The article below is about the Collatz conjecture.
The conjecture is also known as the 3n+1 conjecture, the Ulam conjecture, the
SATOCONOR.COM
Paul Stadfeld ‘Blueprint for Failure:
How to Construct a Counterexample to the Collatz Conjecture’ 5.3. (2006)
Full
paper
Blueprint for Failure
How to Construct a
Counterexample to the
Collatz Conjecture
By Paul Stadfeld
For comments: mensanator@aol.com
Version 1.1 Jun 2, 2006
Abstract:
A proof of the Collatz Conjecture
has so far been elusive. But it may easily be resolved by finding a
counterexample. In this paper we study the 3n+C
extensions of the Collatz Conjecture (most of which fail) to learn what a
counterexample to 3n+1 would look
like if it exists.
The study is based on the
novel approach of creating a data structure (Sequence Vector) based on relative
Collatz sequences without regard to their values. Functions are then derived
from the data structure to allow algebraic analysis of the Collatz sequences.
These functions (Hailstone and
Crossover Point) are the key to understanding why 3n+1 works and why 3n+5
fails. This understanding will then permit the construction of counterexamples
as opposed to simple brute force searching. Furthermore, we will find that
counterexamples must meet stringent structural requirements (the Blueprint)
that will tell us when a counterexample CANNOT be constructed, which has deep
implications concerning proving the Collatz Conjecture.
We will also gain new insight
on the relationship between the positive and negative domain with regard to 3n+1,
that they are not separate problems but are actually a single problem from the
viewpoint of relative Collatz sequences. This, too, has deep implications, for
certain proofs in one domain imply proof in the other which seems to have been
overlooked up to now with startling consequences.
The Collatz Conjecture will
NOT be proved or disproved in this paper. What will be shown are tools and
insights that may further the search for an actual proof.
1. Introduction
The Collatz Conjecture1,2 states that for
any positive integer n, a
sequence created using the rules
1)
successor(n)
-> n/2 if n is even
2)
successor(n)
-> 3*n + 1 if n is odd
always reaches the loop
cycle 4, 2, 1, 4, 2, 1, … . Studied extensively, the
conjecture appears to be true, but no proof has been found. It would, however,
be easy to prove false. A counterexample can be readily verified. The fact that
no counterexample has been located does not necessarily mean one does not
exist. Current thinking is a counterexample in the form of a non-trivial loop
cycle would have hundreds of thousands1 of elements and may very
well be beyond computational power for any foreseeable future.
That is, if we simply do a brute force search. Large
numbers are no barrier if we have some blueprint that tells us how to build
them. For example, the 1st 6th Generation Type [1,2]
Mersenne Hailstone (see Appendix B) has 53,328 decimal digits and
could never be found by simple brute force. Yet, ith, kth Generation
Type [1,2] Mersenne Hailstones have a simple blueprint:

So the smart way to find a counterexample is to discover
a blueprint that will take us directly to a counterexample even if it’s
unimaginably huge.
2. The Blueprint
A blueprint has been found together with the tools
to construct a counterexample. They are based on how one navigates through the
Collatz Tree3. The system presented here has deep implications, not
just for tree navigation, but for the ultimate truth of the entire conjecture.
It all starts with a data structure designed to efficiently describe relative
pathways in the Collatz tree.
2.1. Sequence Vector
To navigate from one node on the Collatz tree to any
other, one follows a sequence of Rules defined by the Collatz Conjecture.
Although proper Collatz sequences always move right and down on the tree, there
is no mathematical reason why one can't also move up and left (provided the
moves are legal) by inverting the standard Collatz Rules.
Thus, we have
Rule1: n/2
Rule2: n*3+1
InverseRule1: n*2
InverseRule2: (n-1)/3
A rule based data structure describing a Collatz
sequence is called a Sequence Vector.
By definition, a Sequence Vector always begins with
an odd number and always ends with at least one iteration of Rule1 (the
terminating number may be even or odd).
For example, the sequence starting on 27 and ending
on 31
27_82
41_124
62
31
is produced by the proper sequence
[Rule2, Rule1, Rule2, Rule1, Rule1]
We can abbreviate this by simply counting how many
consecutive iterations there are of the same rule.
[1*Rule2, 1*Rule1, 1*Rule2, 2*Rule1]
Since a proper Collatz sequence cannot have more
than 1 consecutive iteration of Rule2, we can further abbreviate the sequence
by assuming every block of Rule1 is separated by exactly one Rule2 and only
showing the Rule1 blocks (this assumption is why the Sequence Vector is defined
to start on an odd number and end with a Rule1).
[1*Rule1, 2*Rule1]
Now, since every item shown is a block of Rule1, we
can further abbreviate by just showing the iteration count of each block.
[1, 2]
In the above example, the sequence from 27 to 31 is
just a small portion of the complete sequence from 27 to its normal termination
at 1. The full sequence vector would be
[1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1,
1, 1, 2, 3, 1, 1, 2, 1,
2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 4, 2, 2, 4,
3, 1, 1, 5, 4]
Sequence Vectors can describe a full sequence from n
to 1 or any sub-sequence.
Thus, every proper Collatz sequence can be described
by a simple list of integers >0 (0 is not a valid Sequence Vector element
since the implied Rule2 to the left of each element would result in two
consecutive iterations of Rule2, which is not permitted).
Sequence Vectors are not value-centric. They
describe the structure of the sequence without specifying values. The sequence
from 27 to 31 is a
Type [1, 2] Sequence Vector
since the [1, 2] pattern occurs many times on the
Collatz tree (it occurs seven times in the full sequence from 27 to 1):
27_82
41_124
62
31
. . .
107_322
161_484
242
121
. . .
91_274
137_412
206
103
. . .
155_466
233_700
350
175
. . .
395_1186
593_1780
890
445
. . .
251_754
377_1132
566
283_850
425_1276
638
319
(the last two
occurrences of the seven [1,2] patterns are consecutive)
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
![]()
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
![]()
These will be derived from any arbitrary Sequence
Vector by the Hailstone Function.
2.2. Hailstone Function
For a generic Collatz sequence such as
g_e
d_c
b
a
the Sequence Vector is [1, 2].
We find the Hailstone Function by playing the
Sequence Vector backwards using the InverseRules:

working out all the algebra, we get
![]()
Now we need to solve the equation by finding an
integer a>0 such that g is also an integer:
for a=1
for a=2
for a=3
for a=4
![]()
So 4 is the Hailstone derived from Sequence Vector
[1, 2] whose Seed is 3.
