Abstract
In this paper, we study the bounded trajectories of Collatz-like functions. Fix so that and are coprime. Let so that for each , , is coprime to and , and . We define the function and the sequence a trajectory of . We say that the trajectory of is an integral loop if there exists some in so that . We define the characteristic mapping and the sequence the characteristic trajectory of . Let be a -adic sequence so that . We say that is eventually periodic if it eventually has a purely -adic expansion. We show that the trajectory of eventually enters an integral loop if and only if is eventually periodic.
1. Introduction
The problem, also known as the Collatz conjecture, had its beginnings in the University of Berlin in 1930 [1]. In particular, Dr. Collatz was interested in cycles that could be generated with number-theoretic functions. This study later evolved to what we know today as the problem or Collatz function. It was not until the International Congress of Mathematics of 1950 that the problem began to spread [2]. Eventually, the problem gained sufficient attention and became something akin to American folklore.
Definition 1. What makes this problem initially interesting is the fact that the conjecture itself is so difficult to prove. The problem has proven to be quite evasive to proof by all modern mathematical techniques tried thus far. Insomuch that particular generalizations of the problem have been demonstrated to be arithmetically undecidable [3–5]. The problem stands on the same footing as the Riemann hypothesis being a problem with regard to complexity [4]. It is even attributed to Dr. Erdős to have said, “Mathematics is not yet ripe enough for such questions.”
Conjecture 2. (Collatz). For all , there exists such that .
A resolution to the Collatz conjecture in itself would arguably not be of much practical value. Rather more important would be the method of proof. From a pessimistic standpoint, there may be a solution that only resolves the problem uniquely. This would be a pity, as the problem is only one of a vast collection of number-theoretical functions of its kind. More optimistically, a hypothetical method of proof could be generalized yielding new knowledge and tools to study other topics of mathematics. Such mathematical topics include but are not limited to exponential diophantine equations, discrete dynamical systems, mathematical logic, and Turing machines, among others [6]. The elusiveness of an answer belies the fact the Collatz conjecture touches nearly every branch of mathematics.
In this paper, we study properties of integral loops in a generalized Collatz-like mapping. We make an improvement on the work performed regarding the set of rational cycles for the problem. It has been proven that the parity sequence (characteristic sequence) associated with the trajectory of a starting integer is eventually periodic if and only if the trajectory eventually enters an integral loop [7, 8]. Previous proofs of this fact depended on the 2-adic nature of the parity sequence. Here, we present an alternative proof that is 2-adic invariant. We also provide examples.
2. Preliminary Definitions
2.1. Integral Loops
Definition 3. Fix . Let be defined asLet . Furthermore, for , we denote function composition as . Finally, if , then we write instead of .
Definition 4. Fix . Let be defined asLet . Furthermore, for , we denote function composition as . Finally, if , then we write instead of .
Definition 5. Fix and . Let denote the trajectory of .
Definition 6. Fix and . Let denote the characteristic trajectory of .
Let denote the set of binary sequences of infinite length.
Definition 7. Fix and . We say is associated with the characteristic trajectory of if .
Proposition 8. Fix , , and . Let such that . Let and be the count of the number of zeros and ones, respectively, in for all between zero and inclusive. For , let denote the number of zeros between th and th one.
(1)If n is odd, let and be the remaining zeros after the last one.(2)If n is even, let be the number of leading zeros. Observe that for being odd. So, we obtain by (1) and .
Then, satisfies the following equality:
Proof. Let us assume first that is odd, which implies that . Therefore, equation (4) reduces to the form,By construction of and , we make the substitution . We know that represents compositions of the mapping with starting integer . Therefore, we can advance the trajectory of by one and subtract one step from in the following manner:Both the terms and are odd. Therefore, their sum is even. We can divide by two, number of times yielding the following expression:By hypothesis, this expression is again odd. Therefore, we can apply the map for one more step.In turn, this expression is again even. Therefore, we can divide by powers of two. Following this, we can reduce the expression again,We proceed to follow this process inductively until the length of the entire finite subsequence of is exhausted.On the other hand, we suppose that is even. Then, and by hypothesis for an odd integer . Starting from equation (4), we find by taking steps,Recalling by hypothesis implies that . Substituting this quantity in equation (11) yieldsWe then consider the previous argument starting with , , and unchanged.Therefore, we find that
Proposition 9. Fix and . Let such that . We suppose that is the smallest positive integer such that . Let and be the count of the number of zeros and ones, respectively, in for all between 0 and inclusive. For , let denote the number of zeros between th and th one.
