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# 3n+1 problem

 Posts: 6,138 Add as FriendChallenge to a DebateSend a Message 1/23/2016 8:02:41 PMPosted: 2 years agoThis is a unique problem that gets played with by amateurs a lot. It's known by a few different names. Such as the Collatz, Ulam, or Syracuse Conjecture.The Conjecture has gone unproven despite many mathematicians and minds looking at it. Some become passionate about it's relevance while others are less interested in proving it correct. The Conjecture is given a series constructed from 2 steps: 1. Divide by 2 if Even. 2. multiply by 3 and add 1 if odd. Given those rules any natural integer over 0 will always end in 1.f(n) { n/2 IF n is even OR 3n+1 if odd.}Example: n = 7Series would be:7 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)The idea is that no matter what number you start with you will end up with 1. Series where (n) is 1 to billions have been completed and all end in 1.I'm asking why do people think this has not been proven yet?If the heuristic argument shows that the orbit is shrinking. Take that with every number can be written as the summation of powers of 2. take that step 2 transforms a number to be pushed to step1. (every ascending step will be followed by a descending step) So isn't this proof enough?
 Posts: 13,644 Add as FriendChallenge to a DebateSend a Message 1/23/2016 8:21:28 PMPosted: 2 years agoAt 1/23/2016 8:02:41 PM, Mhykiel wrote:This is a unique problem that gets played with by amateurs a lot. It's known by a few different names. Such as the Collatz, Ulam, or Syracuse Conjecture.The Conjecture has gone unproven despite many mathematicians and minds looking at it. Some become passionate about it's relevance while others are less interested in proving it correct. The Conjecture is given a series constructed from 2 steps: 1. Divide by 2 if Even. 2. multiply by 3 and add 1 if odd. Given those rules any natural integer over 0 will always end in 1.f(n) { n/2 IF n is even OR 3n+1 if odd.}Example: n = 7Series would be:7 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)The idea is that no matter what number you start with you will end up with 1. Series where (n) is 1 to billions have been completed and all end in 1.I'm asking why do people think this has not been proven yet?If the heuristic argument shows that the orbit is shrinking. Take that with every number can be written as the summation of powers of 2. take that step 2 transforms a number to be pushed to step1. (every ascending step will be followed by a descending step) So isn't this proof enough?Uh, doesn't 34/2 = 17?Marrying a 6 year old and waiting until she reaches puberty and maturity before having consensual sex is better than walking up to a stranger in a bar and proceeding to have relations with no valid proof of the intent of the person. Muhammad wins. ~ Fatihah If they don't want to be killed then they have to subdue to the Islamic laws. - Uncung There would be peace if you obeyed us.~Uncung Without God, you are lower than sh!t. ~ SpiritandTruth
 Posts: 304 Add as FriendChallenge to a DebateSend a Message 1/23/2016 9:37:46 PMPosted: 2 years agoAt 1/23/2016 8:21:28 PM, DanneJeRusse wrote:At 1/23/2016 8:02:41 PM, Mhykiel wrote:This is a unique problem that gets played with by amateurs a lot. It's known by a few different names. Such as the Collatz, Ulam, or Syracuse Conjecture.The Conjecture has gone unproven despite many mathematicians and minds looking at it. Some become passionate about it's relevance while others are less interested in proving it correct. The Conjecture is given a series constructed from 2 steps: 1. Divide by 2 if Even. 2. multiply by 3 and add 1 if odd. Given those rules any natural integer over 0 will always end in 1.f(n) { n/2 IF n is even OR 3n+1 if odd.}Example: n = 7Series would be:7 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)The idea is that no matter what number you start with you will end up with 1. Series where (n) is 1 to billions have been completed and all end in 1.I'm asking why do people think this has not been proven yet?If the heuristic argument shows that the orbit is shrinking. Take that with every number can be written as the summation of powers of 2. take that step 2 transforms a number to be pushed to step1. (every ascending step will be followed by a descending step) So isn't this proof enough?Uh, doesn't 34/2 = 17?correctcorrected example for n=77 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)17 (odd so 3(17)+1)52 (even so 52/2)26 (even so 26/2)13 (odd so 3(13)+1)40 (even so 40/2)20 (even so 20/2)10 (even so 10/2)5 (odd so 3(5)+1)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)
