Re: The Collatz discrete primes!

On 24 Mar, 10:57, "mensana...@xxxxxxxxxxx" <mensana...@xxxxxxx> wrote:
On Mar 23, 8:01 am, "Danny" <fasttrac...@xxxxxxxxxxxxx> wrote:

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

That's Pure Hailstoneprimes--- A127928.

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

Great minds think alike, eh?



What interest me more about all the pure hailstone
numbers is the difference pattern they leave.

At first I didn't realize you had gone back to _all_
pure numbers. Changing my program to show all numbers
(not justprimesor odds) I get

(0) 405 [0]
(1) 406 [0]
(2) 407 [0]
(0) 408 [0]
(1) 409 [2]
(2) 410 [0]
(0) 411 [0]
(1) 412 [0]
(2) 413 [0]
(0) 414 [0]
(1) 415 [0]
(2) 416 [0]
(0) 417 [0]
(1) 418 [0]
(2) 419 [2]
(0) 420 [0]
(1) 421 [2]
(2) 422 [0]
(0) 423 [0]
(1) 424 [0]
(2) 425 [0]
(0) 426 [0]
(1) 427 [0]
(2) 428 [0]
(0) 429 [0]
(1) 430 [0]
(2) 431 [2]
(0) 432 [0]
(1) 433 [2]
(2) 434 [0]
(0) 435 [0]
(1) 436 [0]
(2) 437 [0]
(0) 438 [0]

where for (m) n [p]

- m is modulo 3
- n is the number
- p is prime flag (0=composite,2-prime)

pure on the left, impure to the right.

And when I calculate differences, I get something like (not
the same sequence)

1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3 3 3
1 2 3 3 3 3 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3 3 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2 3
1 2 3 3 3
1 2

So I think I'm finally on the same page.

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

Which I concur on.







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?

Ah, the $64 question: why?

There are always rules. Rules have implications. Sometimes
the interaction of these implications is hard to see when they
are all run together. I may not have crossed every i and
dotted every t, but I think it works something like this:

First, every 0(mod3) number is pure and since natural numbers
are always in modulo 3 sequence, we start out with

0(mod3) pure
1(mod3) ?
2(mod3) ?
0(mod3) pure

That means no difference can be greater than 3. If that
were the end of the story we would have only [3333333...]
as a sequence of differences. There is, of course, more
to it than that.

We also know that every 2(mod3) is impure (because its
sub-branch is only one step above). This applies to both
even and odd 2(mod3) numbers. So far, we've got

0(mod3) pure
1(mod3) ?
2(mod3) impure
0(mod3) pure

Which doesn't change our differences. What does change them
is whether or not the 1(mod3) is pure or impure (and it can
be either). That gives us two possibilities

0(mod3) pure
1(mod3) impure
2(mod3) impure
0(mod3) pure

or

0(mod3) pure
1(mod3) pure
2(mod3) impure
0(mod3) pure

For the first case, we still get a 3, but for the second,
we get a 12 instead of a 3 for differences (essentially, a
3 can be replaced by a 12).

So everywhere we see a 12 in the sequence, a 1(mod3) is pure.
Wherever you see a 3, the 1(mod3) inside that block is impure.

Next, every _even_ 1(mod3) is impure. Any 1(mod3) is evenly
divisible by 3 when you subtract 1, but, by the implication
that 3n+1 always results in an even number, you cannot reverse
that operation when the 1(mod3) is odd. You can always reverse
it when 1(mod3) is even, thus, every even 1(mod3) has a smaller
ancestor and thus, every even 1(mod3) is impure.

What does that imply to our sequence? That you cannot have 2
pure 1(mod3) numbers in a row (because at least one will be even).
So a pure 1(mod3) implies

0(mod3) pure
1(mod3) pure
2(mod3) impure
0(mod3) pure
1(mod3) impure
2(mod3) impure
0(mod3) pure

and that means every 12 is followed by a 3, so [123] is the
smallest possible unit and the sum of a unit must be a multiple
of 6

[123] = 6
[12333] = 12
[1233333] = 18

But wait...there's more.

To have two [123] blocks in a row implies that the next numbers
are 12..., in other words

[123][123][12...]

which means we have 3 odd 1(mod3) pure numbers in a row but that
won't happen since to be pure, the first sub-branch can't be a
2(mod3) and out of three consecutive odd 1(mod3) numbers, at least
one will have a 2(mod3) sub-branch. That gives us the major
implication

- there cannot be 3 consecutive odd 1(mod3) pure numbers

Of course, you can skip one, but if you do, you get two 3's,
the one you skipped and the next one because the 1(mod3) number
would be even. So if you don't get [123], you must get at least
[12333].

Likewise, the other major implication is that out of 3 odd
1(mod3) numbers, at least one is pure, so we have

- there cannot be 3 consecutive odd 1(mod3) impure numbers

which means we can't have a unit larger than [1233333].

Let's see, I haven't covered why you don't see [12333][12333],
but my brain is kinda fried right now, I'm sure there's a
similar reason.

The answers are always there if you dig deep enough.





Dan- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Interesting deduction!
Thanks.

So if there was a counter example to the
Collatz conjecture it would have to start
with an odd pure?

Dan

.

Relevant Pages

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