(1)If n is odd, let and be the remaining zeros after the last one.(2)If n is even, let be the number of leading zeros. Observe that for being odd. So, we obtain by (1) and .
Then, satisfies the following equality:
Proof. Equation (15) can be derived from equation (4) by setting .
We will refer to equation (15) as the integral loop formula.
Example 1. We consider the integral loop induced by and . With respect to equation (15), we find that
2.2. Rational Cycles
Definition 10. Let . We say that is a rational 2-adic if it can be written for some integers with odd [8].
Definition 11. Let . We say that is eventually periodic if there exists a positive integer such that for all sufficiently large . If for all , then is called periodic [8].
Theorem 12. There is a one-to-one correspondence between rational numbers , where is odd, and eventually periodic sequences , which associates with each such rational number the bit sequence of its 2-adic expansion. The sequence is strictly periodic if and only if and . [9].
Proof. Let be a strictly periodic sequence of period . Set . Computing in , we findHence,is a negative rational number. We write as a fraction reduced to the lowest terms with positive. Then, is odd, , and .
Conversely, suppose is given in the lowest terms with an odd positive integer, , and . Let be the smallest integer such that . Then, is divisible by , so set and write . Thus, . The calculations leading to equation (18) may be run backward to see that the segment is a single period of a strictly periodic sequence.
Now, we suppose that is an arbitrary rational number. Let be the next largest integer. If , then its 2-adic expansion is finite, ending in a string of zeros. If , then its 2-adic expansion ends in an infinite string of ones. However, , where and . Thus, the 2-adic expansion for is periodic. It follows that the 2-adic expansion for the sum is eventually periodic.
Conversely, an eventually periodic sequence corresponds to a rational number because it is given by a finite transient term for some nonnegative integer plus a periodic term, , both of which are rational numbers.
Proposition 13. Fix and . Let such that . We assume that the trajectory of is bounded above. Then, there exists some contained in the trajectory of and the smallest in such that if and only if is a rational 2-adic.
Proof. We suppose that first in the trajectory of , there exists an so that . We assume further that is the smallest positive integer which satisfies this equality. Then, has a repeating period of length and is rational.
Now, we assume that is a rational 2-adic. Let be the starting value whose characteristic trajectory generates the repeating period of . Let us define the symbol as follows:where . We observe immediately that is finite as both and are finite. We have the following sequence of values by continuous application of the Collatz map generating values We can consolidate the terms into an expression that includes only and as follows:Thus, solving for yieldsWe suppose first that . Following from equation (21), we distribute the first term, yielding the following:Then, there exists an such that for all , . We find a new , which maybe different from such that generates an integral loop coinciding with the original rational loop.
If conversely we suppose we simply reverse equation (21) to solve for in terms of :We suppose first that , then it follows that . This equality is true for all, and thus, we have an integral loop. So, assume instead that . Then, we can write Equation (22) in the form of a ratio exhibited in equation:However, recall and are both fixed values. Thus, the left hand side has a fixed denominator. However, as the right hand side is the quotient of two primes, the denominator expands with irreducible powers of two. If is a trajectory of integers, then there would be some index of which the denominator on the right hand side would be larger than the left hand side. This would be a contradiction.
Finally, as , as a irrational ratio cannot be expressed with a finite binary sequence.
Thus, we have shown that no rational cycle apart from an integral loop exists for any mapping.
3. Generalizations
3.1. Two-State System with Halting Condition
We would like to consider first a generalization of the two-state system given in Definition 3. To begin, we need the following lemma:
Lemma 14. Fix so that , and are pairwise coprimes. Then, there exists a pair of integers satisfying the following linear Diophantine equation:
Proof. A necessary and sufficient condition that a linear Diophantine equation is solvable is that is a multiple of the greatest common denominator of and [10]. By hypothesis, . Thus, is necessarily a multiple. We conclude that the linear Diophantine equation is always solvable.
Proposition 15. Fix so that and are pairwise coprimes. Then, there exists a satisfying the following equation:
Proof. We have shown from Lemma 14 that equation (26) is always solvable for a pair of integers and . Thus, let so that is an integer in .
