SSome considerations on Collatz conjecture
Fabrizio LuccioDepartment of Informatics, Universit`a di [email protected] 7, 2020
Abstract
Some simple facts are proved ruling the Collatz tree and the chains ofvertices appearing in it, leading to the reduction of the number of significantelements appearing in the tree. Although the Collatz conjecture remainsopen, these fact may cast some light on the nature of the problem.
The Collatz Conjecture (1937), also called 3 x + 1 Problem, concerns iteration ofthe map T : Z + → Z + given by the two rules: R1 T ( x ) = x/ x is even R2 T ( x ) = (3 x + 1) / x is oddand asserts that each n ≥ T ( h ) ( n ) = 1. Hence each n wouldeventually converge to the limit cycle 2 → → . . . There is no proof that the conjecture holds, that is for all the integers in Z + no cycle other than 2 → → . . . is formed, or in general that all integersiterate to 1, although experiments conducted on the integers up to 5 × showthat this is the case for them. It is known, however, that some integers followa long chain of transformations before getting to the limit cycle. For example n = 27 takes 80 steps, climbing to 4616 before descending to 1, according to thesequence: , 41, 62, 31, 47, 71, 107, 161, 242, 121, 182, 91, 137, 206, 103, 155, 233, 350,175, 263, 395, 1593, 890, 445, 668, 334, 167, 251, 377, 566, 283, 425, 638, 319,479, 719, 1079, 1619, 2429, 3644, 1822, 911, 1367, 2051, 3077, , 2308, 1154,577, 866, 433, 650, 325, 488, 244, 122, 61, 92, 46, 23, 35, 53, 80, 40, 20, 10, 5, 8,4, 2, The evolution of the integers induced by T can be immediately representedas a directed graph that, according to the conjecture, is in fact a tree (called the Collatz tree ) ending in the limit cycle at the root. A portion of this tree is shownin Figure 1. 1 a r X i v : . [ m a t h . G M ] J u l
2 6 3 10 20 5 16 13 40 8
64 21 32 26 42 128 24 52 17 11 7 34 22 85
Figure 1:
Part of the Collatz tree.
We now state some simple facts ruling the Collatz tree and the chains of verticesappearing in it. To the best of our knowledge these points have not been explicitlyraised before.First note that each vertex x of the graph may have one or two predecessorsderiving from the application of R1 or R2 to a previous vertex. This leads toconsider two inverses of T indicated by P e ( x ) and P o ( x ), where the subscripts e and o indicate if the predecessor is even or odd, hence rule R1 or R2 has beenapplied. Clearly each vertex x has an even predecessor P e ( x ) = 2 x , and mayhave an odd predecessor P o ( x ) = (2 x − / R2 ) if this value is anodd integer. For example, for x = 5 we have P e (5) = 10 and P o (5) = 3, seeFigure 1.In particular, for the non-negative integers consider the residue classes [0],[1], [2] modulo 3, respectively containing the integers 3 k , 3 k + 1, 3 k + 2 with k ≥
0. We have:
Fact 1. (i) each vertex x ∈ [0] has only one predecessor P e ( x ) ∈ [0] ; (ii) each vertex x ∈ [1] has only one predecessor P e ( x ) ∈ [2] ; (ii) each vertex x ∈ [2] has both predecessors P e ( x ) and P o ( x ) , with P e ( x ) ∈ [1] ,and P o ( x ) ∈ [0] if ( x − / ∈ [1] , P o ( x ) ∈ [1] if ( x − / ∈ [0] , P o ( x ) ∈ [2] if ( x − / ∈ [2] . Proof.
Each vertex x has an even predecessor P e ( x ) = 2 x , and also has an oddpredecessor P o ( x ) = (2 x − / R2 applies to that predecessor. In case (i) wehave x = 3 k and P o ( x ) = (6 k − / x = 3 k + 1 and P o ( x ) = (6 k + 1) /
3, however, these two values are not integer and P o ( x ) does2ot exist. In case (iii) we have x = 3 k + 2 and P o ( x ) = (6 k + 3) / k + 1, anodd integer and evolves to 3 k + 2 according to R2 . The residue classes of P o ( x )according to the nature of ( x − / (cid:50) Referring to Figure 1, case (i) of Fact 1 applies to the backward chains 3, 6,12, 24, . . . and 21, 42, . . . with x = 3 k . Case (ii) applies to vertices 1, 4,7, 10, 13, 16 . . . with x = 3 k + 1. Case (iii) applies to vertices 2, 5, 8, 11, 17,32, . . . with x = 3 k + 2. Fact 2.
For each vertex x ∈ [0] we have: (i) T ( x ) ∈ [0] for x/ even; (ii) T ( x ) ∈ [2] for x/ odd. Proof.
Point (i) is immediate. For point (ii) we have x = 3 k and T ( x ) =(9 k + 1) / h + r , with 0 ≤ r ≤
2, hence h = (9 k − r + 1) /
6. Since k is odd, h is integer only for r = 2. (cid:50) Referring to Figure 1, point (i) of Fact 2 applies to vertices 6 and 42, andpoint (ii) applies to vertices 3 and 21.
Fact 3.
For each vertex x ∈ [1] we have T ( x ) ∈ [2] . Proof. x = 3 k + 1. For k even we have: T ( x ) = (9 k + 4) / h + r , with0 ≤ r ≤
2, hence h = (9 k − r + 4) /
6. Since k is even, h is integer only for r = 2.For k odd we have x = 3( k −
1) + 4 even, then T ( x ) = 3( k − / h + 2with h = ( k − / (cid:50) Referring to Figure 1, Fact 3 applies to vertices 7 and 13 for k even, and tovertices 4 and 10 for k odd. In both cases T ( x ) can be even or odd. Fact 4.
