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1 8th November 14:28
30pack
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Posts: 1
Default Primes and the Collatz conjecture!


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
---

3,5,7,11,13,17,19,29,37,53,59,67,89,101,131,149,15 7,179,181,197,241,269,277,
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
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2 8th November 14:29
christian bau
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Posts: 1
Default Primes and the Collatz conjecture!


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 :-)
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3 8th November 14:29
christian bau
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Default Primes and the Collatz conjecture!


Just wanted to say: A novel and interesting idea.
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4 8th November 14:29
larry hammick
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Default Primes and the Collatz conjecture!


"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
LH
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5 9th November 21:40
israel
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Posts: 1
Default Primes and the Collatz conjecture!


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
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6 9th November 21:41
30pack
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Default Primes and the Collatz conjecture!


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
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7 11th November 06:17
30pack
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Posts: 1
Default Primes and the Collatz conjecture!


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
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8 12th November 13:23
mensanator
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Posts: 1
Default Primes and the Collatz conjecture!


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
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9 12th November 13:25
30pack
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Posts: 1
Default Primes and the Collatz conjecture!


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
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10 13th November 21:18
mensanator
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Posts: 1
Default Primes and the Collatz conjecture!


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
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