Check:
3_10
5_16
8
4
The Hailstone Function
![]()
can also be written using powers of 2 and 3:
![]()
Each element i of the Sequence Vector
(sv) applies the successor function
![]()
so if the Hailstone Function of [1, 2] is
![]()
then the Hailstone Function for [3, 1, 2] would be
![]()
If we collect all the numerator terms inside the
parentheses and call the collection Z, then every Sequence Vector has a
Hailstone Function of the form
![]()
There is a relatively simple algorithm to calculate
the values of X,Y,Z from a Sequence
Vector so that we don't have to do all the algebra:
Hailstone Function Parameters Algorithm12
given a Sequence Vector of n elements [s0,
s1, s2, ... sn-1]
example: for [1, 2, 3, 4] n=4
X = 2 raised to the power of the sum of the
elements
![]()
Y = 3 raised to the power of the count of the
elements
![]()
Z is a bit tricky, Z is the sum of n terms,
each of
which is a power of 3 multiplied by a power
of 2
·
create a
template of n terms
![]()
·
the exponents
of the powers of three count from 0 to n-1 from left to right
![]()
·
the
exponents of the powers of 2 starting from the left are
·
the sum
of all the elements except the last 1 (1+2+3)
·
the sum
of all the elements except the last 2 (1+2)
·
the sum
of all the elements except the last 3 (1)
·
the sum
of all the elements except the last 4 (0)

for [1, 2, 3, 4] we get the Hailstone Function
![]()
for which a=1
and g=11
11_34
17_52
26
13_40
20
10
5_16
8
4
2
1
Solving the Hailstone function gives the first of
infinite integer solutions. And if we know the first one (a0, g0), we can find any solution.
We saw in Section 2.1 that for Sequence Vector [1, 2], the ith seed is i*8+3
and the ith hailstone is i*9+4. We can generalize these from the
Hailstone Function:
To find (a0, g0) for any Sequence Vector, we can reformulate the
general Hailstone Function
as a problem in Linear Congruence4,5: if g is an
integer, then (X*a-Z) is divisible by
Y, in other words
Linear congruence problems can be solved by using
the Extended Euclidean Algorithm6,7 to find the modular inverse
of X and Y:
A linear congruence can be solved if and only if the
Greatest Common Divisor10,11 of X and Y divides Z. With X being a power of 2 and Y
a power of 3, GCD(X, Y) is always 1
and will always divide Z. This means
that
·
every Linear Congruence derived from a Hailstone Function is solvable
·
every Hailstone Function has a first solution (a0)
·
every Hailstone Function has an infinite number of solutions (ai)
·
every legal Sequence Vector appears on the Collatz Tree infinitely many
times
And from the Hailstone Function we calculate the
Crossover Point.
2.3. Crossover Point
From any legal Sequence Vector, there exists a
Hailstone Function with parameters X,Y,Z.
From the Hailstone Function Parameters, we can compute the Crossover Point.
First, note that the Hailstone Function is the
equation of a straight line (with slope X/Y
and intercept -Z/Y). Because X is a power of 2 and Y is a power of 3, the slope can never
be 1, and so it cannot be parallel to the identity line whose slope is exactly
1. It must, therefore, intercept the identity line. This interception is called
the Crossover Point.
Second, at the intercept point, since this is on the
identity line, the function of a must
be equal to a, so that
becomes
Thus, we make the following derivation:
But a is the
point of intersection, so it is the Crossover Point (CP), so
Every Sequence Vector/Hailstone Function has a
Crossover Point, but if and only if the Crossover Point is an integer, then the
Sequence Vector is a loop cycle! Every finite Collatz sequence has a first
number which we’ll call the Seed. Every finite sequence has a last number which
we’ll call the Hailstone. A Sequence Vector describes a loop cycle if
Seed=Hailstone (i.e., the sequence returns to its starting value). The
Crossover Point is the rational number Z/(X-Y). If that rational number does
not reduce to an integer, the Sequence Vector cannot be a loop cycle since a
Collatz sequence contains only integers (i.e., the place where Seed=Hailstone
cannot occur in a Collatz sequence). If it is
an integer, then from the fact that this integer is also on the line of the
identity function, that point represents
an integer where Seed=Hailstone and thus, implies that the Sequence Vector is a
loop cycle.
For example, take the Sequence Vector [2]
g_c
b
a
The associated Hailstone Function is
so
which means Sequence Vector [2] is a loop cycle.
1_4
2
1
By convention, for those Sequence Vectors that form
loop cycles, the odd number of smallest magnitude in the loop cycle is called
the Loop Point, so a=1 is the Loop Point of Sequence Vector [2].
If the Collatz Conjecture is true, there is only one
positive Loop Point (1) whose Sequence Vector is [2] and it forms the root of
the Collatz Tree. Any Loop Point/Sequence Vector other than (1)/[2] would be a
counterexample.
2.4. 3n+C Extensions
The Crossover Point becomes our blueprint for
studying how the Collatz Conjecture fails. Unfortunately, the only example of a
Loop Point we have (Sequence Vector [2]) is trivial. But by extending the
conjecture to 3n+C (where C is any odd positive integer), we can
find numerous counterexamples to study via the Crossover Point.
First of all, a quick review of various values of C reveals that:
(See Section 3 for proof of these points)
Second, we need to make the following adjustments:
The
Hailstone Function Parameters X,Y,Z
are invariant under the extensions.
From
the blueprint (Crossover Point), we will be able to determine exactly why the 3n+C systems work and why they fail.
Once we fully understand the failure mechanism, we may be able to extend it
back to 3n+1 to construct a
counterexample.
3. The Essence of Failure
There are two ways the conjecture can fail:
The possibility of a sequence going to infinity is
beyond the scope of this paper and will not be mentioned again. Our blueprint
for construction of a counterexample applies to the second case. But note that
we specify Sequence Vector [2] as the sequence termination instead of 1. Under
the 3n+C extensions, the trivial loop
cycle has an invariant Sequence Vector. But the actual value of the Loop Point
does vary. It is C.
C_4C
C
A terminating value of 1 is a value-centric
specification that applies only to 3n+1.
In
addition, the trivial loop whose Sequence Vector is [1,2] and whose root is C, the tree built from that root is identical
in structure to the 3n+1 tree:
3C_10C
5C_16C
C
Note that the coefficients of
C are the same as the nodes on the 3n+1 tree. So on the trivial loop tree
of any 3n+C system, only multiples of
C are on this tree. For any 3n+C system n=1 is never a multiple
of C, so n=1 is never found on the
trivial tree. Thus, n=1 is always found on a non-trivial tree. This proves the
conjecture is false for any C not a power
of 3 (as observed in Section 2.4).