Definition 16. Fix so that , and are pairwise coprimes. Let satisfying . Let be defined asLet . Furthermore, for , we denote function composition as .
For what follows let be the empty character.
Definition 17. Fix so that and are pairwise coprimes. Let satisfying . Let be defined asLet . Furthermore, for , we denote function composition as .
We define the trajectory and characteristic trajectory of in a similar manner to Definitions 5 and 6, respectively. Furthermore, we associate with in a similar manner to Definition 7.
Proposition 18. Fix so that and are pairwise coprimes. Let satisfying . Fix . Let such that . Let and be the count of the number of zeros and ones, respectively, in for all between zero and inclusive. For , let denote the number of zeros between th and th one. We assume that the halting condition for is not met for all .
(1)If , let and be the remaining zeros after the last one.(2)If , let be the number of leading zeros. We observe that for . So, we obtain by (1) and .
Then, satisfies the following equality:
Proof. By hypothesis, the halting condition is not met. Thus, every step of the characteristic trajectory associated with is nonempty. In that context, the proof is essentially identical to Proposition 8.
Proposition 19. Fix so that , and are pairwise coprimes. Let satisfying . Let such that . Suppose that is the smallest positive integer such that . Let and be the count of the number of zeros and ones, respectively, in for all between 0 and inclusive. For , let denote the number of zeros between th and th one. We assume that the halting condition for is not met for all .
(1)If , let and be the remaining zeros after the last one.(2)If , let be the number of leading zeros. We observe that for . So, we obtain by (1) and .
Then, satisfies the following equality:
Proof. This follows from Proposition 18, where we derive equation (31) from equation (30) by setting .
Example 2. Fix and . We consider that the integral loop is induced by and . The mapping appears asWe find that the trajectory and characteristic trajectory of 109 appear as
Proposition 20. Fix so that , and are pairwise coprimes. Let satisfying . Let such that . We assume that the trajectory of is bounded above. Then, there exists some contained in the trajectory of and the smallest in such that if and only if is a rational 2-adic.
Proof. The proof is similar to the proof of Proposition 13. Here, we replace 3 with and 2 with . Since and are coprimes, then does not vanish and the quotient again has a denominator that expands with higher powers of . Thus, we arrive at the same contradiction.
3.2. Multistate System with No Halting Condition
Definition 21. Fix so that and coprime. Let so that for each , , is coprime to and , and . Let be defined asLet . Furthermore, for , we denote function composition as .
Definition 22. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Let be defined asLet . Furthermore, for , we denote function composition as .
Definition 23. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . Let denote the trajectory of .
Definition 24. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . Let denote the characteristic trajectory of .
Fix . Let denote the set of -adic sequences of infinite length.
Definition 25. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . We say is associated with the characteristic trajectory of if .
Definition 26. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . Fix , and . We suppose further that starts an integral loop of length . Let such that . Let and be the count of the number of zeros and nonzero coefficients in a single period of . Let be the subindexing of so that for all , we have . We call the indexing sequence associated with .
Proposition 27. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . Let such that . Let and be the count of the number of zero and nonzero coefficients, respectively, in for all between zero and inclusive. For , let denote the number of zeros between the th and th nonzero coefficient. Let be the indexing sequence associated with .
(1)If , let and be the remaining zeros after the last nonzero coefficient.(2)If , let be the number of leading zeros. We observe that for . So, we obtain by (1) and .
Then, satisfies the following equality:where for the -th nonzero coefficient in a single period of .
Because Proposition 27 is logically distinct from Proposition 8, it is necessary to restate the proof again.
Proof. Let us assume first that , which implies that . Therefore, equation (36) reduces to the formBy construction of and , we make the substitution . We know that represents compositions of the mapping with starting integer . Therefore, we can advance the trajectory of by one and subtract one step from in the following manner:We recall that . We can divide by , number of times yielding the following expression:By hypothesis, this expression is again nonzero modulo . Therefore, we can apply the map for one more step.In turn, this expression is again zero modulo . Therefore, we can divide by powers of . Following this, we can reduce the expression againWe proceed to follow this process inductively until the length of the entire finite subsequence of is exhausted.On the other hand, we suppose that . Then, , and by hypothesis, for . Starting from equation (36), we find by taking steps, we haveRecalling by hypothesis implies . Substituting this quantity in equation (43) yieldsWe then consider the previous argument starting with , , and unchanged.Therefore, we find that
Example 3. Fix , , , , and . The mapping appears asUnder the same hypothesis, we consider the first nine steps of the characteristic trajectory induced by .We verify equation (36) for .