 Posts: 6,138 Add as FriendChallenge to a DebateSend a Message 1/23/2016 11:15:49 PMPosted: 2 years agoAt 1/23/2016 9:37:46 PM, RainbowDash52 wrote:At 1/23/2016 8:21:28 PM, DanneJeRusse wrote:At 1/23/2016 8:02:41 PM, Mhykiel wrote:This is a unique problem that gets played with by amateurs a lot. It's known by a few different names. Such as the Collatz, Ulam, or Syracuse Conjecture.The Conjecture has gone unproven despite many mathematicians and minds looking at it. Some become passionate about it's relevance while others are less interested in proving it correct. The Conjecture is given a series constructed from 2 steps: 1. Divide by 2 if Even. 2. multiply by 3 and add 1 if odd. Given those rules any natural integer over 0 will always end in 1.f(n) { n/2 IF n is even OR 3n+1 if odd.}Example: n = 7Series would be:7 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)The idea is that no matter what number you start with you will end up with 1. Series where (n) is 1 to billions have been completed and all end in 1.I'm asking why do people think this has not been proven yet?If the heuristic argument shows that the orbit is shrinking. Take that with every number can be written as the summation of powers of 2. take that step 2 transforms a number to be pushed to step1. (every ascending step will be followed by a descending step) So isn't this proof enough?Uh, doesn't 34/2 = 17?correctcorrected example for n=77 (7 is odd so 3(7)+1)22 (even so 22/2)11 (odd so 3(11)+1)34 (even 34/2)17 (odd so 3(17)+1)52 (even so 52/2)26 (even so 26/2)13 (odd so 3(13)+1)40 (even so 40/2)20 (even so 20/2)10 (even so 10/2)5 (odd so 3(5)+1)16 (even 16/2)8 (even 8/2)4 (even 4/2)2 (even 2/2)1 (halt on 1.)Thank you.
 Posts: 6,138 Add as FriendChallenge to a DebateSend a Message 1/24/2016 6:23:32 AMPosted: 2 years agoMy question is since it is true that all even numbers can be written as the sum of distinct powers of two and step 2 creates even numbers and step 1 reduces distinct powers of two then isn't it a proof by induction that the conjecture is true for all numbers?
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 1/24/2016 12:13:49 PMPosted: 2 years agoAt 1/24/2016 6:23:32 AM, Mhykiel wrote:My question is since it is true that all even numbers can be written as the sum of distinct powers of two and step 2 creates even numbers and step 1 reduces distinct powers of two then isn't it a proof by induction that the conjecture is true for all numbers?Here is an algorithm that you may be unaware off:1 Assume you am wrong2 Look for your error3 Find your error4 Learn something newTry converting your word description of your possible solution to algebraic expressions. Your error becomes obvious then.Let's hope "the truth is out there" cos there is bugger all round here.
 Posts: 2,630 Add as FriendChallenge to a DebateSend a Message 1/24/2016 5:32:22 PMPosted: 2 years agoI think this is called the Collatz conjecture - its unproven.https://en.wikipedia.org...This and many similar problems have elegant implementations in functional programming languages, here's a comparison of a C# v and F# implementation.http://www.refactorthis.net...See how terse the F# version is which is much closer to mathematics as a language than C# and other procedural languages - just wanted to point this out.Harry.
 Posts: 2,630 Add as FriendChallenge to a DebateSend a Message 1/24/2016 5:35:18 PMPosted: 2 years agoEven Erdos could not address the problem, one of the most proficient mathematicians of the 20th century.
