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Prime Holdout Problems

This paper introduces prime holdout problems, a problem class related to the Collatz conjecture. After applying a linear function, instead of removing a finite set of prime factors, a holdout problem specifies a set of primes to be retained. A proof that all positive integers converge to 1 is given for both a finite and an infinite holdout problem. It is conjectured that finite holdout problems cannot diverge for any starting value, which has implications for divergent sequences in the Collatz conjecture.


Introduction
The Collatz Conjecture, also known as the 3n + 1 problem, is a simply-formulated problem that is notoriously unproven. It concerns the iterates of a piecewise equation defined on the set of positive integers [1] f (n) = 3n + 1 n ≡ 1 (mod 2) n 2 a n ≡ 0 (mod 2) where a = v 2 (n) is the maximum power of 2 dividing n (also called the 2-adic order of n).
We see that repeated applications of f (n) alternate between removing all factors of 2 from n and applying the 3n + 1 linear map. Lothar Collatz postulated that, in so doing, every starting value would eventually converge to the trivial cycle (1,4,1). This implies that no n exists for which repeatedly applying f (n) diverges or enters a cycle other than the trivial one. The Collatz Conjecture remains unproven.
A more general form of the Collatz conjecture exists wherein a set of prime divisors D are specified for removal in between applications of a linear function. We will refer to these as divisor problems as they take the form and v di represents the d i -adic order of n.
For example, in his empirical verification of the Collatz conjecture [2], Tomas Oliveira e Silva also considers C 5,{2,3} , referring to the 5n + 1 divisor problem in which the divisor set is {2, 3}.
Here we introduce finite prime holdout problems, a new problem class related to divisor problems. Instead of specifying a set of divisors in between applications of a linear function, holdout problems specify a finite set H of prime factors to retain. Any prime factors not appearing in this set are divided out. The general form of holdout problems is Here we focus on the holdout problem H 5,{2,3} to borrow the notation of Silva. We present a series of properties of this problem and prove that, through repeated applications of equation 4, all positive integers n must converge to 1.
An example of application of this equation for n = 28 is as follows 28 = 4 × 7, 4 is retained since it is a power of 2, 7 is removed. 4 is a power of 2, so it will follow the first branch 5 × 4 + 1 = 21. This process continues until 1 is reached, the sequence in its entirety is as follows: 28, 4, 21, 3, 16, 81, 406, 2, 11, 1. This sequence also happens to conveniently illustrate many of the properties established in the H 5,{2,3} proof.
H 5,{2,3} was selected for study because it is the most simple non-trivial example in the problem class. Any holdout problem in which the cardinality of H is 1 must converge because the addition of one in the first branch of equation 3 removes divisibility by the preserved prime factor from any of its powers. Furthermore, the problem is still trivial if c ∈ H since the first branch can never result in a number divisible by c. Therefore H 5,{2,3} is the lowest numbered holdout problem of interest.
In addition to providing a proof of convergence for H 5,{2,3} , we modify the holdout problem definition to consider infinite holdout sets. The distinguishing factor between infinite holdout problems and divisor problems is that the set of divisors is infinite. We present a convergence proof of the following infinite holdout problem: where P denotes the set of all primes and

Definitions and Properties
In order to consolidate the proof of convergence for H 5,{2,3} , we must first establish some basic properties of the problem. In the following discussion, we use "step" to refer to one application of 5n + 1 plus the removal of all prime factors not equal to 2 or 3 from the result. Use of the term convergence will imply convergence to 1.

Lemma 3.1: All n with both 2 and 3 as prime factors converge in 1 step
Let n = 2 a 3 b c for a, b, c ∈ N where N denotes the positive integers. Then We know that none of the prime factors of a number may be present in that number plus 1. Therefore, based on the definition of H 5,{2,3} , numbers of form 2 a 3 b converge Lemma 3.2: All n of the form 2 2a+1 for a ∈ N converge in one step We see that 4 is congruent to 1 modulo 3, 5 is congruent to 2 modulo 3, and 1 is congruent to 1.
We see above that there must be a single power of 2 remaining in n after one step, and 2 always converges to 1 in the next step by Lemma 3.3.

