IInteger patterns in Collatz sequences
Zenon B. Batang
Coastal and Marine Resources Core Laboratory, King Abdullah University of Science andTechnology, Thuwal 23955, Saudi Arabia
Abstract
The Collatz conjecture asserts that repeatedly iterating f ( x ) = (3 x + 1) / a ( x ) ,where a ( x ) is the highest exponent for which 2 a ( x ) exactly divides 3 x + 1, alwayslead to 1 for any odd positive integer x . Here, we present an arborescence graphconstructed from iterations of g ( x ) = (2 e ( x ) x − /
3, which is the inverse of f ( x )and where x (cid:54)≡ [0] and e ( x ) is any positive integer satisfying 2 e ( x ) x − ≡ [0] ,with [0] denoting 0 (mod 3). The integer patterns inferred from the resultingarborescence provide new insights into proving the validity of the conjecture. Keywords:
Collatz conjecture, Arborescence, Covering system, Cycle, Proof
1. Introduction
Denote by N = { , , , . . . } the set of natural numbers and let N := N ∪{ } , E := 2 N and U := 2 N +1. Write [ r ] q for r (mod q ), where r is least nonnegativeresidue modulo q . Every x ∈ U can be stated as x = 3 µ r + r , where µ r ∈ N isthe multiple of x ≡ [ r ] . Note that µ , µ ∈ U and µ ∈ E . Define the mapping f : U → U by f ( x ) = (3 x + 1) / a ( x ) , (1)where a ( x ) ∈ N is the highest exponent for which 2 a ( x ) exactly divides 3 x + 1.The famous Collatz conjecture asserts that for every x ∈ U , there exists k ∈ N such that f k ( x ) = 1, where f k denotes k compositions of f . For an initial input x , let f k ( x ) = x k and any k iterations of f on x generate a sequence of oddintegers, termed Collatz sequence or trajectory , denoted by S k ( x ) = { x , x , . . . , x k } k ∈ N . (2)We have f ( x ) = x as the identity map and f ∞ (1) = 1 forms the trivial cycle ( loop ). Collatz conjecture claims that (1) always yields the trivial cycle for any x ∈ U , i.e. repeatedly iterating (1) always yields a convergent Collatz sequence.
Email address: [email protected] (Zenon B. Batang)
Preprint submitted to TBD July 18, 2019 a r X i v : . [ m a t h . G M ] J u l he Collatz conjecture is an intriguing problem in mathematics that hasremained unsolved for over 80 years despite its apparent simplicity. While prob-abilistic heuristics, stochastic models, and computational verifications suggestthat the conjecture is likely true, it has so far resisted any attempts at a com-plete proof by different mathematical approaches. One can refer to [1], [2] and[5] for an overview of the conjecture, with an annotated bibliography in twoparts provided by Lagarias ([3], [4]). Recognizing the notorious difficulty of theCollatz conjecture, the prolific mathematician Paul Erd˝os famously stated that “Mathematics is not yet ready for such problems” ([1]). In this paper, we presentnew insights into integer patterns that underlie the truth of the conjecture.An arborescence is a directed rooted tree in which there is only one directedpath from the root to any other vertex, thus all edges point away from the root.Here, we construct an infinite arborescence G = ( V, E ), with vertex ( node ) set V ( G ) and edge set E ( G ), based on the inverse mapping g : U p → U defined by g ( x ) = (2 e ( x ) x − / , ∀ x ∈ U p , (3)where U p = U (cid:54)≡ [0] and e ( x ) is any positive integer such that 2 e ( x ) x − ≡ [0] .One easily obtains e ( x ) = 2 n if x ≡ [1] and e ( x ) = 2 n − x ≡ [2] for n ∈ N ,while g ( x ) = ∅ if x ≡ [0] . Obviously, g is a one-to-many association mapping.Ignoring the trivial cycle, each iterate of (3) corresponds to a vertex in V ( G )and a single step of iteration for every n ∈ N represents a directed edge in E ( G ).Collatz conjecture implies that repeatedly iterating (3) forms an arborescence G with a root labeled 1, excluding the trivial cycle since G is inherently acyclic (Figure 1). Any finite sequence of vertices along a directed path of at least oneedge from the root in G corresponds to a Collatz trajectory in reverse. Hence,we call G the inverse Collatz graph . Our aim is to show that each vertex in G ,excluding the trivial cycle, is unique and that V ( G ) = U to validate the Collatzconjecture. k =0 k =1 k =2 trivial cycle 0 (mod 3) 1 (mod 3) or 2 (mod 3) 17 69 277 35 141 565 75 301 1205 2417 9669 38677 151 605 2421 4849 