Re: The Collatz discrete primes!

On 23 Mar, 01:18, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx> wrote:
On Mar 22, 11:20?pm, "Danny" <fasttrac...@xxxxxxxxxxxxx> wrote:





On 22 Mar, 14:14, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx> wrote:

On Mar 22, 12:39 pm, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx>
wrote:

On Mar 22, 1:36 am, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx>
wrote:

On Mar 20, 12:23?am, "Danny" <fasttrac...@xxxxxxxxxxxxx> wrote:

3x+1 revisited.

2,3,7,19,37,43,73,79,97,109,127,151,163,181,199,223,241,271
,277,307,313,331,349,367,379,397,421,439...

The above prime list are primes (p) that are not in any seed (n)
path where any given seed (n) is < (p).

e.g.
For prime 73 to make this list then ---

All seeds (n) where n= (1,2,3,4,.72) 73 does not
appear in any of these seed paths of (n) in the Collatz tree.

Also after the second term in the list they all are...
(p-1)==0(mod 3).

Will there ever be a prime in this list where (p-1) is
not a ?0(mod 3)?

There are many primes (not) in this list where (p-1) is
not a 0(mod 3) and some that are a 0(mod 3) that are
not on this list.

Also, does this list ---->oo?

Dan

This seems to be true of odd numbers in general,
not just primes. Numbers that are not included
in the union of all pathways less than themselves,
include the composites:

25, 55, 115, 133, 145, 169, 187, 217, 235, 259, 289,
295, 343, 361, 385, 403, 451, 469, 475, etc.

all of which are 1(mod3).

I pondered this some more. First, we need some rules on
how branches and sub-branches are organized (I'm going to
use r0, r1, and r2 to represent 0(mod3), 1(mod3) and 2(mod3)
as it's easier to type).

?- all r0 branches have only r0 nodes to infinity and no
? ?sub-branches

?- all r1 & r2 branches alternate r1 & r2 nodes to infinity
? ?and have sub-branches only at r1 nodes (if and only if
? ?r1 is even)

?- sub-branches always appear in modulo 3 order

There are thus, only 7 types of branches:

? ? r0
? ? r0
? ? r0
? ? r0
? ? r0
? ? r0
? ? r0
? __r0

? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r2 ? ? ? ? r1__r0 ? ? ? ? r1__r1
? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r1 ? ? ? ? r1__r2 ? ? ? ? r1__r0
? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r0 ? ? ? ? r1__r1 ? ? ? ? r1__r2
? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? __r1 ? ? ? ? ? __r1 ? ? ? ? ? __r1

? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r2 ? ? ? ? r1__r0 ? ? ? ? r1__r1
? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r1 ? ? ? ? r1__r2 ? ? ? ? r1__r0
? ? r2 ? ? ? ? ? ? r2 ? ? ? ? ? ? r2
? ? r1__r0 ? ? ? ? r1__r1 ? ? ? ? r1__r2
? __r2 ? ? ? ? ? __r2 ? ? ? ? ? __r2

Since an r0 branch can't have any sub-branches, _all_ ancestors
of p must be > p.

For branches where p = r2, we have

? ? a__b
? __p

and doing the algebra, we get b = (2p-1)/3. The -1 isn't so
important. Generally, because of the fraction 2/3, b < p, so
_every_ p = r2 has an ancestor that's smaller than p.

For r1 branches,

? ? b__c
? ? a
? __p

the fraction involved is 4/3, so c > P.

BUT...

What if c is itself an r2?

? ? ? ?d__e
? ? b__c
? ? a
? __p

Then the fraction becomes (4/3)*(2/3) which is 8/9 and thus, e < p.
So an r1 branch _could_ have an ancestor smaller than p. We see this
in the case of 27 to 31:

? ? ? ? ?82__27
? ? 124__41
? ? 62
? __31

Here, 41 > 31 yet 27 < 31.