Lemma 3.4: n must fall below itself after steps where h(5n + 1) = 5n + 1 at the intermediate stage
This lemma concerns cases where division occurs after 5n + 1 is applied. In these cases, the smallest number that may be removed is 7. By the problem definition, prime factors 2 and 3 are preserved. 5 is a factor of 5n, so it must not be a prime factor of 5n + 1. The smallest remaining prime factor that can be removed in the division stage is 7. 5n + 1 a < n f or all a ≥ 7 and n ≥ 1 Lemma 3.5: 3 a+1 is the only power of 3 greater than 3 a and less than 5 × 3 a + 1 for a ∈ N Here we show the convergence to 1 for all starting integers of the simplest non-trivial holdout problem Lemma 4.1: Any power of 4 must descend below itself within 2 steps If there is division in the first step, we are now below the starting number by Lemma 3.5. Otherwise applying the second 5n + 1 will put the number into the form 25 × 4 a + 6. The highest power of 2 dividing this is 1, so it will then fall to 2 after division.
Lemma 4.2: Any power of 3 (that does not create a loop of size 2) must descend below itself in 3 steps.
There are 3 possibilities to consider 1. The first possibility is that a division occurs within the first step. The result is less than the starting number by Lemma 3.4.
2. The next possibility occurs if the number completes the first step without dividing. Now, let the resulting number be a power of 4 since odd powers of 2 directly converge by Lemma 3.2. There are then two cases to consider: (a) The number can grow again without dividing. If this happens, Lemma 4.1 shows that the subsequent step will descend to 2. (a three step descent for the original number) (b) The number divides on the second step. If this is below the starting point, we are finished. If not, then we know from Lemma 3.5 and the fact that we didn't loop that the number is now 3 a+1 . By Lemma 3.3, a must have been odd or 3 a would have fallen to 2 in the first step, so a + 1 must be even. For the same reason, 3 a + 1 must descend to 2 on the next step (also a three step descent).
We will present a proof by induction that each number must enter a cycle (either the trivial one, or another if it exists). 1 returns to 1 by the trivial cycle, so the base case is proven. Let k > 1 be an arbitrary integer, and assume that all numbers less than k eventually enter a cycle. There are 4 cases to consider, k is treated according to the first matching case: 1. k has a prime factor p that is not 2 or 3: k/p < k therefore k eventually cycles. Therefore for all positive integers the 5n + 1 problem cannot diverge.

Corollary 4.4: All loops in H 5,{2,3} must contain exactly two numbers
By Lemmas 4.1 and 4.2, every power of 4 and 3 will fall below themselves within at most three steps unless there is a loop of size 2. Therefore, only loops of size 2 can exist.

Lemma 4.5: No nontrivial loops of size 2 exist.
Let n be the smallest element of the loop, The first step must have no divisions. Otherwise, Lemma 3.4 produces a contradiction (the result will be smaller than n). Since divisions must occur on step two this gives us the equation 5(5n + 1) + 1 = cn, which simplifies to 25n + 6 = cn. This equation can only produce the trivial cycle because if n is a power of 4, or a power of 3 greater than 3 1 , then n divides one side but not the other. And if n is 3, then the equation simplifies further to 27n = cn so c = 27. But c cannot be 27 because powers of 3 do not get divided out. This leaves us with only the trivial cycle as a solution, n = 1 and c = 31. If 2|n, then 4|2n, the only other even remainder mod 4 to occupy is 2.

Theorem 5.2: The 2n+1 problem converges
We proceed by induction, our base case being that 1 = 1. Let n be a positive integer greater than 1. Assuming that all previous numbers converge to 1, there are 2 cases to consider: 1. n is divisible by a prime p congruent to 2 or 3 mod 4. Thus n can be divided, and converges because n/p converges.
2. n is composed only of prime factors congruent to 1 mod 4. Since n is odd, Lemma 5.1 guarantees 2n + 1 is congruent to 3 mod 4. It must have at least one prime factor that is congruent to 3, the smallest of which is 3. For all n greater than 1, (2n + 1)/3 is less than n therefore n converges.

Discussion
In the previous sections, we have demonstrated the truth of both a finite and infinite holdout problems. Here, we examine the provability of these classes, as well as posit a conjecture concerning finite holdout problems.

