20170501, 00:47  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
495_{16} Posts 
this thread is for a Collatz conjecture again
Hi Mersenneforum.org,
Please consider this. It is some Maple code. Maple is a computer algebra system. Regards, Matt 
20170501, 01:57  #2 
"Forget I exist"
Jul 2009
Dumbassville
3×2,801 Posts 

20170503, 21:58  #3 
"Forget I exist"
Jul 2009
Dumbassville
3·2,801 Posts 
best I can think of is maybe reducing it to a statement about the natural numbers ( other than the original one).
Last fiddled with by science_man_88 on 20170503 at 21:59 
20170510, 02:45  #4 
"Matthew Anderson"
Dec 2010
Oregon, USA
3·17·23 Posts 
Hi Mersenneforum,
This is a simple procedure for the Collatz conjecture. It has also been called the hailstone problem. That is all I have to say about that. Regards, Matt 
20170514, 04:34  #5 
Jan 2010
2·43 Posts 
it falls down to 1 because 3n+1=2^m for some n's. Just a trick :)

20170514, 10:44  #6 
"Forget I exist"
Jul 2009
Dumbassville
20D3_{16} Posts 

20170514, 14:37  #7 
Jan 2010
2·43 Posts 
all numbers have formula: odd*2^t, so for odd ones t==0. 3n+1 can be odd only if n is even. in short, this sequence could be steadily increasing iff t ain't greater than 1 for each step. could that condition be possible? Obviously, No. to not fall down to 1 needs to not have 1*2^m at any step.

20170514, 22:05  #8  
"Forget I exist"
Jul 2009
Dumbassville
3·2,801 Posts 
Quote:


20170515, 19:16  #9 
Jan 2010
2×43 Posts 
quite good approx is about lg_{2}(n).
add: it's max bar. Last fiddled with by SarK0Y on 20170515 at 19:24 
20170515, 20:50  #10 
"Forget I exist"
Jul 2009
Dumbassville
3×2,801 Posts 
log_{2}(7) < 3 there are not three steps for 7 it goes:
7>22>11>34>17>52>26>13>40>20>10>5>16>8>4>2>1 of course you probably meant for large n. Last fiddled with by science_man_88 on 20170515 at 21:09 
20170515, 21:33  #11  
Jan 2010
2·43 Posts 
Quote:
7 > 11 > 17 > 13 > 5 > 1. in short, packs count only odds. such scheme is quite reasonable because N/2 == N >>1, it's very cheap op for hardware. 

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