Under 3n+C,
any Sequence Vector other than [2] whose Extended Crossover Point is an integer
is a loop cycle and makes the conjecture false for that value of C. An integer implies that the
denominator of the Extended Crossover Point
is a unit, either because
3.1. Trivial Loops
When X-Y
is a unit (as is the case for Sequence Vector [2]), CP is trivially an integer
and forms a trivial loop cycle. Trivial loop cycles are determined solely by
the Sequence Vector and are common to all 3n+C
systems.
Thus, every 3n+C
system has a trivial Loop Point at C.
And no others since 4-3 is the only pair of powers of 2 and powers of 3 whose
difference is +1.
3.2. Non-Trivial Loops
That leaves factor canceling as the only means of
failure. And there’s plenty of it.
Example 3n+5:
trivial loop at 5: 5, 20, 10, 5
non-trivial loop at 1: 1, 8, 4, 2, 1
non-trivial loop at 19: 19, 62, 31, 98, 49, 152, 76, 38,19
non-trivial loop at 23: 23, 74, 37, 116, 58, 29, 92, 46, 23
non-trivial loop at 187: 187, 566, 283, 854, 427, 1286, 643, 1934,
967, 2906, 1453,
4364, 2182, 1091, 3278,
1639, 4922,
2461, 7388, 3694, 1847, 5546,
2773, 8324,
4162, 2081, 6248, 3124, 1562,
781, 2348, 1174,
587, 1766, 883, 2654, 1327,
3986, 1993,
5984, 2992, 1496, 748, 374, 187
non-trivial loop at 347: 347, 1046, 523, 1574, 787, 2366, 1183, 3554,
1777, 5336, 2668, 1334, 667, 2006, 1003,
3014, 1507,
4526, 2263, 6794, 3397, 10196,
5098, 2549,
7652, 3826, 1913, 5744, 2872,
1436, 718, 359,
1082, 541, 1628, 814, 407,
1226, 613, 1844,
922, 461, 1388, 694, 347
3n+5 has 5 non-trivial Loop Points: 1, 19, 23, 187 and
347. Picking one of these Loop Points, say 187, its Sequence Vector is
[1,1,1,1,1,2,1,1,2,1,2,3,2,1,1,1,5]
From which we
get the Extended Hailstone Function
And Extended Crossover Point (C is kept
separate in the numerator to illustrate factor canceling)
When the common factors 5, 71 and 14303 are
cancelled, we are left with 11*17=187 which is the Loop Point for this loop
cycle. But since the Sequence Vector is a loop cycle, every cyclic permutation
of the Sequence Vector must also have an integer Extended Crossover Point. And
since in a cyclic permutation, the count and sum of the elements are invariant,
only Z changes in the Extended Crossover Point.
[1,1,1,1,2,1,1,2,1,2,3,2,1,1,1,5,1] (283*71*14303)*(5)/(5*71*14303)
[1,1,1,2,1,1,2,1,2,3,2,1,1,1,5,1,1] (7*61*71*14303)*(5)/(5*71*14303)
[1,1,2,1,1,2,1,2,3,2,1,1,1,5,1,1,1] (643*71*14303)*(5)/(5*71*14303)
[1,2,1,1,2,1,2,3,2,1,1,1,5,1,1,1,1] (967*71*14303)*(5)/(5*71*14303)
[2,1,1,2,1,2,3,2,1,1,1,5,1,1,1,1,1] (1453*71*14303)*(5)/(5*71*14303)
[1,1,2,1,2,3,2,1,1,1,5,1,1,1,1,1,2] (1091*71*14303)*(5)/(5*71*14303)
[1,2,1,2,3,2,1,1,1,5,1,1,1,1,1,2,1]
(11*149*71*14303)*(5)/(5*71*14303)
[2,1,2,3,2,1,1,1,5,1,1,1,1,1,2,1,1]
(23*107*71*14303)*(5)/(5*71*14303)
[1,2,3,2,1,1,1,5,1,1,1,1,1,2,1,1,2] (1847*71*14303)*(5)/(5*71*14303)
[2,3,2,1,1,1,5,1,1,1,1,1,2,1,1,2,1] (47*59*71*14303)*(5)/(5*71*14303)
[3,2,1,1,1,5,1,1,1,1,1,2,1,1,2,1,2] (2081*71*14303)*(5)/(5*71*14303)
[2,1,1,1,5,1,1,1,1,1,2,1,1,2,1,2,3] (11*71*71*14303)*(5)/(5*71*14303)
[1,1,1,5,1,1,1,1,1,2,1,1,2,1,2,3,2] (587*71*14303)*(5)/(5*71*14303)
[1,1,5,1,1,1,1,1,2,1,1,2,1,2,3,2,1] (883*71*14303)*(5)/(5*71*14303)
[1,5,1,1,1,1,1,2,1,1,2,1,2,3,2,1,1] (1327*71*14303)*(5)/(5*71*14303)
[5,1,1,1,1,1,2,1,1,2,1,2,3,2,1,1,1] (1993*71*14303)*(5)/(5*71*14303)
Although Z changes for every Sequence Vector,
note that every Z contains the factors 71 and 14303. Combined with the 5
factor from C, every denominator becomes 1 and every Extended Crossover
Point becomes an integer. And if you check, you’ll find that the 17 integer
Extended Crossover Points are the 17 odd numbers in the original sequence. The
convention of taking the smallest magnitude number of the loop cycle as the
Loop Point is so that we don’t have to constantly list all the cyclic
permutations. When we refer to a Loop Point and its Sequence Vector, it is
assumed that we are collectively referring to all the cyclic permutations
unless otherwise noted.
Equally important, Z never has 5 as a factor
(otherwise, this Sequence Vector would be a loop in 3n+1 also). It is
the combined factors of Z and C that create non-trivial loops.
This immediately leads to the
Z Conjecture The factors of Z alone cannot cancel
all the factors of the denominator,
C must contribute at least one
useful factor (where “useful” means >3).
A factor must be “useful”
because the denominator, being the difference between a power of 2 and a power
of 3 can never have a 3 as a factor. This explains why when C is a power
of 3, it doesn’t fail the conjecture. Or perhaps we should say it has exactly
the same loop cycles as 3n+1 because 1 has no useful factors either. If the
conjecture is true for 3n+1, it’s true for all C that are a power
of 3 (as observed in section 2.4).
3.3. Factor Congruence
If we are given C, we can now construct
Sequence Vectors that are non-trivial loops. This is done via Factor Congruence
matching. For example, suppose C is a prime (41). Being prime, C
has only one factor: 41. This factor is the only one available to cancel the
factors uncancelled by Z. So for CP to be an integer, it is necessary
that X-Y has a factor of
From The Hailstone Function, we know that X is a
power of 2 and Y is a power of 3. So that
Also from the Hailstone Function, we have that
where sv
is the Sequence Vector. Thus, only those Sequence Vectors where a power of 2 is
the same congruence class mod 41 as a power of 3 can be a potential loop
(potential because Z must supply the rest of the factors).
