance follows that if we KNOW A 'and' B work, then we KNOW all work. Because of B's location next to (in rows), A in next column. The chart above, knowing 4 to 54 go down to 1 provides the same needed information as knowing A and B works here. This again, negates the necessity for any further checking beyond B's equivalent in the Collatz diagram, 54. 7. Knowing anything further than A and B working serves no purpose. Knowing anything further than 54 going down to 1 serves no purpose either, on the Collatz chart. Just look where your 'proved 4 to 54' from column 1 reside right of, when viewed from the second column? 64, 224, 384, 544, 704, 864, in the first column!. Just 4, 14, 24, 34, 44 and 54, each multiplied by 16. Four spaces backwards from each 4, 14, 24, 34, 44 and 54 respectively. Therefore the process is complete because ANY 'therefore only...?...needs to be proved' is/are all multiples of 4 steps (doubling) backwards from either 4, 14, 24, 34, 44 or 54. 8. How about?... A B C D E F G A H I J K L M N O P Q R S T U V W B X Y etc(downwards and outwards. Next A, beyond seen here, will be in column 3, and same as 4, will be eight to the right of Z + 77 rows (102 rows down from A, column 1. What would be 1024 (column1) for 4 (column 9) on Collatz set up above. . If you know A to F 'work', do you know everything works? Yes! This is just a stretching of the previous 2. F, being the last 'known' serves the same function as B in previous and A in previous to that. THIS is EXACTLY THE SAME AS the (4 + 10n) Collatz diagram above. Just change A - F to 4 - 54 and W to 224 for the all information needed. These are the only numbers we need to KNOW go down to 1. 9. As long as 'known information' (4 14 24 34 44 54 working) has its last known term (54) either one row above the first term of 'known information' (4), in next column, or anywhere downwards (between 4 and 54, in next column) ""AND"" the subsequent separations of 4 ,14, 24 etc have a common factor increase all is good! In this case it is 16. Column 1 separation between 4, 14, 24 etc terms is 1. Column 5 separation between 4, 14, 24 etc terms is 16. Column 9 separation of terms 4, 14, 24 etc is 16 x 16. Column 13 is (16 × 16 × 16) etc. Every 4 columns, × 16 for separation of terms. 9. All (4 + 10n) go down to 1 and every odd number is hit on the way. PROBLEM SOLVED! 10. Just look how disorganised it looks, lining up from 1 3 5 7 9 11 etc. Can't see the wood for the trees. 11. I know another way but this one is easier.
2018-03-02, 15:57   #2
CRGreathouse

Aug 2006

10111011000112 Posts

Quote:
 Originally Posted by Steve One 3. As can be seen in the truncated section below, within 4 operations of 3n + 1 rules, every number, odd and even, (1 + n5) (2 + n5) (3 + n5) (4 + n5) and (5 + n10), is 'hit' and passed through on the way hopefully, to 1. (5 + 10n) occurs down 3rd column from (16 + n30). (16 - 1)/3 = 5. (45 - 1)/3 = 15 etc. Remember, it is only necessary to KNOW 5 15 25 35 45 go down to 1. (5 15 25 35) etc occur elsewhere but l wanted them within the truncated space so as to have 'proof' right up front. No rule is broken. 4 2 1 14 7 24 12 6 3 34 17 44 22 11 54 27 64 32 16 8 [[[[[[[[[4]]]]] 74 37 84 42 21 94 47 104 52 26 13 114 57 124 62 31 134 67 144 72 36 18 [[[[[[[9] 154 77 164 82 41 174 87 184 92 46 23 194 97 204 102 51 214 107 224 112 56 28 [[[[[14]]]]]
I don't see how this proves that all numbers which are 0, 1, 2, 3, 5, 6, 7, 8, or 9 mod 10 end in 1. Would you make this more explicit, please?

2018-03-02, 16:15   #3
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by science_man_88 My question is what does this say about Goldbach's conjecture etc. They can be related partially at very least. What implications will your "proof" of the Collatz conjecture say about other conjectures.
Let's avoid leading him further astray...

2018-03-02, 17:50   #4
Steve One

Feb 2018

22×32 Posts

Quote:
 Originally Posted by CRGreathouse I don't see how this proves that all numbers which are 0, 1, 2, 3, 5, 6, 7, 8, or 9 mod 10 end in 1. Would you make this more explicit, please?
Within 4 operations of (3n+1) rules, although in this formation only halving occurs up to that point, every number eg(1+n5)3rd column, (2+n5)2nd column, (3+n5)4th column, (4+n5)5th column, and (5+n10)3rd column from 16, 46, 76 etc is hit, before hopefully going down to 1.
Therefore if all 4+n10 go down to one then all numbers go down to 1.
Pointless saying anything else until this is accepted.

