August 28th, 2018, 04:18 PM  #1 
Newbie Joined: Aug 2018 From: Spain Posts: 4 Thanks: 0  Collatz Conjecture
Hello, I am from Spain and I am asking in some forums for my ideas about Collatz Conjecture, the next text are translated by google, and I expect that you can understand me. Thank you. DRAFT: IN SEARCH OF THE DEMONSTRATION OF THE COLLATZ CONJECT I apologize if the ideas are not very clear and if the notation or words is not of the scientific rigor that in this forum should have. Comment that this document is an attempt to demonstrate the Collatz conjecture with another vision that I found on the internet. Everything has emerged as a cluster of ideas that have been given in my head, therefore, maybe it is wrong and do not make sense or believe that it is a good starting point and can be a demonstration mode. We start from the fact that given a positive integer N we make it grow if it is odd (3N + 1), and decrease if N is even (N / 2). The conjecture says that any number ends up with a value of 1. More information about the conjecture in: https://es.wikipedia.org/wiki/Conject_of_Collatz To try to prove the birth conjecture that when performing the iterations (in a limit vision) the decrease when it takes even numbers is greater than the increase when it takes odd values, with this I try to show that it is a decreasing function. Once this has been demonstrated, it would be demonstrated that it "converges" to cycle 1,4,2. We start from the base of the following formula that compares the incremental force: *3N + 1≤2 ^ K N ∀K> 1 * Note that equality is only fulfilled if N = 1 for K = 2. This means that, if we increase a positive integer with 3N + 1 it will be less than if we decrease it N / 2 ^ K for K> 1 (this reduction is what happens in the function if the number is even until it reaches to be another odd one). Therefore, if we show that the amount of a number that increases 3N + 1 is less than the amount of decreasing N / 2 ^ K (in absolute value) in infinite iterations, it would be shown that this function is decreasing. Therefore, we will study in what quantity a number increases or decreases as a function of K, and then the distribution and probability of K in the even numbers to show that the pairs will decrease in greater quantity than the odd ones. Given a number N we have: ***A = 3N + 1 ***B = 2 ^ K N From here you have to: AB = (2 ^ K3) N + 1 That is, if K = 1 the number would increase N + 1 more units (A) than if we do it by (B). For example, for k = 1: *If N = 1 > A = 4 and B = 2 (the difference is N + 1 = 2). *If N = 2 > A = 7 and B = 4 (the difference is N + 1 = 3). If we now make k = 2 the number would decrease N +1: If N = 1 > A = 4 and B = 4 (the difference is N + 1 = 0). If N = 2 > A = 7 and B = 8 (the difference is N + 1 = 1). If we now do K = 3 then it would decrease 5N + 1: If N = 1 > A = 4 and B = 8 (the difference is 5N + 1 = 4). If N = 2 > A = 7 and B = 16 (the difference is 5N + 1 = 9). Now, given an even number, what is the probability that K is greater than or equal to 1 (in N / 2 ^ K)? This is interesting to know to detect if the iteration in our function represents more force than the iteration. of increase 3N + 1. To do this, we must study the distribution of even numbers, more specifically the number of times that is divisible by 2, in other words, the distribution of K in even numbers. Let's see the even numbers up to 90: N K N K N K 2 1 32 4 62 1 4 2 34 1 64 5 6 1 36 2 66 1 8 2 38 1 68 2 10 1 40 3 70 1 12 2 42 1 72 3 14 1 44 2 74 1 16 3 46 1 76 2 18 1 48 4 78 1 20 2 50 1 80 4 22 1 52 2 82 1 24 3 54 1 84 2 26 1 56 3 86 1 28 2 58 1 88 3 30 1 60 2 90 1 We observe that K oscillates between 1 and a number greater than 1, that is, the probability of k = 1 is 50% and that K> 1 of 50%. I do not know whether there is a function of this distribution, which would make the calculations more accurate, and may be the subject of my study in the future, however, we observe a cyclic behavior of k of the form 1,2,1, X that repeat from par 10, with X greater than or equal to 3. Therefore, and in order to show that the function is decreasing and putting us in the worst case it would be that the cycle was 1,2,1,3 the theoretical increase with respect to an increase of 3N + 1 each time it goes out odd would: + N + 1 N + 1 + N + 1 5N + 1 =  4N + 4 * Note that the forces are identical only if N = 1. Therefore, by probability, at infinity we are faced with a decreasing function since the rise of 3N + 1 will be less than the decrease of N / 2 ^ K. Now we know that it is a decreasing function, but how are we sure that it reaches number 1? It is a function that is jumping with decreasing trend, however, the only way that it does not reach 1 would be to enter a cycle iterating to infinity, which implies that some number is repeated, however, it is not possible that are repeated by the difference in the strength of increase with respect to the decrease, only this can only happen when the equality is met: 3N + 1 = 2 ^ K N That is, the strength of increasing an odd is equal to that of increasing a pair, and this only happens if N = 1 and K = 2, in line with the function of the beginning of the document. Or put another way, when the pair is 4 and the odd 1. For this, and to be decreasing to N = 1, this series will always end cyclically in 1,4,2,1. Thank you very much for your attention, I await your opinions and advice. Note that I have written this same text in several forums in case someone thinks it is interesting and can demonstrate the conjecture. Last edited by skipjack; August 28th, 2018 at 04:41 PM. 
August 28th, 2018, 05:01 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,790 Thanks: 629 Math Focus: Yet to find out. 
Surely you can't start by taking an algebraic approach.. The conjecture belongs to dynamical systems/iterated maps which, sure, have connections to number theory and algebra but why not use the tools of DS to explore first.. It seems to be a common approach for all those starting on this problem, but then again I've no idea myself.

August 28th, 2018, 10:59 PM  #3 
Newbie Joined: Aug 2018 From: Spain Posts: 4 Thanks: 0 
Thanks for the answer, of course, it's not a complex answer to the conjecture, and I do not know how to explain it using the ds tools since I'm not an expert on the subject. However, the equation of equality of intensities of increase and decrease resolves at what point the series ends and begins to repeat a cycle, that is, that explains why this function ends in that cycle 1,4,2 If for example the function were: When odd we increase 3N + 3 and when it is even we divide by 2. It also shows that it is decreasing, and also now the only way that the cycle is repeated is only if it meets equality: 3N + 3 = 2 ^ K N This only happens if K = 2 and N = 3 (the cycle is repeated, 3 , 12 , 6) for example: If N = 1 we would have 1 6 3 12 6. . . 3 12 6 If N = 2 5 18 9 30 15 48 24 12 6 ... 3 12 6 If N = 10 33 102 51 156 78 39 120 60 30 15 48 24 12 6 ... 3 12 6 Sorry if I'm really wasting your time. Regards, 
August 29th, 2018, 12:15 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
It might be helpful to consider a conjecture of this type that happens to be false.

August 29th, 2018, 09:04 AM  #5 
Newbie Joined: Aug 2018 From: Spain Posts: 4 Thanks: 0  
August 29th, 2018, 12:15 PM  #6 
Newbie Joined: Aug 2018 From: Spain Posts: 4 Thanks: 0 
sorr, edit: Given a number N we have: ***A = 3N + 1 ***B = 2 ^ K N From here you have to: AB = ((2^K)+3) N + 1 
August 29th, 2018, 12:29 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,255 Thanks: 681 
Isn't this the same (false) argument of another poster that went on for multiple pages? Half the time you multiply by 3 and half the time you divide by 2 and then half the time you divide by 2 again so on average the results keep getting smaller therefore Collatz is true.


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