With our example 41, the congruence classes are
The only classes they have in common are:
{1,9,32,40}
|
p or q |
2p (mod 41) |
3q (mod 41) |
|
|
(pattern repeats every 20) |
(pattern repeats every 8) |
|
0 |
1 |
1 |
|
1 |
2 |
3 |
|
2 |
4 |
9 |
|
3 |
8 |
27 |
|
4 |
16 |
40 |
|
5 |
32 |
38 |
|
6 |
23 |
32 |
|
7 |
5 |
14 |
|
8 |
10 |
1 |
|
9 |
20 |
3 |
|
10 |
40 |
9 |
|
11 |
39 |
27 |
|
12 |
37 |
40 |
|
13 |
33 |
38 |
|
14 |
25 |
32 |
|
15 |
9 |
14 |
|
16 |
18 |
1 |
|
17 |
36 |
3 |
|
18 |
31 |
9 |
|
19 |
21 |
27 |
|
20 |
1 |
40 |
|
21 |
2 |
38 |
|
22 |
4 |
32 |
|
23 |
8 |
14 |
Table 1: Congruence classes of
powers of 2 & 3 mod 41.
So p and q must be chosen so that the
congruence classes match. The first such match is p=0 and q=0 match.
But a Sequence Vector cannot have 0 length or have a 0 sum, so that pairing is
disallowed. To be a legal Sequence Vector p must be >= q. And
to be a positive loop, it must be that p>q*(log 3/log 2). Under those restrictions, the first available
pair is pq(15,9), where both congruences classes are 9. The next
available one is pq(10,4) for congruence class 40. The pair pq(5,6)
cannot form a loop even though they are both congruence class 32 because p
must be greater than q (in other words, there are no legal Sequence
Vectors with more 3n+C steps than n/2 steps). And any loop in pq(20,16)
would be in the negative domain since p< q*(log 3/log 2). A
good candidate would be pq(20,8).
But there are many different Sequence Vectors where p=20
and q=8. Constructing them is equivalent to partitioning 20 objects into
8 containers such that each container has at least 1 object. Using an algorithm13 for counting and generating such partitions, we can
search every partition for an integer Crossover Point. For (20,8) the number of
partitions is 50,388 and we have to check them all because Factor Congruence
matching is a necessary but not sufficient condition for loop cycle formation.
Running the Partition Generator for p=20 q=8
and testing for integer Crossover Points yields
[1,1,9,2,1,3,1,2] 47
[1,2,1,1,9,2,1,3] 19
[1,3,1,2,1,1,9,2] 11
[1,9,2,1,3,1,2,1] 91
[2,1,1,9,2,1,3,1] 49
[2,1,3,1,2,1,1,9] 1
[3,1,2,1,1,9,2,1] 37
[9,2,1,3,1,2,1,1] 157
So there are 8 Sequence Vectors that produce loops. But
upon close inspection, we see that all 8 are cyclic permutations of the same
numbers. That means the 8 integer Crossover Points are the 8 odd numbers of a
single loop cycle:
1_44
22
11_74
37_152
76
38
19_98
49_188
94
47_182
91_314
157_512
256
128
64
32
16
8
4
2
1
1 is a Loop Point in 3n+41 whose Sequence
Vector [2,1,3,1,2,1,1,9] is a non-trivial loop cycle.
See Appendix A
for an analysis of a composite C.
Using the Extended Crossover Point as our blueprint
and Factor Congruence matching as our tools, we now have the means to construct
counterexamples to the 3n+C extensions. And although the lack of useable
factors prevents us from constructing any counterexample to 3n+1, it
would be premature to leap to the conclusion that 3n+1 must be true. Up
to this point, we have been assuming the Z Conjecture has been true.
Actually, it doesn’t matter to Factor Congruence matching whether the Z
Conjecture is true or not. If it’s false, the loop cycles constructed by
Factor Congruence matching are still loop cycles, it’s just that they won’t be
the only non-trivial loop cycles. For 3n+1 to be true, we must show that
the Z Conjecture is true so that when we say no loop cycles can be
constructed by Factor Congruence matching, then no loop cycles can be
constructed period.
For that we will need to make one more extension.
4. The Negative Domain
The Collatz Conjecture is specified as a problem in
the positive domain. We can, of course, extend it to the negative domain just
as we extended 3n+1 to 3n+C. We quickly discover that the
negative domain is rather uninteresting in that the conjecture fails for every C
in 3n+C. Every 3n+C (including 3n+1) has Loop Points at –C,
But we’re looking to understand why it fails, that’s
how we were able to derive the Blueprint for Failure in the preceding sections.
And although there is nothing in the negative domain that teaches us anything
about trivial loops or Factor Congruence that we don’t already know, it does
contain this bombshell:
Proof that the Z Conjecture is false!
The Loop Points at –C and –5C (with
Sequence Vectors [1] and [1,2] respectively) are trivial because from their
Crossover Points, (X-Y) starts out as a unit (-1). There is no need for
factor canceling.
But –17C is not trivial and forms a
loop cycle even when C=1:
g_s
r_q
p_o
n_m
l
k_j
i_h
f_e
d
c
b
a
![]()
-17_-50
-25_-74
-37_-110
-55_-164
-82
-41_-122
-61_-182
-91_-272
-136
-68
-34
-17
The Crossover Point is an integer because the
factors of Z cancel all the factors of X-Y proving that the Z
Conjecture is actually false. It is possible for all the factors of X-Y
to be cancelled without C contributing any useful factors. So even though
the counterexample –17 is in the negative domain, the fact that it exists at
all has deep implications for the positive domain. We still are unable to
construct a counterexample to 3n+1 via Factor Congruence matching, but
we cannot rule out that a construction based on the Z Conjecture being
false is possible.
5. Conclusions
As we have seen, a counterexample must be one of
three types:
The Trivial loops for 3n+C are +C, -C
and –5C. Factor Congruence loop cycles have been proven impossible to
construct for 3n+1. That leaves the Z Conjecture as the only way 3n+1
can be false.
Unfortunately, we have no blueprint that can exploit
the Z Conjecture being false, so we cannot construct counterexamples
even knowing such a counterexample exists. And the fact that the only known
counterexample to the Z Conjecture is in the negative domain doesn’t
mean there can’t be one in the positive domain. Factor canceling is independent
of the domain, so we can’t say it’s impossible for there to be a counterexample
in the positive domain of 3n+1.
Thus, the Collatz Conjecture continues to be as elusive
as ever.
-o0o-
3n+
From Section 3.3, we found the
non-trivial Loop Point by searching only those Sequence Vectors whose Hailstone
Function parameters had matching congruence classes mod C.
