30pack2009-11-08 14:28:31

For any Collatz sequence path find only the prime start numbers that

creates an odd prime for each of its odd integers except 1.

Here are the first few prime start numbers that meet these criteria

—

3,5,7,11,13,17,19,29,37,53,59,67,89,101,131,149,157,179,181,197,241,269,277,

349,397,739,853,1109,1237,1429,1621,1861,1877,2161,2389,2531,2957,3413,3797,4549…

I believe I found them all up too prime 4549.

What is interesting also, only primes from the (above list) are

possible in any sequence where all the odd integers are prime except 1

and where this prime list above ——>oo.

If you do not believe this to be true, please show a counter example.

Also, the all odd integer = prime paths (except start # 3 and 5) all

seem to merge at this point 40,20,10,5,16,8,4,2,1

Thanks,

Dan

Christian bau2009-11-08 14:29:03

I think k = 22,369,621 is a prime, and 3k+1 = 2^26. The odd numbers on

the path are 22,369,621 -> 1 ðŸ™‚

Christian bau2009-11-08 14:29:06

Just wanted to say: A novel and interesting idea.

Larry hammick2009-11-08 14:29:54

”Christian Bau”

Er, 22,369,621 = 8191*2731 (in hex, 0xAAB * 0x1FFF). But I get the idea.

Anyhow, the Collatz conjecture is not really about arithmetic ðŸ™‚

LH

Israel2009-11-09 21:40:03

It is not a prime: it is 8191 * 2731. More generally, 2^(2j)-1 is

always divisible by 2^j-1, so (2^(2j)-1)/3 is never prime for j >= 3.

Of course what this means is that if the sequence starting at k gets

to 1, and all odd numbers in the path until 1 are prime, the last part

of the path must be 5 -> 16 -> 8 -> 4 -> 2 -> 1.

Robert Israel israel@math.ubc.ca

Department of Mathematics http://www.math.ubc.ca/~israel

University of British Columbia

Vancouver, BC, Canada V6T 1Z2

30pack2009-11-09 21:41:42

Another interesting find is —

Starting with powers of 2 where

n =

(((2^4)*10)-1)/3 = 53

(((2^8)*10)-1)/3 = 853

(((2^10)*10)-1)/3 = 3413

(((2^16)*10)-1)/3 = 218453

(((2^20)*10)-1)/3 = 3495253

(((2^22)*10)-1)/3 = 13981013 where n= a prime starting numbers in a

Collatz path.

and continuing on — avoiding 2^6, 2^12, 2^18, 2^24, 2^30,… these

will not produce a prime because

n == 0 mod 3.

These prime starting # (seed) produce the following similar sequences

—

E.g.

53, 160, 80, 40, 20, 10 , 5, 16, 8, 4, 2, 1

853, 2560, 1280, 640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

13981013, 41943040, 20971520, 10485760, 5242880, 2621440, 1310720,

655360, 327680, 163840, 81920, 40960, 20480, 10240, 5120, 2560, 1280,

640, 320, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

I am checking to see if there are numerous primes for larger powers of

2.

If there is then these primes could be called Collatz seed primes.

The next one is at (((2^40)*10)-1)/3 = 3665038759253 is also prime.

Dan

30pack2009-11-11 06:17:11

Correction—

Starting with (even) powers of 2.

These two should have been 1st on the list for these special primes.

(((2^0)*10)-1)/3 = 3

(((2^2)*10)-1)/3 = 13

Mensanator2009-11-12 13:23:23

2^4 * 10 is the same as 2^5 * 5. Why the distinction? Because 5 is the

root of the branch. When you look at a branch in binary, the root is

101 and every number higher up is formed by simply appending a 0:

10100000000000

1010000000000

101000000000

10100000000

1010000000

101000000

10100000

1010000

101000

10100

1010

101

The root pattern 101 is preserved across all members of the branch.

Since 5 == 2 (mod 3) and under multiplication by 2, the

successor of 2 (mod 3) is 1 (mod 3) and the

successor of 1 (mod 3) is 2 (mod 3)

the numbers in the branch alternate between 2 (mod 3) and 1 (mod 3).

Sub-branches only attach at 1 (mod 3), so every alternate member of

the 5 branch has a sub-branch attached, starting with the second.

10100000000000_

1010000000000

101000000000_

10100000000

1010000000_

101000000

10100000_

1010000

101000_

10100

1010_

101

Each successive sub-branch is related to the previous one by 4n+1, and

the successor rules for 4n+1 are

– the binary pattern appends 01

– n (mod 3) is succeeded by n+1 (mod 3)

Once we know that the first sub-branch is 3 (0 (mod 3)), we know all the

rest

10100000000000_110101010101 2 (mod 3)

1010000000000

101000000000_1101010101 1 (mod 3)

10100000000

1010000000_11010101 0 (mod 3)

101000000

10100000_110101 2 (mod 3)

1010000

101000_1101 1 (mod 3)

10100

1010_11 0 (mod 3)

101

Each new sub-branch is going to pass through all the numbers of the

previous sub-branch. That’s why the sequences are similar.

Where I thought this was going was some insight into the relationship

between prime numbers and binary patterns. But when you list the

COMPLETE tree structure, there is no apparent pattern. It looks like

you just got lucky in that a bunch of the lower sub-branches happened to

be prime. That pattern evaporates as you move up the branch, even

allowing for the fact that all 0 (mod 3) are not prime.

n (n – 1)/3 prime?