For each vertex x ∈ [2] we have: (i) T ( x ) ∈ [1] , for x even; (ii) T ( x ) ∈ [2] , for x odd. Proof. x = 3 k + 2. For x even we have k even and T ( x ) = 3 k/ x odd we have k odd and T ( x ) = (9 k + 7) / h + r , with 0 ≤ r ≤
2, hence h = (9 k − r + 7) /
6. Since k is odd, h is integer only for r = 2. (cid:50) Referring to Figure 1, point (i) of Fact 4 applies to vertices 8 and 20, andpoint (ii) applies to vertices 5 and 11.
Fact 5.
The Collatz graph contains no cycle of length one and exactly only cycleof length two consisting of vertices 1 and 2.
Proof.
Obviously T ( x ) (cid:54) = x hence no cycle of length one exists. In a cycle oflength two we have T (2) ( x ) = x and both rules R1 , R2 must apply to avoid that x iterates in two steps to a value that is certainly smaller or certainly larger than x . For x even we have T (2) ( x ) = (3 x/ / x + 2) /
4, and the equality(3 x + 2) / x implies x = 2. For x odd we have T ( x ) = (3 x + 1) /
2. In this case,if T ( x ) is even we have T (2) ( x ) = ((3 x + 1) / / x + 1) /
4, and the equality(3 x + 1) / x implies x = 1; if T ( x ) is odd we have T (2) ( x ) > x since rule R2 applies twice. (cid:50) simple cycles where these iterations have been eliminated. Having said that, it isnot known whether cycles exist in the Colatz graph, other than the one composedof vertices 1 and 2. As far as the existence of longer cycles, we can only statesome partial results. Fact 6.
If the rules R1 and R2 are respectively applied r and r times in acycle C , any vertex x of C must fulfil the equation: x = A/ (2 r + r − r ) , where A is a positive integer and r > . r . Proof.
Staring from vertex x , a sequence of k = r + r applications of R1 and R2 must occur in C until x is reached again, that is T ( k ) ( x ) = x . It isstraightforward to verify that this leads to the equation: (3 r x + A ) / r + r = x ,that is x = A/ (2 r + r − r ). As x must be a positive integer we have 2 r + r > r hence r > r ( log − (cid:50) We can easily prove that no cycle of length three exists. In fact, by Fact 6such a cycle should require applying R2 once and R1 twice, in one of the threeorderings R1-R1-R2 , R1-R2-R1 , R2-R1-R1 , and in all cases the relation T (3) ( x ) = x cannot be satisfied. By the same reasoning, to have a cycle of lengthfour one should apply R1 two or three times, and only the sequences R1-R2-R1-R2 and
R2-R1-R2-R1 satisfy T (4) ( x ) = x with x = 2 and x = 1 respectively,corresponding to the non-simple cycles 2-1-2-1 and 1-2-1-2. Fact 7.
No vertex v ∈ [0] can belong to a finite cycle. Proof.
All the vertices in [0] belong to some infinite chain C of vertices 2 i × k with k ≥ i ≥
0, whose last vertex is 3 k and each vertex has exactlyone predecessor by Fact 1(i). A finite cycle including one or more vertices of C should include vertex 3 k and procede through a chain of vertices not in C , oneof which, say x , has a successor y in C to form the cycle. Then y should havetwo predecessors, one in C and the other not in C , against Fact 1(i) . (cid:50) For example see the backwards chains 3, 6, 12, 24, . . . and 21, 42, . . . inFigure 1.
Fact 8.
If a vertex x ∈ [1] belongs to a finite cycle of length at least two, theexistence of a cycle is preserved by eliminating x from the graph and connecting P e ( x ) to T ( x ) directly. Proof.
By Facts 1 and 3 each vertex x ∈ [1] has exactly one predecessor and onesuccessor in [2]. If x belongs to a cycle C of length (cid:96) ≥
2, a new cycle of length (cid:96) − P e ( x ) to T ( x ) directly . (cid:50) For example, in Figure 1 vertices 4 and 13 could be eliminated connecting 8to 2, and 26 with 20, respectively.As a consequence of Facts 7 and 8, the existence of cycles can be studiedmaintaining only the vertices in [2] with the insertion of new edges as indicatedin Fact 8, and reformulating the Collatz conjecture as the iteration of the map T (cid:48) : [2] → [2] given by the rules: 4 ’1 T (cid:48) ( x ) = x/ x and x/ R’2 T (cid:48) ( x ) = (3 x + 2) / x is even and x/ R’3 T (cid:48) ( x ) = (3 x + 1) / x is oddand asserting that each n ∈ [2] has some iterate T ( h ) ( n ) = 2.It can be easily seen that if x ∈ [2] also T (cid:48) ( x ) ∈ [2]. In particular, comparedwith the original map T , rules R’1 and
R’2 perform two steps while
R’3 coincideswith R2 of T , since in the original graph vertex x is followed by another vertexin [2] (see Fact 4). Figure 1 then reduces to Figure 2. Note that R’1 is appliedto vertices 8, 20, 32, and 128;
R’2 is applied to vertices 2 and 26;
R’3 is appliedto vertices 5, 11, and 17. The limit cycle 2 - 1 reduces to a cycle of length onecontaining the sole vertex 2.
20 5 8
32 26 128 17 11
Figure 2:
The reduced Collatz tree.3 Bibliography