But not all matching parameters
generated Loop Points. And what if C were composite, say 85085. Why 85085?
Because it is the composite of the 5 primes > 3: 85085 = 5 * 7 * 11 * 13 *
17. Thus, there will be a lot of factors available to the Crossover Point
Function to form Loop Points.
Running the collatz_C function14
for C=85085 over the range -2000000 to +
When C is composite, some of its
factors may not be necessary to form a Loop Point. In such cases, the unused
factors will appear as the greatest common divisor of the odd numbers in the
loop sequence. So the gcd will tell us the divisor whose congruence classes
must be matched:
divisor = 85085/gcd
And just as factor congruence
matching is necessary but not sufficient for Loop Points, there can be any
number of Loop Points found for any given divisor.
Since there are 5 factors in
85085 any of which may or may not be cancelled, there are 32 possible
gcd/divisor pairs:
fac5 fac7
fac11 fac13 fac17
divisor gcd
1 1
1 1 1
1 85085
1 1
1 1 17
17 5005
1 1
1 13 1
13 6545
1 1
1 13 17
221 385
1 1
11 1 1
11 7735
1 1
11 1 17
187 455
1 1
11 13 1
143 595
1 1
11 13 17
2431 35
1 7
1 1 1
7 12155
1 7
1 1 17
119 715
1 7
1 13 1
91 935
1 7
1 13 17
1547 55
1 7
11 1 1
77 1105
1 7
11 1 17
1309 65
1 7
11 13 1
1001 85
1 7
11 13 17
17017 5
5 1
1 1 1
5 17017
5 1
1 1 17
85 1001
5 1
1 13 1
65 1309
5 1
1 13
17 1105 77
5 1
11 1 1
55 1547
5 1
11 1 17
935 91
5 1
11 13 1
715 119
5 1
11 13 17
12155 7
5 7
1 1 1
35 2431
5 7
1 1 17
595 143
5 7
1 13 1
455 187
5 7
1 13 17
7735 11
5 7
11 1 1
385 221
5 7
11 1 17
6545 13
5 7
11 13 1
5005 17
5 7
11 13 17
85085 1
The 156 Loop Points are
distributed such that all possible gcds are represented. So we have to
calculate matches for each of the 32 divisors.
But we're just trying to verify
that Factor Congruence matching actually works, so instead, we'll take the
known Loop Points and see if they hold up to the congruence matching scrutiny.
See Table A1.
Which they do. That means we can
verify the Loop Points with the partition generator program. Generating every
partition of P items in Q groups is
equivalent to creating every possible Sequence Vector whose sum is P and count
is Q.
So, grabbing a P,Q pair from the
stats table
python
partition_generator.py 4 3
searching: 3 Sequence
Vectors
[1,1,2] -146965 *
[1,2,1] -177905
[2,1,1] -224315
we get three Sequence Vectors
with integer Crossover Points (the one marked * is the Loop Point shown on the
stats table).
But note that the three vectors
are all cyclic permutations of the same numbers. that means they are all part
of the same loop:
-146965__-355810
-177905__-448630
-224315__-587860
-293930
-146965
So the partition generator will
return exactly Q Sequence Vectors for each Loop Point found.
Exactly?
python
partition_generator.py 4 2
searching: 3 Sequence
Vectors
[1,3] 60775
[3,1] 133705
[2,2] 85085
Here Q is 2 but we got 3 vectors.
But that's because the third vector is a different loop cycle. But it only has
1 vector. That's because it's actually a second generation type [2] vector, so
we'll mark these false positives with an x. The multi-generation Loop Points
will always have less than Q vectors.
python
partition_generator.py 6 4
searching: 10 Sequence Vectors
[1,1,1,3] -325325 *
[1,1,3,1] -445445
[1,3,1,1] -625625
[3,1,1,1] -895895
[1,1,2,2] -365365 *
[1,2,2,1] -505505
[2,2,1,1] -715715
[2,1,1,2] -515515
[1,2,1,2] -425425 x
[2,1,2,1] -595595
Here the stats table shows two Loop
Points for P=6 Q=4 and that's what we got, not counting the multi-generation
vector.
More interesting is the set of Loop
Points found for P=8 Q=5
python
partition_generator.py 8 5
searching: 35 Sequence
Vectors
[1,1,1,1,4] 1380995 *
[1,1,1,4,1] 2114035
[1,1,4,1,1] 3213595
[1,4,1,1,1] 4862935
[4,1,1,1,1] 7336945
[1,1,1,2,3] 1485715 *
[1,1,2,3,1] 2271115
[1,2,3,1,1] 3449215
[2,3,1,1,1] 5216365
[3,1,1,1,2] 3933545
[2,1,1,1,3] 2231845
[1,1,1,3,2] 1695155 *
[1,1,3,2,1] 2585275
[1,3,2,1,1] 3920455
[3,2,1,1,1] 5923225
[1,1,2,1,3] 1642795 *
[1,2,1,3,1] 2506735
[2,1,3,1,1] 3802645
[1,3,1,1,2] 2873255
[3,1,1,2,1] 4352425
[1,1,2,2,2] 1852235 *
[1,2,2,2,1] 2820895
[2,2,2,1,1] 4273885
[2,2,1,1,2] 3226685
[2,1,1,2,2] 2441285
[1,2,1,1,3] 1878415 *
[2,1,1,3,1] 2860165
[1,1,3,1,2] 2166395
[1,3,1,2,1] 3292135
[3,1,2,1,1] 4980745
[1,2,1,2,2] 2087855 *
[2,1,2,2,1] 3174325
[1,2,2,1,2] 2402015
[2,2,1,2,1] 3645565
[2,1,2,1,2] 2755445
Once separated into groups of
cyclic permutations, we find we have seven groups - seven Loop Points. But the
stats table only lists six. Look at the Loop Point on the last loop: 2087855.
The original Loop Point search
only extended to 2000000, so the partition generator found one missed in the
original search. The Sequence Vector search will always find all the Loop
Points...
...provided we know what P and Q
to test. Doubling the collatz_C test range, we find yet another Loop Point:
3182179.