 Posts: 210 Add as FriendChallenge to a DebateSend a Message 1/26/2016 8:26:00 PMPosted: 2 years agoAt 1/26/2016 7:15:39 PM, Mhykiel wrote:I was asking why it wasn't a proof. The real quedtion for me that would make a proof is how high does the series go based on the shape of the number before it nose dives down. If one can prove n becomes double or triple even then plummets below n. Well then every cycle will end in 1.It only strongly suggests that it is true which is why most believe that it is. However, for it to be a proof it has to be written in a way that applies to all integers. Unless you have a solid proof about all natural numbers there exists a possibility of say a prime number with an unknown property that could make it false. Unlikely, but still possible without the proof otherwise. There is some work on upper level mathematics that treats numbers as Objects that has been said to be making headway on some of these difficult to formulate proofs. But it's hard to look it up since you'll end up with a lot of programming articles when you try to search for it. It's been used to make some headway on Fermat's Last Theorem, but these things take so long to debate that I don't know what came of it.
 Posts: 6,138 Add as FriendChallenge to a DebateSend a Message 1/26/2016 9:14:51 PMPosted: 2 years agoAt 1/26/2016 8:26:00 PM, medv4380 wrote:At 1/26/2016 7:15:39 PM, Mhykiel wrote:I was asking why it wasn't a proof. The real quedtion for me that would make a proof is how high does the series go based on the shape of the number before it nose dives down. If one can prove n becomes double or triple even then plummets below n. Well then every cycle will end in 1.It only strongly suggests that it is true which is why most believe that it is. However, for it to be a proof it has to be written in a way that applies to all integers. Unless you have a solid proof about all natural numbers there exists a possibility of say a prime number with an unknown property that could make it false. Unlikely, but still possible without the proof otherwise. There is some work on upper level mathematics that treats numbers as Objects that has been said to be making headway on some of these difficult to formulate proofs. But it's hard to look it up since you'll end up with a lot of programming articles when you try to search for it. It's been used to make some headway on Fermat's Last Theorem, but these things take so long to debate that I don't know what came of it.I did make a claim about all natural numbers. They all, even primes, can be written as the summation of base 2s.Dividing by 2 is the same as subtracting 1 from the lowest exponent of base 2.No matter how many times the number is duplicated if enough 1s are added to even out any odd compents eventually the number will be divisble by 2 till it reduces to one.If n is odd then 3n+1 makes to even components. After division any odd component comes out again.I was trying to start some interesting conversation I found the puzzle interesting.As for proofing it I don't think is just in the domain of professionals. I yhink anyone with an interest can add to a field.
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 1/27/2016 10:40:20 AMPosted: 2 years agoAt 1/26/2016 9:14:51 PM, Mhykiel wrote:At 1/26/2016 8:26:00 PM, medv4380 wrote:At 1/26/2016 7:15:39 PM, Mhykiel wrote:I was asking why it wasn't a proof. The real quedtion for me that would make a proof is how high does the series go based on the shape of the number before it nose dives down. If one can prove n becomes double or triple even then plummets below n. Well then every cycle will end in 1.It only strongly suggests that it is true which is why most believe that it is. However, for it to be a proof it has to be written in a way that applies to all integers. Unless you have a solid proof about all natural numbers there exists a possibility of say a prime number with an unknown property that could make it false. Unlikely, but still possible without the proof otherwise. There is some work on upper level mathematics that treats numbers as Objects that has been said to be making headway on some of these difficult to formulate proofs. But it's hard to look it up since you'll end up with a lot of programming articles when you try to search for it. It's been used to make some headway on Fermat's Last Theorem, but these things take so long to debate that I don't know what came of it.I did make a claim about all natural numbers. They all, even primes, can be written as the summation of base 2s.Dividing by 2 is the same as subtracting 1 from the lowest exponent of base 2.No it is not.No matter how many times the number is duplicated if enough 1s are added to even out any odd compents eventually the number will be divisble by 2 till it reduces to one.If n is odd then 3n+1 makes to even components. After division any odd component comes out again.I was trying to start some interesting conversation I found the puzzle interesting.As for proofing it I don't think is just in the domain of professionals. I yhink anyone with an interest can add to a field.Let's hope "the truth is out there" cos there is bugger all round here.