005629499534213120

002814749767106560 00000938249922368853 == 0 (mod 3)

001407374883553280

000703687441776640 00000234562480592213 No 1009 * 232470248357

000351843720888320

000175921860444160 00000058640620148053 No 733 * 80000846041

000087960930222080

000043980465111040 00000014660155037013 == 0 (mod 3)

000021990232555520

000010995116277760 00000003665038759253 Yes

000005497558138880

000002748779069440 00000000916259689813 No 13 * 70481514601

000001374389534720

000000687194767360 00000000229064922453 == 0 (mod 3)

000000343597383680

000000171798691840 00000000057266230613 No proven composite

000000085899345920

000000042949672960 00000000014316557653 No 41 * 349184333

000000021474836480

000000010737418240 00000000003579139413 == 0 (mod 3)

000000005368709120

000000002684354560 00000000000894784853 No proven composite

000000001342177280

000000000671088640 00000000000223696213 No 13 * 17207401

000000000335544320

000000000167772160 00000000000055924053 == 0 (mod 3)

000000000083886080

000000000041943040 00000000000013981013 Yes

000000000020971520

000000000010485760 00000000000003495253 Yes

000000000005242880

000000000002621440 00000000000000873813 == 0 (mod 3)

000000000001310720

000000000000655360 00000000000000218453 Yes

000000000000327680

000000000000163840 00000000000000054613 No 13 * 4201

000000000000081920

000000000000040960 00000000000000013653 == 0 (mod 3)

000000000000020480

000000000000010240 00000000000000003413 Yes

000000000000005120

000000000000002560 00000000000000000853 Yes

000000000000001280

000000000000000640 00000000000000000213 == 0 (mod 3)

000000000000000320

000000000000000160 00000000000000000053 Yes

000000000000000080

000000000000000040 00000000000000000013 Yes

000000000000000020

000000000000000010 00000000000000000003 Yes, == 0 (mod 3)

000000000000000005

But what is the significance of that? Can you predict which of the

next 100 sub-branches are prime? I thought you had something

interesting until I noticed you only mentioned the successes and

glossed over the failures. To properly understand something, you

should show ALL the data. The patterns formed by the failures can

be just as significant as those formed by the successes.

—

Mensanator

2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm

30pack2009-11-12 13:25:05

Thanks for you’re in depth reply. You have much more expertise then I

on any math involved here or any other math-related subject I am sure.

My only objective here was basically to show these special primes that

reside in this special sequence as the starting (seed) integer for

this one path in the Collatz tree. Where each node off this path is

only one and that is where all of these special odd primes and 0 (mod)

3’s and other odd composites with prime factors reside thus creating

only 3 odd integers, the (seed), prime 5, and 1 in their entire path

starting with (seed) 3. I am just picking the primes related by one

node to this one path. Also there is no 0 (mod) 5 in any of these odd

composites (seeds).

As in the Mersenne primes, not all-prime exponents of 2 create another

prime. The same is true here, but with a twist, not all-even exponents

(n) where (2^n*10-1)/3 will create another prime. You also have to

consider the *10 factor of the Collatz (seed) primes when comparing

densities of the two.

Sorry if I misled anyone here into thinking there was some kind of

pattern, but I never had that intention.

I am doing a comparison table of the Mersenne primes and the Collatz

(seed) primes. Yes, there is many more even exponents creating the

Collatz (seed) prime then the Mersenne prime exponents, so that will

explain the density factor between the two.

A question also remains, is they’re a Mersenne prime or Mersenne

primes member(s) of this path in the Collatz tree?

Thanks for your interesting input.

Dan

Mensanator2009-11-13 21:18:03

When using tree structures to represent Collatz sequences, the ORDER

is the number of branches a number is from the trunk (the branch whose

root is 1). The ORDER is simply the number of odd integers (or the count

of 3x+1 iterations) in the sequence, where branch 1 is considered ORDER 0.

In your list, all your primes are sub-branches of branch 5 . Each new prime

on your list attaches higher and higher up on branch 5 requiring more and

more itertions of x/2, but they all are ORDER 2.

Now a Mersenne number (2^n – 1), whether prime or not, never has an

ORDER less than n. Typically, the ORDER is n*4.819. That formula is not

exact, although the error gets proportionally smaller as n increases. See

http://members.aol.com/mensanator666/Page.htm

(In that chart, “Cycles” is used as a synonym for ORDER)

So 3, which is 2^2 – 1, is the only possible Mersenne Number that can

(and does) appear on your list.

Other Mersenne numbers will have sequences that pass through 5 on their

way to 1. 31 passes through the prime 53, but it is ORDER 39 (and along

with 27) is one of the only two numbers whose ORDER is larger than itself.

—

Mensanator

2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm

Mensanator2009-11-13 21:19:13

Also, as I mentioned before, every 2nd through nth sub-branch appends a 01 onto

the binary pattern of the first sub-branch:

11

1101

110101

11010101

..

..

..

Since Mersenne numbers are always all 1s in binary

1

11

111

1111

11111

..

..

..

if a Mersenne number appears as a sub-branch, it must be the first one and

there can be no further ones attached to that branch.

—

Mensanator

2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm

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