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Collatz conjecture: is someone working with it? Do you know someone who is?
This is the conjecture:
Take a natural number n, if n is even, divide it by 2
If n is odd, find 3n+1
Examples:
5 -> 3*5 + 1 = 16 -> 8 -> 4 -> 2 -> 1
7-> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 1
The conjecture states that all n -> 1
I am working with it, and I'd like to get in touch with people that is or that would like to work on it with me.
Ana
These are some of my findings
1) 1 ->1
5= 1+4 -> 16 -> 1
21 =1+4+16 -> 64 -> 1
85 =1 +4+16+64 -> 256 -> 1
....
They are all sum [1 to n] 2^i
I will call them X numbers.
2) These X numbers generate numbers like 5^n
13 -> 5
133 -> 25
1333 -> 125
..
53 ->5
533 -> 25
5333 -> 125
....
3) N and 4n+1 have the same behaviour
So do 16N+5, 64N+21, 256N+85, etc
(A power of 2 N multiplied by N and added with the correspondant X number)
Yes, I am trying to prove that the conjecture is true or, at least, find as many particular cases as it is possible to me.
Ana
2 Antworten
- Anonymvor 1 JahrzehntBeste Antwort
Hmmmm... interesting problem.
Define
f(n) = n/2 if n is even
f(n) = 3n + 1 if n is odd
g (n) = f(n)
0
g (n) = f(g (n))
i + 1 i
The conjecture states that there exists i for which
g (n) = 1
i
The only way to get 1 is n = 2^k so you should prove that 3n + 1 eventually result in an even number.
I also state that i is bounded by k log(n) for some k.
- UnknownDLv 6vor 1 Jahrzehnt
You showed us. Well at least our group.... And uhh... what are you working with?
-----
EDIT: Are you trying to prove that it (the conjecture) works?
