JOURNAL OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 00, Number 0, Pages 000–000
S 0894-0347(XX)0000-0
A SORTING ALGORITHM FOR (2,4) MOD 6. THE SYRACUSE
SOLUTION
ANDREW S. BOUNDS
Part 1. Introduction
DOMAINS
The Domain of x=positive integers, Domain of n=natural numbers for odds and
positive integers for evens.
Gives: 3n+1=y where: if n=2x, n
2 , and if n=2x+1, 3n+1
This domain definition yields two trivial loops at x=0, and another at x=1 for
y=2x as it then feeds into the odd trivial loop of x=0. It also yields a single domain
for all x, giving a process for negating any dependent linear expressions.
Every power of two (2a) will collapse to the trivial odd loop if we divide by 2,
so we need to ensure our algorithm can cycle through all possible evens that can
have powers of two, divide by two if it is a power of two, and kick out of our ”slot
machine” if it is not.
We need to account for every even and odd input, and every possible output.
Our function cannot have solutions that repeat, but the more values it covers the
more likely we are to find a power of two. Linear function solutions are preferred
as they ensure no repeat solutions.
Ideally, we would also be able to extend the approach to the domain of negative
integers. If there is a linear solution, or set of solutions, then we should be able to
extend those to the negative integer domain while maintaining the same range.
COPRIME NOTATION
With the realization that only powers of two will collapse to the trivial loop
of x=0 for all 2(x) and is the only point in which they crossover, we can use the
expanded equation
3(2x+1)+1=6x+4
(0.1)
to begin understanding the possible combinations. This would lead to using the
6-coprime expansion, also a 6 mod c notation. This also leads to the expansion of
all 6-coprimes listed in the next section. this coprime expansion will also be the
basis for expanding to the negative integers.As this expansion is inclusive of all of
2010 Mathematics Subject Classification. Primary .
1
2 HAILSTONE PROBLEM
the coprime expansions of 2 and 3, all possibilities for the Collatz expansion are
represented.
Part 2. Coprime Expansion
6-COPRIMES
2(3x) 2(3x+1) 2(3x+2)
6x+0 6x+1 6x+2 6x+3 6x+4 6x+5
3(2x) 3(2x)+1 3(2x)+2 3(2x+1) 3(2x+1)+1 3(2x+1)+2
As is illustrated in the above diagram, the odd and even coprime multiples of 3
and their related doublets as even numbers are completely covered. This represents
every possible permutation of coprime linear equations for 3’s. This also gives a
structure for separating out that which can be a power of two and that which can’t.
While this is not specific, it does limit the amount of possibilities significantly. This
also has an impact on the sequencing of every possibility.
With the above table it is trivial to see that the odd and even outputs of the
algorithm 3n+1 have unique values. What is not so trivial to see is that the function
3n + 2 can also be used as a substitute for finding twos with a little tweaking to
the algorithm, but it can be done by setting the expression equal to the expression
for odds and taking the resultant root as the starting value for the next iteration.
3n + 2 = 2x + 1
3x + 2 = 2×1 + 1
and so on, using the convention of dividing by 2 until you arrive at 1 or an odd
number other than 1. Also, after subtracting 1, if 2x is a power of two, it would
also be one in the 3n+1, so you can divide down to 1. That is where it breaks down
however, as there can be no zero value that is odd in that substitution function.
The inference of dividing to a trivial loop is able to be realized with the divide rule
mentioned above.
This also gives a maximum slope of 3(2x+1)+2 or 6x+5 for any iteration path
that would be climbing towards infinity. This would also require that every seed
value produce an odd value for the seed value x and there only be one division by
2 every iteration. The distribution of evens and odds in the seed value x supports
all previous work that took the probability as abstract. This applies to all previous
work on 3n+1 and 3n+2 on the Collatz Conjecture as these are the only outputs,
and therefore secondary seed values, for the Collatz Conjecture.
Part 3. Negative Integer Domain
NEGATIVE INTEGERS
If we change our domain, we need to continue the negation across all steps
2x + 1 = 2x − 1 = n
3n + 1 = 3n − 1 = 3(2x − 1) − 1 = 6x − 4
AMS ARTICLE TEMPLATE 3
Which gives us the following expansion like the positive integers:
2(3x) 2(3x-1) 2(3x-2)
6x-0 6x-1 6x-2 6x-3 6x-4 6x-5
3(2x) 3(2x)-1 3(2x)-2 3(2x-1) 3(2x-1)-1 3(2x-1)-2
This allows us to see the direct expansion to negative integers and that every
iteration following the form outlined above will give the same result for the domain
of the negative integers. All of the above shows that previous findings of other and
in this paper apply to the negative integers when using this form.
Part 4. Conclusion
CONCLUSION
With the above information we can determine, for any given x:
(1) The maximum slope for any iteration of this algorithm is 6x + 5 which
implies a minimum logarithmic density.
(2) The only outputs possible from any given input are divisions by 2a and the
parity of a tells you if it is 6x + 1 or 6x + 5 and anything else results in
another division by 2.
(3) the only way out of the loop is to find a power of 2.
(4) the even numbers that correlate to 6x+2 and 6x+4 can be sorted by this
process into 3 distinct outputs, the odd roots correspond to 6x+5 and the
even roots that are not a power of 2 will correspond to 6x + 1. The zero
value being the only identity crossover point. all other values are distinct.
(5) If changing the domain of x to the negative integers, then all operations
must be properly negated in order to show the same results for that domain,
but the algorithm can be completely mapped to the negative integers.
With the above points we can determine that the 6-coprime expansion encompasses
all of the behaviors and examples of the algorithm output possibilities. The expansion
also shows the differentiation by parity of the input value of x and by the
number of divisions by 2, or 2a. The parity of a is also determinant to which output
is achieved, pairing even values of a with even values of x. The only other
outputs available are the divisions by 2, which yield the trivial loop when there are
no odd factors.
Abstract. ABSTRACT: In this it is shown that there is an algorithm for
sorting (2,4) mod 6 (a.k.a. Syracuse problem, Collatz conjecture, Hailstone
problem, etc.) is true through the use of 6 coprime notation and seeing the
problem as an algorithm for sorting roots of even numbers to find powers of
two from integers-reduced-to-odd inputs, a robust algorithm in fact. Previous
work focused on the 3n+1 structure, but this reasoning shows that it only
encompassed half of the solution. 3n+2 being the other half.
CA 92630