"Blueprint for Failure"

[bill]

"If we were to list all the Type [1, 2] Sequence Vector starting
points,

3, 11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91, . . .

the ith occurrence (for i: 0, 1, 2, 3, ...) starts at i * 8 + 3

And if we were to list all the Type [1, 2] Sequence Vector ending
points,

4, 13, 22, 31, 40, 49, 58, 67, 76, 85, 94, . . .

the ith occurrence (for i: 0, 1, 2, 3, ...) ends at i * 9 + 4. "

I am assuming that 3 , 11 , 19 , etc are the starting "seeds".
Then only 11, 27, 43, 67, ... 16 * i + 3 have 'true' Type {1,2)
vectors.
The others have (1, x); x > 2. For example; Seed 35 has a
SV =(1,5).

I am not sure what you gain by starting the .Type [1, 2] 's at i * 8
+ 3,
rather than at " i * 16 + 11"? . If you use my suggestion,
then
all of the {1.2) SV's end with " i * 18 + 13".

Reading further on (in Blueprint) , I get the impression that a
Hailstone does NOT have to be an odd number?

Bill J

[mensa...@aol.com]

Yes, the Hailstone does NOT have to be odd, but the
Seed does. The data structure, a simple list [1,2],
has an implied (3n+1) operation preceeding each
integer in the list. Thus, the Seed must be odd.

However, the structure implies nothing about what
comes after the last (n/2) operation of the last
number in the list. It can (and often does) have
one or more (n/2) operations. This system was
developed from the viewpoint of the Hailstone
looking back towards the Seed. So even though
you can say Seed 3 has a [1,4] sequence to 1,
it remains true that when looking backwards from
4 to 3, the vector is [1,2].

What you say above would be true IF we required
that the Hailstone be odd. I could also require
that the sequence end at 1 and take however many
numers end up in the list. But I was looking for
a more general usage than specific. If I required
ending at 1, I couldn't find arbitrary Seeds with
Mersenne Numbers as Hailstones.

By my definition, a [1,5] Sequence Vector is also
a [1,4] Sequence Vector with the same Seed (which
are not required to be unique). It's also a [1,3]
and a [1,2] and a [1,1] Sequence Vector. Now here's
the interesting part: the Z term in the Hailstone
Function Parameters is calculated WITHOUT using
the value of last integer in the list. So any
Sequence Vector of Type [1,n] has an invariant
Z term.

It is certainly possible to build a system where
the sequence can start with an even number, see
the Collatz article in Wikipedia where the structure
in question is called a Parity Sequence. So far, I
have not needed to do so. My updating the section
on "Iterating over the Rationals with Common
Denominator" was based on the similarities between
Sequence Vectors and Parity Sequences.

Another place where a related issue comes up is
the concept of "prime" Sequence Vectors (applies to
Parity Sequences also). Here, "prime" means the
sequence can't be partitioned into identicle
sub-sequences. Thus, [1,2,1,2] isn't "prime" because
it can be partitioned into [1,2][1,2]. In this case,
we say [1,2,1,2] is a second generation [1,2]. And just
as we can choose to only go [1,2] on a sequence that's
really [1,5] doesn't make that choice invalid. Likewise,
we can go bacwards one generation on a sequence that's
2nd or 3rd generation. What we CAN'T do is exceed the
maximum. We can't go forwards [1,7] on a sequence
that's actually [1,5]. We can consider an nth generation
sequence to be any generation from 1 to n but no more.

One place where this matters is the Crossover Point,
the rational that determines whether the Sequence Vector
is a loop cycle. The Crossover Point is invariant by
generation. Thus, [1,2] and [1,2,1,2] and [1,2,1,2,1,2]
and [1,2,1,2,1,2,1,2] ... all have the same Crossover
Point.


>
> Bill J

[bill]

In the following [1,2,1,2] SV, where is the CP?

251 754 377 1,132 566 283 850 425 1,276 638 319

I get X= 64, Y= 81. Z = 85. Then the

CP = 85 / (64 - 81) = - 5 , which is a negative number.

A sequence starting with -5 does indeed loop. But so does
the sequence starting with -7.

The sequence starting with -17, has -25, -37, -41, -55, -61
and -91 as CP's.

In Appendix B, you calculate "a 6th generation Type [1,2] Mersenne
Hailstone".

Calculating backwards from "a", I get the
following value for "d" ;

a = 2 ^ 177149 - 1
b = 2 * (2^177149) - 2
c = 4 * (2^177149) - 4
d = (4/3) * (2^177149) - 1.
If d is an integer, then (2^177149) mod 3 = 0.

But 2^n mod 3 <> 0 for all integers
Therefore, "d" cannot be an integer.