19397 77589 k =3 Figure 1: Structure of G . . Constructing G Given adjacent vertices u and v , denote by e = ( u, v ) the directed edge from u to v , which we call parent and child , respectively. Children of the same parentare siblings and form a sibling set . Applying (3) leads to the transition rule as g n ( u ) = v n = ∅ , if u ≡ [0] , n u − , if u ≡ [1] , n − u − , if u ≡ [2] , (4)for n ∈ N . Hence, each u (cid:54)≡ [0] at depth k ∈ N in G is locally infinite since g n ( u ) yields infinitely many children { v n } n ∈ N . It follows that every u (cid:54)≡ [0] is a branching ( inner ) node, with infinitely many outgoing edges { e n = ( u, v n ) } n ∈ N ,while u ≡ [0] is a leaf ( terminal ) node, without outgoing edges. Denote bydeg − ( u ) and deg + ( u ) the indegree (number of incoming edges) and outdegree (number of outgoing edges), respectively, of any u ∈ V ( G ). Excluding the trivialcycle, we have for u ∈ U thatdeg − ( u ) = (cid:40) u = 1 , u > , deg + ( u ) = (cid:40) , if u ≡ [0] , ℵ , if u (cid:54)≡ [0] , (5)where ℵ denotes the cardinality of N . For n ∈ N , let z n := 2 n −
13 = n (cid:88) i =1 i − = 1 + 4 + · · · + 4 n − (6)to form the infinite strictly ascending sequence denoted by Z = { z n } n ∈ N = { , , , , , . . . } . (7)Then for u ∈ U , write u ≡ [ r ] as u = 3 µ r + r to restate (4) as g n ( u ) = v n = ∅ , if u ≡ [0] ,z n + 2 n µ , if u ≡ [1] ,z n + 2 n − µ , if u ≡ [2] , (8)for n ∈ N . Also note the recurrence relation h ( v n ) = v n +1 = 1 + 4 v n ,h n ( v ) = v n +1 = z n + 4 n v . (9)With (8) or (9), one can specify the sequence of vertices in the sibling set of anyparent u ∈ U p in G .Denote by H k ( u ) the sibling set at k ∈ N arising from u ∈ U p at k − k be the union of all vertices at any k ∈ N in G . Suppose τ is the root of3 , which is a lone vertex at k = 0, then Π = { τ } is a singleton subset of V ( G ).Applying (4) or (8) on τ yields H ( τ ) = { g n ( τ ) } n ∈ N = { v n } n ∈ N = Z (10)at k = 1, where one should note that e = ( τ, v ) forms the trivial cycle. Sincethe root is a lone vertex at k = 0, only a single sibling set with infinitely manyelements exists at k = 1. Excluding the trivial cycle, we have thatΠ := H ( τ ) \ { } = Z \ { } , (11)such that Π ∪ Π = Z . Note that | Π | = 1 and | Π | = ℵ , where | A | denotes thecardinality of A . In turn, every v n (cid:54)≡ [0] in Π generates an infinite sibling setat k = 2, i.e. | H ( v n ) | = ℵ for every v n ∈ U p at k = 1. Since |{ v n ∈ Π : v n ∈ U p }| = ℵ , it follows that a countably infinite number of siblings sets exist at k = 2, where | Π | = ℵ given that ℵ × ℵ = ℵ . In general, if u k ∈ V ( G ) isany vertex at k ∈ N in G , i.e. u k ∈ Π k , then we have thatΠ k +1 = (cid:91) u k ∈ U p H k +1 ( u k ) . (12)One can also restate (12) using a multi-index that specifically identifies eachsibling set at any k , but such a notation becomes complicated as k increases in G . We thus ignore any specific reference to every sibling set at higher k , withoutloss of generality. Note that | Π | = 1 | Π | = ℵ , | Π | = ℵ × ℵ = ℵ , . . . | Π k | = k (cid:122) (cid:125)(cid:124) (cid:123) ℵ × ℵ × · · · × ℵ = ℵ . The above pattern holds from the root to infinitely increasing k , such that | Π k | = (cid:40) , if k = 0 , ℵ , if k > . (13)Hence, from the root, G recursively assembles into a nested hierarchy of infinitelymany sibling sets, each having a cardinality ℵ (Figure 1). As shown below, eachsibling set uniquely represents an infinite sequence of odd positive integers. It isimplicit from (13) that G is both vertically infinite from the root and horizontallyinfinite except for the root.
3. Patterns in G Let u ∈ V ( G ) and define P := { u ∈ V ( G ) : u ∈ U p } as the subset containingall parent nodes in G , such that P ⊂ V ( G ). From (12), it follows that | P | = ℵ ,which implies that there exist infinitely many sibling sets in G . Observe that (8)and (9) permute the residue class (mod 3) of the vertices in every sibling set.4he notation { ( x, y, z ) } indicates that ( x, y, z ) forms a cycle in the sequence,i.e { ( x, y, z ) } = { x, y, z, x, y, z, . . . } and { u, ( x, y, z ) } = { u, x, y, z, x, y, z, . . . } . Itis understood that u ∈ U p at k − H k ( u ) = { v n } n ∈ N . Werecord the fact that Lemma 1.