And it need not be the first subranch. Take for example

? ?e__f
? ?d
? ?b__c
? ?a
? _p

Here f is 16/3 p. But if we string together a big enough r2 chain,
we can overcome the 16/3 and eventually reach a fraction with a
smaller ancestor:

? ?(16/3)*(2/3) = ?32/9
? ?(32/9)*(2/3) = ?64/27
? (64/27)*(2/3) = 128/81
?(128/81)*(2/3) = 256/243
(256/243)*(2/3) = 512/729 ?<-- fraction is now less than 1

So, in the case of

? ? ? ? ? ? ? ?m__n
? ? ? ? ? ? k__l
? ? ? ? ?i__j
? ? ? g__h
? ?e__f
? ?d
? ?b__c
? ?a
? _p

Oops, made a mistake there, should be

? ? ? ? ? ? ? ? ? ?o__q
? ? ? ? ? ? ? ? m__n
? ? ? ? ? ? ?k__l
? ? ? ? ? i__j
? ? ? ?g__h
? ? e__f
? ? d
? ? b__c
? ? a
? ?_p

so it's q < p.

n < p. A p = r1 joins the r2's as having an ancestor smaller than
p _if_ a long enough chain of r2's can form. And obviously, the
higher up the branch you go, the longer the r2 chain has to be
(of course, they don't have to be consecutive, just that the r2's
eventually overtake the net effect of the r1's).

In the case of consecutive r2 chains, it should be noted that r2
chains occur in 3-adic sequence on any given branch. The 3-adic
sequence is

1,1,2,1,1,3,1,1,2,1,1,4,1,1,2,1,1,3,1,1,2,1,1,5,
1,1,2,1,1,3,1,1,2,1,1,4,1,1,2,1,1,3,1,1,2,1,1,6,
1,1,2,1,1,3,1,1,2,1,1,4,1,1,2,1,1,3,1,1,2,1,1,5,
1,1,2,1,1,3,1,1,2,1,1,4,1,1,2,1,1,3,1,1,2,1,1,7...

thus our r2 sub-branch chains could look like

? ? r2
? ? r1__r2 ? <-- chain is length 3
? ? r2
? ? r1__r1
? ? r2
? ? r1__r0
? ? r2
? ? r1__r2 ? <-- chain is length 1
? ? r2
? ? r1__r1
? ? r2
? ? r1__r0
? ? r2
? ? r1__r2 ? <-- chain is length 1
? ? r2
? ? r1__r1
? ? r2
? ? r1__r0
? ? r2
? ? r1__r2 ? <-- chain is length 2
? ? r2
? ? r1__r1
? ? r2
? ? r1__r0
? ? r2
? ? r1__r2 ? <-- chain is length 1
? ? r2
? ? r1__r1
? ? r2
? ? r1__r0
? ? r2
? ? r1__r2 ? <-- chain is length 1
? ? r2
? __r1

But also note that we could enter the 3-adic sequence at a random
point, so the first non-1 number could be anything. I exploit this
in my tree-crawler algorithm. I do a 3-level look-ahead (out of
every 3 chains, at least one has a length>1) to see if I found a
chain that is long enough to compensate for the extra cost of
reaching it. Doesn't happen very often, but it _does_ happen.

In conclusion, I'll state, the numbers on your list

?- have nothing to do with primes

?- always on r0 branches, never on r2 branches, and possibly
? ?on r1 branches

?- if the r1 branch can'r find a long enough r2 chain
? ?(consecutive or aggregate)

?- the list goes to infinity- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Just spotted this In OEIS --- A127781
"The Hailstone pure numbers"
So my special list of primes could be called
the Hailstone pure primes!

That's Pure Hailstone primes --- A127928.



I missed that one, I guess I just reinvented
the wheel!

What interest me more about all the pure hailstone
numbers is the difference pattern they leave.
I belive we discussed this on an earlier post
but delving into it more I discovered this --

After the initial start these are the only difference
patterns in the hailstone pure numbers.

1,2,3
1,2,3,3,3
1,2,3,3,3,3,3

There never is a sequential repeat of the first
two patterns (1,2,3) and (1,2,3,3,3) but
(1,2,3,3,3,3,3) at times will repeat once.

These repeats occur between hailstone pure @
405 - 438
891 - 924
1863 - 1896
3159 - 3192
3321 - 3354
4779 - 4812
5265 - 5298
6237 - 6270
6705 - 6738
-- etc.

Most of the time the general pattern will be --
[123][12333][123][12333][123][1233333][12333][123]---
and once in awhile the repeat of [1233333][1233333]
but never [123][123] or [12333][12333].
I wonder why that is?

Dan

.

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