Proposition 28. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Fix . Let such that . We suppose that is the smallest positive integer such that . Let and be the count of the number of zeros and ones, respectively, in for all between 0 and inclusive. For , let denote the number of zeros between the th and th nonzero coefficient. Let be the indexing sequence associated with .
(1)If , let and be the remaining zeros after the last nonzero coefficient.(2)If , let be the number of leading zeros. We observe that for . So, we obtain by (1) and .
Then, satisfies the following equality:
Proof. Equation (50) can be derived from equation (36) by setting .
Example 4. Fix and . We consider the integral loop induced by , , and . The mapping appears asWe find that the trajectory and characteristic trajectory of 358 appear as
Definition 29. Fix . Let . We say that is rational-adic if eventually has a purely periodic -adic expansion.
Proposition 30. Fix so that and are coprimes. Let so that for each , , is coprime to and , and . Let such that . Let be the indexing sequence associated to . We assume that the trajectory of is bounded above. Then, there exists some contained in the trajectory of and the smallest in such that if and only if is a rational -adic.
Proof. We suppose first that in the trajectory of , there exists an so that . We assume further that is the smallest positive integer which satisfies this equality. Then, has a repeating period of length and is rational.
Now, we assume that is a rational -adic. Let be the starting value whose characteristic trajectory generates the repeating period of . Let us define the symbol as follows:where . We observe immediately that is finite as both and are finite. We have the following sequence of values by continuous application of the map generating values We can consolidate the terms into an expression that includes only and as follows:Thus, solving for yieldsWe suppose first that . Following from equation (55), we distribute the first term, yielding the following:Then, there exists an such that for all , . We find a new , which maybe different from such that generates an integral loop coinciding with the original rational loop.
If conversely we suppose , then we simply reverse equation (55) to solve for in terms of :Suppose first that , then it follows that . This equality is true for all , and thus, we have an integral loop. So, we assume instead that . Then, we can write equation (56) in the form of a ratio exhibited in equation:However, we recall that and are both fixed values. Thus, the left hand side has a fixed denominator. However, as the right hand side is the quotient of two coprime integers, the denominator expands with irreducible powers of two. If is a trajectory of integers, then there would be some index of which the denominator on the right hand side would be larger than the left hand side. This would be a contradiction. Finally, as , as a irrational ratio cannot be expressed with a finite -adic sequence. Thus, we have shown that no rational cycle apart from an integral loop exists for any mapping.
Example 5. Fix , , , and . The mapping appears asWe find that the trajectory and characteristic trajectory starting at appears asThat is to say, 25 starts an integral loop which induces a rational 3-adic sequence. Let us attempt to construct a trajectory, different from the integral loop starting at 25, based on the 3-adic sequence . We will proceed inductively by considering the smallest starting integer corresponding to repetitions of the finite sequence (see Table 1) .
We find that the sequence of starting integers diverges. We conclude that the only trajectory corresponding to the rational 3-adic sequence for all iterations is the integral loop starting at .
4. Conclusion
We have shown that the characteristic trajectory of a starting integer is eventually periodic if and only if the trajectory eventually enters an integral loop. This is not a new idea, but the proof presented here is new. The proof contained in Corollary 2.1a [7] employs a direct proof comparing parity vectors (characteristic trajectories). That proof would also apply to any two state system with halting condition (see Section 3.1). In that case, the nonzero symbol is replaced with one. The proof contained in Proposition 4 [8] demonstrates that if a characteristic trajectory is terminally periodic, then it represents a 2-adic number. The proof here does not require detailed knowledge of the 2-adics. Apart from the previous two proofs, this new proof demonstrates the claim is 2-adic invariant. In this way, we have presented a generalization of a known result.
Data Availability
The data used to support the findings of this study can be obtained from the corresponding author upon reasonable request.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research was sponsored by People’s Republic of China under the Chinese Government Scholarship via the Chinese Scholarship Council (CSC NO.: 2019ZMX001377). We are immensely grateful for their continued and unwavering support of the natural sciences. The authors would like to thank the following people for their kindness, support, and guidance: Mr. Yueheng Bao, Ms. Emily, Dr. Omer Hassan, Dr. Xu Bin, and B.K.