This one didn't get picked up in
the Sequence Vector search because its loop has P=27 Q=17 and that combination
hadn't appeared before in the stats table. That warrants another run of the
partition generator.
python
partition_generator.py 27 17
searching: 5311735
Sequence Vectors
[1,1,1,1,1,2,1,1,2,1,2,3,2,1,1,1,5]
3182179 *
[1,1,1,1,2,1,1,2,1,2,3,2,1,1,1,5,1]
4815811
[1,1,1,2,1,1,2,1,2,3,2,1,1,1,5,1,1]
7266259
[1,1,2,1,1,2,1,2,3,2,1,1,1,5,1,1,1]
10941931
[1,2,1,1,2,1,2,3,2,1,1,1,5,1,1,1,1]
16455439
[2,1,1,2,1,2,3,2,1,1,1,5,1,1,1,1,1]
24725701
[1,1,2,1,2,3,2,1,1,1,5,1,1,1,1,1,2]
18565547
[1,2,1,2,3,2,1,1,1,5,1,1,1,1,1,2,1]
27890863
[2,1,2,3,2,1,1,1,5,1,1,1,1,1,2,1,1]
41878837
[1,2,3,2,1,1,1,5,1,1,1,1,1,2,1,1,2]
31430399
[2,3,2,1,1,1,5,1,1,1,1,1,2,1,1,2,1]
47188141
[3,2,1,1,1,5,1,1,1,1,1,2,1,1,2,1,2]
35412377
[2,1,1,1,5,1,1,1,1,1,2,1,1,2,1,2,3]
13290277
[1,1,1,5,1,1,1,1,1,2,1,1,2,1,2,3,2]
9988979
[1,1,5,1,1,1,1,1,2,1,1,2,1,2,3,2,1]
15026011
[1,5,1,1,1,1,1,2,1,1,2,1,2,3,2,1,1]
22581559
[5,1,1,1,1,1,2,1,1,2,1,2,3,2,1,1,1]
33914881
[1,2,1,2,2,1,1,1,1,3,1,1,1,1,2,2,4]
6109103
[2,1,2,2,1,1,1,1,3,1,1,1,1,2,2,4,1]
9206197
[1,2,2,1,1,1,1,3,1,1,1,1,2,2,4,1,2]
6925919
[2,2,1,1,1,1,3,1,1,1,1,2,2,4,1,2,1]
10431421
[2,1,1,1,1,3,1,1,1,1,2,2,4,1,2,1,2]
7844837
[1,1,1,1,3,1,1,1,1,2,2,4,1,2,1,2,2]
5904899 *
[1,1,1,3,1,1,1,1,2,2,4,1,2,1,2,2,1]
8899891
[1,1,3,1,1,1,1,2,2,4,1,2,1,2,2,1,1]
13392379
[1,3,1,1,1,1,2,2,4,1,2,1,2,2,1,1,1]
20131111
[3,1,1,1,1,2,2,4,1,2,1,2,2,1,1,1,1]
30239209
[1,1,1,1,2,2,4,1,2,1,2,2,1,1,1,1,3]
11350339
[1,1,1,2,2,4,1,2,1,2,2,1,1,1,1,3,1]
17068051
[1,1,2,2,4,1,2,1,2,2,1,1,1,1,3,1,1]
25644619
[1,2,2,4,1,2,1,2,2,1,1,1,1,3,1,1,1]
38509471
[2,2,4,1,2,1,2,2,1,1,1,1,3,1,1,1,1]
57806749
[2,4,1,2,1,2,2,1,1,1,1,3,1,1,1,1,2]
43376333
[4,1,2,1,2,2,1,1,1,1,3,1,1,1,1,2,2]
32553521
And, lo and behold, another Loop
Point is found: 5904899.
Unfortunately, verification was
halted before the end of the stats table was reached. For the simple reason
that as P and Q get large, Sequence Vector searching becomes intractable:
python
partition_generator.py 48 24
searching: 16123801841550
Sequence Vectors
python
partition_generator.py 52 24
searching: 196793068630200
Sequence Vectors
python
partition_generator.py 56 32
searching:
2488589544741300 Sequence Vectors
python
partition_generator.py 60 30
searching:
59132290782430712 Sequence Vectors
.
.
.
python
partition_generator.py 492 264
searching:
66189415264331559482776409694993032407028709677550596291300192890141937773498314175433116122939513634124491233746912456893016976209252459301489030
Sequence Vectors
And there may be some more hiding
in other congruence match pairs that aren't on the original stats table. Noting
that the congruence generations reach as high as 13, 4 there may be others.
And if there are, they'll fall
into their proper place in the Factor Congruence table.
-------------------------------------------------------------
Table A1: Factor Congruence Statistics for C
= 85085
Column Definition
------
---------------------------------------------------------------
Loop Point Smallest (magnitude) odd number in the loop
sequence.
gcd Greatest Common Divisor of the odd
numbers in the loop sequence.
Composite of the factors of C that
DO NOT cancel anything in the
denominator of the Crossover Point
Function (X-Y).
Gcd*divisor = 85085.
divisor Composite of the factors of C that DO
cancel factors in the
denominator of the Crossover Point
Function (X-Y).
Gcd*divisor = 85085.
P From the Hailstone Function, X is
always a power of 2.
P is the exponent, i.e., X = 2**P.
P is the sum of elements in the
Sequence Vector, for [1,2,3,4], P = 10
Q From the Hailstone Function, Y is
always a power of 3.
Q is the exponent, i.e., Y = 3**Q.
Q is the count of elements in the
Sequence Vector, for [1,2,3,4], Q = 4
ccP Congruence class of 2**P (mod
divisor).
For Factor Congruence matching, ccP
must equal ccQ.
ccQ Congruence class of 3**Q (mod
divisor).
For Factor Congruence matching, ccP
must equal ccQ.
cclenP Length of the set of congruence class of
2**P (mod divisor).
Congruence class repeats at this
frequency.
ccP = cclenP*n + gen1P for n=0,1,2,3...
cclenQ Length of the set of congruence class of
3**Q (mod divisor).
Congruence class repeats at this
frequency.
ccQ = cclenQ*n + gen1Q for n=0,1,2,3...
gen1P The first legal power of 2 that is
congruence class ccP.
(P=0 is invalid for Sequence
Vectors)
gen1Q The first legal power of 3 that is
congruence class ccQ.
(Q=0 is invalid for Sequence
Vectors)
genP Congruence class generation:
IF P>cclenP THEN
genP = (P - gen1P)/cclenP
ELSE
genP = 1
genQ Congruence class generation:
IF Q>cclenQ THEN
genQ = (Q - gen1Q)/cclenQ
ELSE
genQ = 1
notes Where noted, Factor Congruence matching
does not apply.
These are the loops (+C -C
No factors are cancelled in the
Crossover Point Function and as a result,
the gcd for these will be 85085.