Connections to Theory of Computability
A particular generalization of the Collatz conjecture posed by John Horton Conway is known to be undecidable. It concerns functions of the form: g(n) = a i n + b i where n ≡ i (mod P ) and a 0 , b 0 , ..., a P −1 , b P −1 are rational coefficients chosen so that g(n) is always a positive integer [3]. Conway establishes that each of these problems corresponds to a Turing machine so that there is no general way to determine if the procedure of repeatedly applying a periodic piecewise linear function until the result is 1 will always halt. The Collatz conjecture can be represented this way by assigning P = 2, a 0 = 1/2, b 0 = 0, a 1 = 3, and b 1 = 1. The main idea in the proof is that in the case when all b i = 0, the powers in the prime factorization of n (i.e. a,b,and c... in 2 a , 3 b , 5 c ...) correspond to registers that store a program's state of execution, and the rational coefficients a i act as program instructions, incrementing and decrementing the registers.
Divisor problems are a subset of Conway's periodic linear functions. For example, the Collatz conjecture could be modified such that P = 4, a 0 = 1/2, b 0 = 0, and a i = 3, b i = 1 for i ∈ {1, 2, 3}. In this case, if a number is divisible by 2 but not 4, the factor of 2 is not removed. This example is a member of Conway's problem class, but is not divisor problem. Thus, Conway's result does not show the undecidability of the halting problem over this set. Divisor problems seem to be meaningfully less computationally expressive as they rely solely on the difficult-to-predict action of adding 1 to increase the value of any given register. Each prime in D can only be removed, decrementing a register, and although multiplication by c can increase the value of some registers, adding 1 will always zero them out. It should be noted that addition is explicitly avoided in Conway's proof by setting b = 0, and that the Collatz conjecture contains a +1 term and is suspected to not be universal [1, p. 20] Unlike either of these Collatz conjecture generalizations, both finite and infinite holdout problems are not periodically linear. This is because there are infinitely many primes in the set of divisors. If a finite modulus P were chosen, there will always be a prime number in D that is larger than P whose division is needed to properly specify the operations of the holdout problem. This means that the possibility for a general algorithm that can predict the convergence of holdout problems remains an open question.
The sets of all finite holdout problems and all divisor problems are both countably infinite since you could assign each a binary integer for the primes included in their finite set, each value indicating the inclusion status of the n-th prime, such as 1101 for {2, 3, 7} where 0 indicates the omission of 5. Specifying an infinite holdout set requires an infinite string of bits, so there are an uncountably infinite number of them. This, of course, means that there are a great multitude of uncomputable infinite holdout problems since the set of computable functions is countable. However, there exist potentially meaningful computable subclasses, such as using remainders of primes in a particular modulus, an example of which is proven in this paper.

The Finite Holdout Conjecture and its Implications
After a brief computational survey of the holdout problem space, we posit the finite holdout conjecture -that no finite prime holdout problems can support divergent trajectories. Furthermore, we suspect this conjecture is provable. Truth of the finite holdout conjecture would imply that there cannot be a holdout problem which doesn't reach 1 for all starting integers, but for which this is not provable. This is because without diverging, the only way to not reach 1 is to loop, and a loop can be encountered in a finite number of steps.
Additionally, the finite holdout conjecture has the direct implication that the potential of the Collatz conjecture to diverge lies in its infinite set of preserved primes. Notably, it's possible to construct holdout problems that behave identically to the Collatz conjecture for an arbitrarily large range of integers 0 -k. This can be done by including all prime numbers under 3k + 1 in the holdout set (excluding 2). But this is a finite set, so if the Collatz conjecture had a divergent trajectory, then all of these holdout problems would eventually have to stop following it.
By a similar reasoning, the finite holdout conjecture implies that, if the Collatz conjecture diverges on a particular input, the union of the sets of prime factors for every element of the divergent trajectory must be infinite, otherwise the finite prime holdout problem that preserves all the prime factors present in the trajectory can diverge.
The contrapositive of the finite holdout conjecture (that divergence of a problem implies the holdout set is infinite) could also be of interest should the construction of provably divergent trajectory prove easy given a particular holdout set. The contrapositive could then be used to show that the holdout set in question is indeed infinite.
While the holdout problems discussed in this paper are fascinating, we note that the exploration of this problem space is valuable for its potential insights into the behaviors of the Collatz conjecture, particularly the connections between the finite holdout conjecture and the behaviors of divergent trajectories.