this thread is for a Collatz conjecture again

[MattcAnderson]

Hi Mersenneforum.org,

Please consider this.

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

Regards,

Matt

[science_man_88]

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 ...

[science_man_88]

best I can think of is maybe reducing it to a statement about the natural numbers ( other than the original one).

[MattcAnderson]

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

[SarK0Y]

it falls down to 1 because 3n+1=2^m for some n's. Just a trick :)

[science_man_88]

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 ?

[SarK0Y]

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.

[science_man_88]

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 ?

[SarK0Y]

Originally Posted by science_man_88

and how many steps do you expect for a given n ?

quite good approx is about lg₂(n).

add: it's max bar.

[science_man_88]

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.

[SarK0Y]

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.