Loop Point gcd
divisor P Q
ccP ccQ cclenP cclenQ gen1P gen1Q genP genQ notes
-85085 85085
1 1 1 2 3 -C
85085 85085
1 2 1 0 3 +C
17017 17017
5 3 1 3
3 4 4
3 1 1
1
-425425 85085
1 3 2 0 0
6545 6545
13 4 1 3 3
12 3 4
1 1 1
60775 12155
7 4 2 2 2
3 6 1
2 2 1
-146965 7735
11 4 3 5 5
10 5 4
3 1 1
391391 17017
5 5 3 2 2
4 4 1
3 2 1
323323 17017
5 5 3 2 2
4 4
1 3 2
1
7735 7735
11 6 2 9 9
10 5 6
2 1 1
10829 1547
55 6 2 9 9
20 20 6
2 1 1
-365365 5005
17 6 4 13 13
8 16 6
4 1 1
-325325 5005
17 6 4 13 13
8 16 6
4 1 1
5005 5005
17 7 2 9 9
8 16 7
2 1 1
3575 715
119 7 2 9 9
24 48 7
2 1
1
7865 715
119 7 2 9 9
24 48 7
2 1 1
41327 2431
35 8 4 11 11
12 12 8
4 1 1
31603 2431
35 8 4 11 11
12 12 8
4 1 1
1485715 6545
13 8 5 9 9
12 3 8
2 1 2
1695155 6545
13 8 5 9 9
12 3 8
2 1 2
1878415 6545
13 8 5 9 9
12 3 8
2 1 2
1642795 6545
13 8 5 9 9
12 3 8
2 1 2
1852235 6545
13 8 5 9 9
12 3 8
2 1 2
1380995 6545
13 8 5
9 9 12
3 8 2
1 2
-1446445 85085
1 11 7 0 0
1547 1547
55 12 4 26 26
20 20 12
4 1 1
23375 935
91 12 6
1 1 12
6 12 6
1 1
55165 935
91 12 6 1 1
12 6 12
6 1 1
-293293 1001
85 12 8 16 16
8 16 4
8 2 1
100555 7735
11 14
8 5 5
10 5 4
3 2 2
10115 595
143 16 7 42 42
60 15 16
7 1 1
17255 595
143 16 7 42 42
60 15 16
7 1 1
3757 221
385 18 6
344 344 60
60 18 6
1 1
-222365 715
119 18 12 106 106
24 48 18
12 1 1
2387 77
1105 20 8
1036 1036 24
48 20 8
1 1
1925 385
221 20 8 152 152
24 48 20
8 1 1
1771 77
1105 20 8
1036 1036 24
48 20 8
1 1
-1209533 221
385 22 14 114 114
60 60 22
14 1 1
-763997 221
385 22 14 114 114 60
60 22 14
1 1
-1274065 1105
77 22 14 37 37
30 30 22
14 1 1
-1092845 1105
77 22 14 37 37
30 30 22
14 1 1
-1153841 221
385 22 14 114 114
60 60 22
14 1 1
-937261 221
385 22 14 114 114
60 60 22
14 1 1
-1287325 1105
77 22 14 37 37
30 30 22
14 1 1
19261 187
455 24 12 1 1
12 12 12
12 2
1
24871 1309
65 24 12 1 1
12 12 12
12 2 1
857395 6545
13 24 15 1 1
12 3 12
3 2 5
89375 715
119 25 14 2 2
24 48 1
14 2 1
2275 455
187 26 12 174 174
40 80 26
12 1 1
19565 455
187 26 12 174 174
40 80 26
12 1 1
46291 119
715 28 16 146 146
60 60 28
16 1 1
63427 1547
55 28 16
36 36 20
20 8 16
2 1
115115 5005
17 31 18 9 9
8 16 7
2 4 2
19591 143
595 32 16 256 256
24 48 8
16 2 1
4147 143 595
32 16 256
256 24 48
8 16 2
1
697 17
5005 36 12 911 911
60 60 36
12 1 1
833 119
715 36 12 196 196
60 60 36
12 1 1
1751 17
5005 36
12 911 911
60 60 36
12 1 1
4369 17
5005 36 12 911 911
60 60 36
12 1 1
1105 1105
77 38 16 25 25
30 30 8
16 2 1
16445 715
119 39 18 43 43
24 48 15
18 2 1
5083 221
385 40 20 331 331
60 60 40
20 1 1
7 7 12155
44 8 6561
6561 120 240
44 8 1
1
8041 187
455 48 24 1 1 12
12 12 12
4 2
5797 187
455 48 24 1 1
12 12 12
12 4 2
935 935
91 48 24 1 1
12 6 12
6 4 4
6643 91
935 52 24 356 356
40 80 12
24 2 1
51947 7
12155 56 32
11306 11306 120 240
56 32 1
1
37835 35
2431 56 32
1582 1582 120
240 56 32
1 1
45157 7
12155 56 32
11306 11306 120 240
56 32 1
1
24059 7
12155 56 32
11306 11306 120 240
56 32 1
1
36197 7
12155 56 32
11306 11306 120 240
56 32 1
1
28651 7
12155 56 32
11306 11306 120 240
56 32 1
1
12803 7
12155 56 32
11306 11306 120 240
56 32 1
1
44765 35
2431 56 32
1582 1582 120
240 56 32
1 1
11515 35
2431 56 32
1582 1582 120
240 56 32
1 1
73339 7
12155 56 32
11306 11306 120 240
56 32 1
1
38171 7
12155 56 32
11306 11306 120 240
56 32 1
1
6307 119
715 56 32 581 581
60 60 56
32 1 1
12047 7 12155
56 32 11306 11306
120 240 56
32 1 1
51023 7
12155 56 32
11306 11306 120 240
56 32 1
1
42595 35
2431 56 32
1582 1582 120
240 56 32
1 1
34741 7
12155 56 32
11306 11306 120 240
56 32 1
1
15385 85
1001 60 30 1 1
60 30 60
30 1 1
6035 85
1001 60 30 1 1
60 30 60
30 1 1
1859 143
595 64 32
86 86 24
48 16 32
3 1
715 715
119 75 42 8 8
24 48 3
42 4 1
10549 77
1105 76 40 16 16
24 48 4
40 4 1
11765 65
1309 82 44 191 191
120 240 82
44 1 1
3179 187
455 84 48 1 1
12 12 12
12 7 4
23647 221
385 84 48 71 71
60 60 24
48 2 1
1331 11
7735 96 48 1 1
24 48
24 48 4
1
8327 11
7735 96 48 1 1
24 48 24
48 4 1
9449 11
7735 96 48 1 1
24 48
24 48 4
1
9515 55
1547 96 48 1 1
24 48 24
48 4 1
5027 11
7735 96 48 1 1
24 48 24
48 4 1
3091 11
7735 96 48 1 1
24 48 24
48 4 1
2365 55
1547 96 48 1 1
24 48 24
48 4 1
605 55
1547 96 48 1 1
24 48 24
48 4 1
889 7
12155 100 40 8856
8856 120 240
100 40 1
1
7007 1001
85 100 56 16
16 8 16
4 8 13
4
8015 35
2431 112 64 1225
1225 120 240
112 64 1
1
7021 119
715 112 64 81
81 60 60
52 4 2
2
9667 7
12155 112 64 3656
3656 120 240
112 64 1
1
19859 7
12155 112 64 3656
3656 120 240
112 64 1
1
1463 77
1105 116 56 1036
1036 24 48
20 8 5
2
781 11
7735 120 48 1
1 24 48
24 48 5
1
9823 11
7735 132 72 4096
4096 24 48
12 24 6
2
13321 77
1105 132 72 781
781 24 48
12 24 6
2
4165 595
143 140 80 100
100 60 15
20 5 3
6
4009 1
