Proof that there are only 4 possible Collatz cycles Andrea Frasca June 4, 2023 In this paper I’ll prove that there are only 4 possible Collatz cycles in the extension of the Collatz conjecture to all integers, and excluding the 0 → 0 cycle. Preface The proof makes use of the ”Initial Partial Collatz Path”: Let d be a binary string representing the initial partial (legal) path that a positive integer can follow (1s being the 3 c + 1 step, 0s being the division by 2), and x an odd integer, for example: 101 is a legal initial collatz partial path for 3, as its path is 3 → 10 → 5 → ... 1001 is not a legal initial partial collatz path for 5, as its path is 5 → 16 → 8 → 4 → ... 10011001 is not a legal initial partial collatz path for any x , as the Collatz algorithm doesn’t allows consecutive 3 c + 1 steps. Follows that d can be whatever binary string that starts with a 1, doesn’t have consecutive 1s in its representation, and ends with a 1. Tools and Formalizations ⌢ : sequence concatenation; d { n } → 1 000 ... ︸ ︷︷ ︸ n 1 , n ∈ N d { n 1 ,n 2 ,n 3 ,...,n m } → 1 000 ... ︸ ︷︷ ︸ n 1 1 000 ... ︸ ︷︷ ︸ n 2 1 000 ... ︸ ︷︷ ︸ n 3 1 ... 1 000 ... ︸ ︷︷ ︸ n m 1 , n 1 , n 2 , n 3 , ..., n m ∈ N 1 ∆ = { n 1 , n 2 , n 3 , ..., n m } δ 0 (∆) is the sum of every term of ∆, informally the sum of the number of divi- sions occurring in the initial partial Collatz path: δ 0 (∆) = ∑ n m ∈ ∆ n m σ (∆ , k ) is the sum of the first k terms of ∆: σ (∆ , k ) = 0 , k = 0 k ∑ m =1 n m , k > 0 We now formalize the main function: d ∆ ( x ) → 3 | ∆ | +1 x + | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) 2 δ 0(∆) This is just a formalization of the recursive 3 x +1 2 ? process that depends on how many and which type of divisions must occur. Example: d { 2 , 2 , 2 } ( x ) = 3 4 x +3 3 +2 2 · 3 2 +2 4 · 3+2 6 2 6 = 81 x +175 64 In this case, ∆ = { 2 , 2 , 2 } , and, in fact, 1 is an odd integer for which d { 2 , 2 , 2 } (1) is an even number (The infinite path of 1 does, in fact, start with 3 consecutive divisions by 4). We can now proceed with the first step: if from d ∆ ( x ) = d ∆ ⌢ ∆ ( x ) follows that x ∈ N ∧ x ≡ 1 (mod 2) ∧ d ∆ ( x ) ≡ 0 (mod 2), then d ∆ is a Collatz cycle, and x is one of its odd elements; ——————————————————————————————————— Let’s expand d ∆ ( x ) = d ∆ ⌢ ∆ ( x ): 3 | ∆ | +1 x + | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) 2 δ 0(∆) = 3 2 | ∆ | +1 x + 2 | ∆ | ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) 2 2 δ 0(∆) Multiplying each side by 2 2 δ 0 (∆) : 2 δ 0 (∆) (3 | ∆ | +1 x + | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) ) = 3 2 | ∆ | +1 x + 2 | ∆ | ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) 2 Expanding: 2 δ 0 (∆) 3 | ∆ | +1 x +2 δ 0 (∆) | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) = 3 2 | ∆ | +1 x + 2 | ∆ | ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) Collecting the x s: 2 δ 0 (∆) 3 | ∆ | +1 x − 3 2 | ∆ | +1 x +2 δ 0 (∆) | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) = 2 | ∆ | ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) 3 | ∆ | +1 (2 δ 0 (∆) − 3 | ∆ | ) x = 2 | ∆ | ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) − 2 δ 0 (∆) | ∆ | ∑ k =0 3 | ∆ |− k 2 σ (∆ ,k ) In this step, I’m removing the elements of the first summation that coincide with elements of the second summation: 3 | ∆ | +1 (2 δ 0 (∆) − 3 | ∆ | ) x = | ∆ |− 1 ∑ k =0 3 2 | ∆ |− k 2 σ (∆ ⌢ ∆ ,k ) Dividing each term by 3 | ∆ | +1 : (2 δ 0 (∆) − 3 | ∆ | ) x = | ∆ |− 1 ∑ k =0 3 2 | ∆ |− k −| ∆ |− 1 2 σ (∆ ⌢ ∆ ,k ) (2 δ 0 (∆) − 3 | ∆ | ) x = | ∆ |− 1 ∑ k =0 3 | ∆ |− k − 1 2 σ (∆ ⌢ ∆ ,k ) Because k ≤ | ∆ | − 1, σ (∆ ⌢ ∆ , k ) now collapses to σ (∆ , k ): (2 δ 0 (∆) − 3 | ∆ | ) x = | ∆ |− 1 ∑ k =0 3 | ∆ |− k − 1 2 σ (∆ ,k ) x = | ∆ |− 1 ∑ k =0 3 | ∆ |− k − 1 2 σ (∆ ,k ) 2 δ 0(∆) − 3 | ∆ | ——————————————————————————————————— We now need to find for which ∆s, x ∈ N . This is next to impossible, so I am instead going to restrict the domain D ⊂ Q to which x belongs. This way, I can just test all odd integers in D to find possible cycles. To do this, we need to make D as big as possible without collapsing to Q and with reasonable min( D ) and max( D ): We need a way to restrict the possible ways a