mersenneforum.org this thread is for a Collatz conjecture again
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 2017-05-01, 00:47 #1 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 23×149 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
2017-05-01, 01:57   #2
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by MattcAnderson Hi Mersenneforum.org, Please consider this. It is some Maple code. Maple is a computer algebra system. Regards, Matt
no attached code ...

 2017-05-03, 21:58 #3 science_man_88     "Forget I exist" Jul 2009 Dartmouth NS 2·3·23·61 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 2017-05-03 at 21:59
2017-05-10, 02:45   #4
MattcAnderson

"Matthew Anderson"
Dec 2010
Oregon, USA

23·149 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
Attached Files
 20170311090423.pdf (22.3 KB, 278 views)

 2017-05-14, 04:34 #5 SarK0Y     Jan 2010 2·43 Posts it falls down to 1 because 3n+1=2^m for some n's. Just a trick :)
2017-05-14, 10:44   #6
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by SarK0Y it falls down to 1 because 3n+1=2^m for some n's. Just a trick :)
and what makes you certain that all odd numbers are connected ?

2017-05-14, 14:37   #7
SarK0Y

Jan 2010

2·43 Posts

Quote:
 Originally Posted by science_man_88 and what makes you certain that all odd numbers are connected ?
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.

2017-05-14, 22:05   #8
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2·3·23·61 Posts

Quote:
 Originally Posted by SarK0Y 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.
and how many steps do you expect for a given n ?

2017-05-15, 19:16   #9
SarK0Y

Jan 2010

2×43 Posts

Quote:
 Originally Posted by science_man_88 and how many steps do you expect for a given n ?
quite good approx is about lg2(n).

Last fiddled with by SarK0Y on 2017-05-15 at 19:24

2017-05-15, 20:50   #10
science_man_88

"Forget I exist"
Jul 2009
Dartmouth NS

2×3×23×61 Posts

Quote:
 Originally Posted by SarK0Y quite good approx is about lg2(n). add: it's max bar.
log2(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 2017-05-15 at 21:09

2017-05-15, 21:33   #11
SarK0Y

Jan 2010

2×43 Posts

Quote:
 Originally Posted by science_man_88 log2(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.
hmm.. here is disputable moment how to count steps: you can count each one or packs.

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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