85085 156 72 65976 65976
120 240 36
72 2 1
1469 13
6545 156 72 526
526 120 240
36 72 2
1
685 5
17017 156 72 14925 14925
120 240 36
72 2 1
2063 1
85085 156 72 65976 65976
120 240 36
72 2 1
1355 5
17017 156 72 14925 14925
120 240 36
72 2 1
1567 1
85085 156 72 65976 65976
120 240 36
72 2 1
1223 1
85085 156 72 65976 65976
120 240 36
72 2 1
2297 1
85085 156 72 65976 65976
120 240 36
72 2 1
3605 35
2431 156 72 339
339 120 240
36 72 2
1
4313 1
85085 156 72 65976 65976
120 240 36
72 2 1
1357 1
85085 156 72 65976 65976
120 240 36
72 2 1
1853 17
5005 156 72 911
911 60 60
36 12 3
2
3529 1
85085 156 72 65976 65976
120 240 36
72 2 1
1933 1
85085 156 72 65976 65976
120 240 36
72 2 1
409 1
85085 156 72 65976 65976
120 240 36
72 2 1
1927 1
85085 156 72 65976 65976
120 240 36
72 2 1
5785 65
1309 156 72 526
526 120 240
36 72 2
1
1267 7
12155 156 72 5201
5201 120 240
36 72 2
1
1327 1
85085 156 72 65976 65976
120 240 36
72 2 1
529 1
85085 156 72 65976 65976
120 240 36
72 2 1
3475 5
17017 156 72 14925 14925
120 240 36
72 2 1
3961 17
5005 156 72 911
911 60 60
36 12 3
2
5447 13
6545 156 72 526
526 120 240
36 72 2
1
1 1 85085
156 72 65976 65976 120
240 36 72
2 1
3731 91
935 160 80 1
1 40 80
40 80 4
1
3029 13
6545 160 80 1871
1871 120 240
40 80 2
1
10855 65
1309 164 88 1138
1138 120 240
44 88 2
1
12563 17
5005 168 96 2731
2731 60 60
48 36 3
2
1699 1
85085 168 96 7736
7736 120 240
48 96 2
1
21347 1
85085 168 96 7736
7736 120 240
48 96 2
1
27637 1
85085 168 96 7736 7736
120 240 48
96 2 1
539 77
1105 188 104 1036 1036
24 48 20
8 8 3
3839 11
7735 192 96 1 1 24
48 24 48
8 2
1003 17
5005 240 120 1 1
60 60 60
60 4 2
3479 7
12155 268 136 4436 4436
120 240 28
136 3 1
341 11
7735 288 144 1 1
24 48 24
48 12 3
793 13
6545 320 160 5611 5611
120 240 80
160 3 1
2881 1
85085 324 168 50506 50506 120
240 84 168
3 1
3005 5
17017 324 168 16472 16472 120
240 84 168
3 1
457 1
85085 324 168 50506 50506 120
240 84 168
3 1
3305 5
17017 324 168 16472 16472 120
240 84 168
3 1
1721 1
85085 324 168 50506 50506 120
240 84 168
3 1
4447 1
85085 324 168 50506 50506 120
240 84 168
3 1
1193 1
85085 324 168 50506 50506 120
240 84 168
3 1
899 1
85085 492 264 4096 4096
120 240 12
24 5 2
ith, kth Generation Type [1,2] Mersenne
Hailstones
In the following diagram, a is a
6th generation Type [1,2] Mersenne Hailstone.
ae_ad
ac_ab
aa
generation 1 --> z_y
x_w
v
generation 2 --> u_t
s_r
q
generation 3 --> p_o
n_m
l
generation 4 --> k_j
i_h
g
generation 5 --> f_e
<-----|
d_c | Type [1,2]
b |
generation 6 --> a <-----|
The 1st such hailstone, found by plugging (1,6) into the
blueprint results in:
a = 2177149-1
In binary, a has 177,149 contiguous 1 bits. In decimal,
it has 53,328 digits.
Starting from the hailstone, the Collatz sequence takes 2,386,509
iterations to reach 1.
Because a is a Mersenne Number, the sequence is predicted to contain m + (1.585 * m)/0.415 or 4.819 * m iterations of 3n+1, where m is the number of bits in the hailstone.
Therefore, the
sequence should have
4.819*177,149 or
853,681
iterations of 3n+1.
It actually has
854,697
making the
prediction accuracy 99.88%
Notes & References:
1) Weisstein E. ‘Collatz problem’ Mathworld
http://mathworld.wolfram.com/CollatzProblem.html
2)
Anonymous ‘Collatz conjecture’ Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Collatz_conjecture
3)
Stadfeld P. ‘The Collatz tree’
http://members.aol.com/mensanator666/treemap/mainmap.htm
4)
Weisstein E. ‘Linear congruence’ Mathworld
http://mathworld.wolfram.com/LinearCongruenceEquation.html
5)
Anonymous ‘Linear congruence’ Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Linear_congruence
6) Weisstein E. ‘Euclidean algorithm’ Mathworld
http://mathworld.wolfram.com/EuclideanAlgorithm.html
7)
Anonymous ‘Extended Euclidean algorithm’ Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Extended_Euclidean_Algorithm
8) Weisstein E. ‘Modular inverse’ Mathworld
http://mathworld.wolfram.com/ModularInverse.html
9)
Anonymous ‘Multiplicative inverse’ Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Multiplicative_inverse
10)
Weisstein E. ‘Greatest common divisor’ Mathworld
http://mathworld.wolfram.com/GreatestCommonDivisor.html
11)
Anonymous ‘Greatest common divisor’ Wikipedia, the free encyclopedia
http://en.wikipedia.org/wiki/Greatest_common_divisor
12) Stadfeld P. ‘Hailstone function parameters
algorithm’
http://members.aol.com/mensanator666/collatz_utilities/python_code.htm#hfp
13) Stadfeld P. ‘Partition generator’
http://members.aol.com/mensanator666/collatz_utilities/python_code.htm
- partitions
14) Stadfeld P. ‘Collatz_C program’
http://members.aol.com/mensanator666/collatz_utilities/python_code.htm#collatz_c