cycle can happen. My idea is to focus on the smallest integer that can belong to a cycle, that so happen to necessarily be an odd number. We’ll call this number α As the smallest odd integer of a cycle, we can play over the fact that after each Collatz step, every newly reached value must be ≥ α 3 How does a potential α behave? Our test is 5. Let’s perform the Collatz algorithm on it, but, this time, after every 3 c +1 steps, the number can be divided by 2 up to the point of not making it < 5: A simple Python program gives the following result: 5 → [16 , 8 , 25 , 6 25 , 19 75 , 9 875 , 30 625 , 7 65625 , ... ] Converting, the equivalent ∆ would be: ∆ = { 1 , 2 , 1 , 2 , 2 , 1 , 2 , 2 , ... } Follows that each value reached during a potential α ≥ 5 path can never be divided by more than 4 or would end up smaller than α , impossible by definition. Proof 1. Suppose, after an indefinite number of 3 c + 1 and division steps, to end up with a rational number just below 2 α . This will be the greatest rational that can’t be divided by 2 as it would go below α After the 3 c + 1 step, we get a rational number just below 6 α + 1, and because we need at least 8 α to be divisible by 8 and be greater than α , we need to prove the following inequality: 6 α + 1 < 8 α → α > 1 2 This concludes the proof. You could make the path ”climb” more, followed by a greater division, but the incoming Proof 2 will prove this is not possible. Sadly, because ln 2 ln 3 ∈ R \ Q , we still have an unpredictable behaviour of the sequence, alternating between division by 2 and 4; Luckily, just replacing the sequence with [1 , 2 , 2 , 1 , 2 , 2 , 1 , 2 , 2 , 1 , 2 , 2 ... ] (Substi- tuting every [ ..., 1 , 2 , ... ] with [ ..., 1 , 2 , 2 , ... ]) is enough of a compromise by excess that will make D a little bigger, but still not enough to cause min( D ) and max( D ) to diverge as | ∆ | grows. Simply put, I’m compromising in accepting greater values potentially belonging to a cycle instead of just focusing on α alone in favour of a periodic and more manageable sequence. We now prove that, on equal terms of δ 0 (∆), this is the best sequence of divisions for α (Why is { 1 , 2 , 2 , 1 , 2 , 2 , ... } a better option than, say, { 1 , 1 , 1 , 3 , 2 , 2 , ... } , in maximizing D ?): Proof 2. In { 1 , 2 , 2 , 1 , 2 , 2 ... } case, the first terms of the numerator are: 3 | ∆ |− 1 + 3 | ∆ |− 2 2 1 + 3 | ∆ |− 3 2 1+2 + 3 | ∆ |− 4 2 1+2+2 + 3 | ∆ |− 5 2 1+2+2+1 + ... While in { 1 , 1 , 1 , 3 , 2 , 2 ... } case: 3 | ∆ |− 1 + 3 | ∆ |− 2 2 1 + 3 | ∆ |− 3 2 1+1 + 3 | ∆ |− 4 2 1+1+1 + 3 | ∆ |− 5 2 1+1+1+3 + ... Follows that, ignoring the first two terms because they coincide, we need to prove: 4 3 t − 3 2 1+2 + 3 t − 4 2 1+2+2 + 3 t − 5 2 1+2+2+1 > 3 t − 3 2 1+1 + 3 t − 4 2 1+1+1 + 3 t − 5 2 1+1+1+3 3 t − 3 2 3 + 3 t − 4 2 5 + 3 t − 5 2 6 > 3 t − 3 2 2 + 3 t − 4 2 3 + 3 t − 5 2 6 That is just proving 3 t > 0. You can’t increase D by shifting divisions down the line to obtain a single great division. This would cause the numerator of x to shrink. This is because the exponent of each 2 is obtained by summation of consecutive values, and not by picking each value alone. This concludes the proof. ——————————————————————————————————— From now on, ∆ will just refer to sequences of the form: ∆ = { 1 , 2 , 2 , 1 , 2 , 2 , 1 , 2 , 2 , ... } We can now simplify the definition of x : x = | ∆ |− 1 ∑ k =0 3 | ∆ |− k − 1 2 a ( k ) 2 a ( | ∆ |− 1) − 3 | ∆ | where a ( β ) = 5 β 3 + 1 , β ≡ 0 (mod 3) 5 ⌊ β 3 ⌋ + 3 , β ≡ 1 (mod 3) 5 ⌊ β 3 ⌋ + 5 , β ≡ 2 (mod 3) As x now finds itself in a modular equations, we have 3 results (These are the smallest and biggest values that x reached before converging to, respectively, to 23 5 , 37 5 and 29 5 ): x ∈ { 23 5 } , | ∆ | ≡ 0 (mod 3) ( − 5095 139 , 213341 6487 ) , | ∆ | ≡ 1 (mod 3) ( − 5 , 367 13 ) , | ∆ | ≡ 2 (mod 3) restricting to integers we have: x ∈ { ( − 36 , 32) , | ∆ | ≡ 1 (mod 3) [ − 5 , 28) , | ∆ | ≡ 2 (mod 3) Merging the sets, follows that: α ∈ ( − 36 , 32) Meaning that all the smallest odd numbers of all possible cycles belongs to this set. Testing every odd integer in that set, we get that: α ∈ {− 17 , − 5 , − 1 , 1 } , follows that only 4 cycles exist. 5