Bill J












>
>
>
> > Bill J

[mensa...@aol.com]

Correct, the Crossover Point just happens to be in the
negative domain. And because it is an integer, it is a
loop cycle in the negative domain. A Hailstone Function
can have only one Crossover Point so no nth generation
type [1,2] can ever loop in the positive domain.

>
> A sequence starting with -5 does indeed loop. But so does
> the sequence starting with -7.

Correct. And if you look closely, you will that they
are the same loop cycle: -5 -> -14 -> -7 -> -20 -> -10 -> -5

By convention, a loop cycle is named for the smallest
magnitude element, so this is called the loop cycle at -5.

>
> The sequence starting with -17, has -25, -37, -41, -55, -61
> and -91 as CP's.

The interesting thing about a loop cycle, regardless of
which number you start with, you always return to that
number. But it is confusing to have multiple names for
the same thing, that's why the "smallest magnitude"
convention was adopted.

A cyclic permutation of a loop is the same loop.

Due to the way X, Y & Z are calculated, different cyclic
permutations result in different Hailstone Functions with
different Crossover Points. Now, if ANY of the Crossover
Points is an integer (regardless of domain), then ALL
the Crossover Points of the cyclic permutations must be
integers also.

All those Crossover Points above represent cyclic
permutations of the same Sequence Vector:

SV CP
[1, 1, 1, 2, 1, 1, 4] -17
[1, 1, 2, 1, 1, 4, 1] -25
[1, 2, 1, 1, 4, 1, 1] -37
[2, 1, 1, 4, 1, 1, 1] -55
[1, 1, 4, 1, 1, 1, 2] -41
[1, 4, 1, 1, 1, 2, 1] -61
[4, 1, 1, 1, 2, 1, 1] -91
[1, 1, 1, 2, 1, 1, 4] -17
[1, 1, 2, 1, 1, 4, 1] -25
[1, 2, 1, 1, 4, 1, 1] -37
[2, 1, 1, 4, 1, 1, 1] -55
[1, 1, 4, 1, 1, 1, 2] -41
[1, 4, 1, 1, 1, 2, 1] -61
[4, 1, 1, 1, 2, 1, 1] -91
[1, 1, 1, 2, 1, 1, 4] -17

And in a loop cycle, the set of Crossover Points is
the set of odd numbers in the Collatz Sequence.

>
> In Appendix B, you calculate "a 6th generation Type [1,2] Mersenne
> Hailstone".
>
> Calculating backwards from "a", I get the
> following value for "d" ;
>
> a = 2 ^ 177149 - 1
> b = 2 * (2^177149) - 2
> c = 4 * (2^177149) - 4
> d = (4/3) * (2^177149) - 1.
> If d is an integer, then (2^177149) mod 3 = 0.
>
> But 2^n mod 3 <> 0 for all integers
> Therefore, "d" cannot be an integer.

There's something wrong there. Let's try it again
using x for the exponent of 2.

a = (2**x - 1)

b = 2*(2**x - 1)

c = 4*(2**x - 1)

d = 4*(2**x - 1) - 1
----------------
3

= 2**(x+2) - 4 - 1
----------------
3

= 2**(x+2) - 5
------------
3

Note: x was an odd number, so x+2 will be an odd number
and any power of 2 to an odd power is 2(mod 3), so in
(mod 3), we have

d = 2(mod 3) - 2(mod 3)
-------------------
0(mod 3)

= 0(mod 3)
--------
0(mod 3)

Therefore, d is an integer.

And x cannot be an even number since 2**x-1 is always
0(mod 3) when x is even and therefore cannot be the
Hailstone of any Sequence Vector since a number 0(mod 3)
cannot have an odd ancestor (Seed).


>
> Bill J
>
>
>
>
>
> > > Bill J

[mensa...@aol.com]

Antoher way to look at this.

Given the graph structure:

d_c
b
a

c MUST be 1(mod 3) since sub-branches (d) can only connect
to 1(mod 3) nodes. So,

d = 1(mod 3) - 1
------------
3

= 0(mod 3)
--------
3

so d is ALWAYS an integer.

Because of the laws of multiplication (or division) by 2
in modulo 3:

succ(0(mod 3)*2) -> 0(mod 3)
succ(1(mod 3)*2) -> 2(mod 3)
succ(2(mod 3)*2) -> 1(mod 3)

EITHER all nodes of a branch are 0(mod 3) OR they alternate
between 1(mod 3) and 2(mod 3).

Because of this and the fact that c MUST be 1(mod 3),
'a' MUST also be 1(mod 3) whenever you see this structure

d_c
b
a

so, ANY Sequence Vector whose last number is even has a
Hailstone that's 1(mod 3).

(If the last number were odd, the Hailstone must be 2(mod 3)
and the structure must be

c
d_b
a

And, by definition, there are NO Hailstones that are 0(mod 3).)


>
>
>
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> > Bill J
>
> > > > Bill J