2018-03-02, 17:56   #5
Steve One

Feb 2018

22×32 Posts

Quote:
 Originally Posted by CRGreathouse Let's avoid leading him further astray...
Would you like to see how to COUNT Goldbach prime pairs, not just prove they go on forever?
Sadly though, it will show all the numbers necessary to prove it, with a correct answer to how many there are up to a certain number. SO: CORRECT NUMBERS, CORRECT ANSWER, BUT YOU WILL STILL DISBELIEVE IT. BECAUSE YOU """WANT""" TO. I can't change that. It's not my weakness.

2018-03-02, 18:39   #6
CRGreathouse

Aug 2006

598710 Posts

Quote:
 Originally Posted by Steve One Would you like to see how to COUNT Goldbach prime pairs, not just prove they go on forever? Sadly though, it will show all the numbers necessary to prove it, with a correct answer to how many there are up to a certain number. SO: CORRECT NUMBERS, CORRECT ANSWER, BUT YOU WILL STILL DISBELIEVE IT. BECAUSE YOU """WANT""" TO. I can't change that. It's not my weakness.
I'd like to avoid talking about the Goldbach conjecture here -- please stay on topic. You can start another thread (if you don't have one already) for Goldbach.

2018-03-02, 18:52   #7
CRGreathouse

Aug 2006

135438 Posts

Quote:
 Originally Posted by Steve One Within 4 operations of (3n+1) rules, although in this formation only halving occurs up to that point, every number eg(1+n5)3rd column, (2+n5)2nd column, (3+n5)4th column, (4+n5)5th column, and (5+n10)3rd column from 16, 46, 76 etc is hit, before hopefully going down to 1. Therefore if all 4+n10 go down to one then all numbers go down to 1. Pointless saying anything else until this is accepted.
You should clarify what you mean by an operation of the 3n+1 rules (is that A006370, A139391, A075677, or something else?), (2+n5) and the like (does this mean the integers which are 2 mod 5?), the columns, a "hit", etc.

Perhaps you should explain how, for example, 13179928405231 is "hit" within 4 "operations". That might help explain what you mean.

 2018-03-02, 19:44 #8 10metreh     Nov 2008 2×33×43 Posts Am I correct in thinking you're saying that knowing 4, 14, 24, 34, 44 and 54 go down to 1 proves everything goes down to 1? If so, how does your proof show that 13 goes down to 1?
 2018-03-02, 19:44 #9 Collag3n   Feb 2018 23 Posts All you do here is eventually show that every even number end-up in an odd number (althought we don't know what you do with $5n$ numbers): you just build segments of even numbers which later will correspond to 1 full branch (all even attached to 1 odd number) in the collatz tree (where children of an odd number $n$ are $n\cdot 2^i$) e.g. with $n=7$: you first have $14=7\cdot 2^1$, then later you add another segment of the branch $224=7\cdot 2^5$, $112=7\cdot 2^4$ ...down to $14=7\cdot 2^1$, then later you add the segment $3584=7\cdot 2^9$ down to $224$....which in the end will give you the full $7$ branch (it is only 1 odd in the tree !!!) Of course they all end-up in a $5n+x$ form. You probably can show that they end-up on every odd numbers (well...don't we already know that even numbers goes down to an odd?) I don't see how knowing that (line 4 to 54) or (odd numbers 1,7,3,17,11,27) tell you anything about $37$ (next odd in your building). Even if you could show that all odds are covered, I don't see how you show that there is no cycles somewhere.
2018-03-03, 11:19   #10
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

203428 Posts

Quote:
 Originally Posted by Collag3n All you do here is eventually show that every even number end-up in an odd number (althought we don't know what you do with $5n$ numbers): you just build segments of even numbers which later will correspond to 1 full branch (all even attached to 1 odd number) in the collatz tree (where children of an odd number $n$ are $n\cdot 2^i$) e.g. with $n=7$: you first have $14=7\cdot 2^1$, then later you add another segment of the branch $224=7\cdot 2^5$, $112=7\cdot 2^4$ ...down to $14=7\cdot 2^1$, then later you add the segment $3584=7\cdot 2^9$ down to $224$....which in the end will give you the full $7$ branch (it is only 1 odd in the tree !!!) Of course they all end-up in a $5n+x$ form. You probably can show that they end-up on every odd numbers (well...don't we already know that even numbers goes down to an odd?) I don't see how knowing that (line 4 to 54) or (odd numbers 1,7,3,17,11,27) tell you anything about $37$ (next odd in your building). Even if you could show that all odds are covered, I don't see how you show that there is no cycles somewhere.
His claim seems weaker, than a claim of all odd numbers go to one. This is because 4n+10 only covers 4y+2 greater than 8 with n positive so it misses the line with 6 on it.

2018-03-03, 13:53   #11
Steve One

Feb 2018

22·32 Posts

Quote:
 Originally Posted by science_man_88 His claim seems weaker, than a claim of all odd numbers go to one. This is because 4n+10 only covers 4y+2 greater than 8 with n positive so it misses the line with 6 on it.
You epitomise 'lack of reading'. 4+10n, not 4n+10

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