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# 3n+1 problem

 Posts: 6,127 Add as FriendChallenge to a DebateSend a Message 1/23/2016 8:02:41 PMPosted: 1 year 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: 1 year 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: 1 year 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: 6,127 Add as FriendChallenge to a DebateSend a Message 1/23/2016 11:15:49 PMPosted: 1 year 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: 6,127 Add as FriendChallenge to a DebateSend a Message 1/24/2016 6:23:32 AMPosted: 1 year 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: 1 year 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: 2,419 Add as FriendChallenge to a DebateSend a Message 1/24/2016 5:32:22 PMPosted: 1 year 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: 2,419 Add as FriendChallenge to a DebateSend a Message 1/24/2016 5:35:18 PMPosted: 1 year agoEven Erdos could not address the problem, one of the most proficient mathematicians of the 20th century.