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Subject: "algebraic opertations"     Previous Topic | Next Topic
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elgr53
Member since Oct-12-04
Oct-12-04, 08:31 PM (EST)
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"algebraic opertations"
 
   years ago i read in a math puzzle book, the name of which i forgot, that was translated from russian that if you choose any positive integer greater than 1 and perform the following procedure you'll allways reach 1.
procedure of operations:
if the integer you chose is even - keep dividing it by 2 until the quotient is 1 or until the quotient is an odd number greater than 1.
if the ouotient of division by 2 is an odd number greater than 1 - multiply it by 3 and than add 1 to the result and divide it by 2 repeating the above procedure.

if you initially choose an odd integer multiply it by 3,add 1 to the result and continue as described in the previous paragraph.

example 1: integer chosen: 17

17*3 1=52, 52/2=26, 26/2=13, 13*3 1=40, 40/2=20, 20/2=10, 10/2=5,

5*3 1=16, 16/2=8, 8/2=4, 4/2=2, 2/2=1.

example 2: integer chosen:31

31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479,
1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154,
577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1.

according to the above mentioned book there was no proof of the generality of this rule but nobody has found a contrary example.also there was no known rule concerning the number of steps needed to reach the final result (1) as a function of the initial (chosen) integer.

i would be very grateful if you could kindly tell me the name of the book as well as if there was found a proof for the above rule or if anyone managed to find a contrary example.


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  Subject     Author     Message Date     ID  
algebraic opertations elgr53 Oct-12-04 TOP
  RE: algebraic opertations alexb Oct-12-04 1
     RE: algebraic opertations Kyle Jun-02-06 2
         RE: algebraic opertations mr_homm Jun-02-06 3
     RE: algebraic opertations kfom Jul-06-06 4
         RE: algebraic opertations sfwc Jul-07-06 5
     RE: algebraic opertations Pierre Charland Jul-09-06 6
         RE: algebraic opertations Gino Aug-01-06 7

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alexb
Charter Member
1866 posts
Oct-12-04, 08:35 PM (EST)
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1. "RE: algebraic opertations"
In response to message #0
 
   You are looking for what is commonly known as the Collatz sequence. Search the web - there's plenty information around. However, do not expect to find answers to your questions. These are still open.


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Kyle
guest
Jun-02-06, 06:02 AM (EST)
 
2. "RE: algebraic opertations"
In response to message #1
 
   I don't think this question is "open"...
say your number is
p_1^e_1 * p_2^e^2 * ...
the product of prime numbers p_1,p_2 = 2,3,5,7,...

then dividing by 2^kkills p_1.. so the number is now odd...
multiplying by 3 and then adding 1 will give you an even number. you then are allowed to divide out that 2 again...
continuing on in this fashion we see that at some point we can't divide by 2 again. so the proceduce does give us 1 (this is i think called Fermat's method of descent, but i'm not sure)


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mr_homm
Member since May-22-05
Jun-02-06, 09:29 AM (EST)
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3. "RE: algebraic opertations"
In response to message #2
 
   Hi Kyle,

I don't think this part is right:

>continuing on in this fashion we see that at some point we
>can't divide by 2 again. so the proceduce does give us 1

When you can't divide by 2 again, all you know is that the value is now odd, but it could still be a large odd number. The trouble is that when you triple and add 1, you of course make make the number larger and guarantee that it is even, but you do not know how many factors of 2 it now has. Suppose there is only a single factor of 2 in the new number; then the sequence of operations n --> 3n+1 --> (3n+1)/2 = m yields m>n, with m odd again. Since you don't know how often this case will occur, the value could possibly keep growing indefinitely.

The conjecture is that the division by 2 eventually always "wins" and decreases the value to 1. Notice that once you reach 1 the value will cycle 1 --> 4 --> 2 --> 1. This raises the possibility that the sequence might also fall into a cycle that does not include n=1. So there are three possibilities: an unbounded (but random looking and non-repeating) growth, a cycle that avoids n=1, or a cycle that includes n=1. As far as I know, no one has either proven than the process must cycle through 1 or even that there is some kind of upper bound on the maximum value the sequence will hit before returning to n=1.

Perhaps I am misinterpreting what you said, but I think these questions are still open (as of the last time I checked, which was admittedly not all that recently).

--Stuart Anderson


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kfom
guest
Jul-06-06, 08:18 PM (EST)
 
4. "RE: algebraic opertations"
In response to message #1
 
   Is this also known as the hailstone sequence or is that something different?


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sfwc
Member since Jun-19-03
Jul-07-06, 07:33 AM (EST)
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5. "RE: algebraic opertations"
In response to message #4
 
   >Is this also known as the hailstone sequence or is that
>something different?
Yes, this is the hailstone sequence.

Thankyou

sfwc
<><


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Pierre Charland
Member since Dec-22-05
Jul-09-06, 03:06 PM (EST)
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6. "RE: algebraic opertations"
In response to message #1
 
   Yes, this is also known as the 3n+1 sequence.
I have been collecting references about it.

You may want to check:

https://www.cut-the-knot.org/Curriculum/Arithmetic/Collatz.shtml

https://www.ieeta.pt/~tos/3x+1.html

https://www.numbertheory.org/php/collatz.html

https://www.numbertheory.org/php/markov.html

https://home.earthlink.net/~mrob/pub/math/seq-wondrous.html

The attached file is a list of record values (which need many steps to reach 1)

AlphaChapMtl

Attachments
https://www.cut-the-knot.org/htdocs/dcforum/User_files/44b14c4a09401fef.txt

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Gino
guest
Aug-01-06, 08:29 AM (EST)
 
7. "RE: algebraic opertations"
In response to message #6
 
   This problem is one of the most important open problems in number theory. It is called the Collatz or the 3n+1 conjecture. Some people here are obviously confused as to what the problem actually states. The problem says:

Let n be an integer. If n is even, divide n by 2.
If n is odd, multiply it by 3 and add 1 (3n+1)

Question: will the sequence ever get back to 1 for any and every choice of n? The Collatz conjecture is the conjecture that the answer is yes, although (it being a conjecture) there is no proof.

The number e^PI will never get back to 1, of course. But the problem is stated for integers only. There is another version of the Collatz conjecture for either rational or irrational numbers (I forget which) and that one says something different.

Although computers have tried running this for numbers as big as 2^60 or so, the numbers seem to always return to 1. Most mathematicians believe that the conjecture is true; they believe that the sequence will return to 1 for any choice of n. What is missing is a proof that would categorically show that no matter what number you pick, you always get back to 1.

I had a good friend who worked on this problem. At one point, he thought he discovered a number that would violate the conjecture, thus giving a counterexample and proving the conjecture false. He later realized he was wrong and that he had made no progress on the problem.


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