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Old 2017-05-01, 00:47   #1
MattcAnderson
 
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"Matthew Anderson"
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Default 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
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Old 2017-05-01, 01:57   #2
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Quote:
Originally Posted by MattcAnderson View Post
Hi Mersenneforum.org,

Please consider this.

It is some Maple code. Maple is a computer algebra system.

Regards,

Matt
no attached code ...
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Old 2017-05-03, 21:58   #3
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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
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Old 2017-05-10, 02:45   #4
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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
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File Type: pdf 20170311090423.pdf (22.3 KB, 175 views)
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Old 2017-05-14, 04:34   #5
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it falls down to 1 because 3n+1=2^m for some n's. Just a trick :)
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Old 2017-05-14, 10:44   #6
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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 ?
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Old 2017-05-14, 14:37   #7
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Quote:
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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.
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Old 2017-05-14, 22:05   #8
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Quote:
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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 ?
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Old 2017-05-15, 19:16   #9
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and how many steps do you expect for a given n ?
quite good approx is about lg2(n).

add: it's max bar.

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Old 2017-05-15, 20:50   #10
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Quote:
Originally Posted by SarK0Y View Post
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
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Old 2017-05-15, 21:33   #11
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Quote:
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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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