The residue class (mod 3) of the vertices in H k ( u ) = { v n } n ∈ N for k > takes the cycle { ( r, r + 1 , r + 2) } (mod 3) , where v ≡ [ r ] .Proof. Write v ≡ [ r ] as v = 3 µ r + r for r = 0 , , v ≡ [0] , v = 3(4 µ ) + 1 , v = 3(4 µ + 1) + 2 , . . .v ≡ [1] , v = 3(4 µ + 1) + 2 , v = 3(4 µ + 7) , . . .v ≡ [2] , v = 3(4 µ + 3) , v = 3(4 µ + 12) + 1 , . . . (14)which correspond to the claim of the lemma.Let w n be the multiple of z n ≡ [ r ] . Write z n = 3 µ r + r for n ∈ N , such that w n = z n − r n/ (cid:88) i =1 i − , if n ≡ [0] , ( n − / (cid:88) i =1 i − , if n ≡ [1] , ( n − / (cid:88) i =1 i − , if n ≡ [2] . (15)From (15), it is easy to see that w n = µ r for n ≡ [ r ] in z n . Denote by W theset of multiples corresponding to the elements in Z . Then W = { w n } n ∈ N = { , , , , , , . . . } . (16)Similarly, let M k ( u ) = { m n } n ∈ N be the set of multiples corresponding to thevertices in H k ( u ) = { v n } n ∈ N at k >
0, such that for any v n ≡ [ r ] we have that m n = v n − r . (17)Note that M ( τ ) = W \ { } given that H ( τ ) = Z \ { } in (11), where τ is theroot of G . A general pattern emerges as Lemma 2.
Every M k ( u ) = { m n } n ∈ N at k > is an infinite strictly ascendingsequence as M k ( u ) = { µ , w n − + 4 n − µ } n> , if v ≡ [0] , { µ , w n + 4 n − µ } n> , if v ≡ [1] , { µ , w n +1 + 4 n − ( µ − } n> , if v ≡ [2] , (18) where m = µ r is the multiple of v ≡ [ r ] and w n is as above. roof. One can easily check that the sequences in (18) results from successivelyiterating (9) on v ≡ [ r ] written as v = 3 µ r + r . Lemma 3.
For every H k ( u ) = { v n } n ∈ N , write u = 3 µ r + r , where r = 1 , .Given that g n ( u ) = v n , we have that u ≡ [1] : v = u + µ , v = 2 u + v , v = 2 u + v , . . .v n = 2 n − u + v n − = u n − (cid:88) i =1 i + v ,u ≡ [2] : v = ( u + µ ) / , v = 2 u + v , v = 2 u + v , . . .v n = 2 n − u + v n − = u n − (cid:88) i =1 i − + v . (19) Proof.
From (4), we obtain v = (3 µ + 1) −
13 = u + µ , if u ≡ [1] , µ + 2) −
13 = u + µ , if u ≡ [2] . Then repeatedly applying (9) to find v n given u ≡ [ r ] leads to (19).Both (18) and (19) also provide alternative ways to enumerate the sequenceof vertices in H k ( u ) = { v n } n ∈ N arising from any u ∈ U p . Call v the initialvertex in H k ( u ). From Lemma 2, it is clear that the sequence of initial verticesin all sibling sets at k > µ ∈ E and µ , µ ∈ U , thus assuring that v n ∈ U for all n ∈ N given u ≡ [ r ] ,where r = 1 ,
2. From (19), we have that v n +1 − v n = (cid:40) n u, if u ≡ [1] , n − u, if u ≡ [2] , (20)which implies that the interval between consecutive vertices in any sibling set ismonotone increasing to infinity. Along with Lemma 3, the importance of (20)to our argument shall be evident subsequently. At this point, it is also relevantto note that Lemma 4.
Let u i , u i +1 ∈ U p be successive nodes in a sibling set at k > in G .If u i ≡ [1] for some i ∈ N , then g ( u i ) = v ( u i ) = 1 + 4 µ = µ ,g ( u i +1 ) = v ( u i +1 ) = 1 + 2 v ( u i ) = 3 + 8 µ , (21) where v ( u i ) and v ( u i +1 ) are the initial vertices in H k +1 ( u i ) and H k +1 ( u i +1 ) ,and µ and µ are the multiples of u i and u i +1 (mod 3) , respectively. roof. From Lemma 1, if u i ≡ [1] with multiple µ , then u i +1 ≡ [2] with µ =1 + 4 µ . Applying (8) yields g ( u i ) = v ( u i ) = 1 + 4 µ and g ( u i +1 ) = v ( u i +1 ) =1 + 2 µ . By substitution, v ( u i ) = µ and v ( u i +1 ) = 1 + 2 v ( u i ) = 3 + 8 µ , asstated in (21).Given the monotone increasing sequence of infinitely many vertices in everysibling set, the least element in Π k is the initial vertex of the first (leftmost)sibling set at any k ∈ N in G . Lemma 4 also implies that the least parent nodein Π k always yields the least element in Π k +1 .
4. Collatz is Right
Call u ∈ U p > non-trivial parent in G . One easily infers from (8) that nosuch u iterates to itself, i.e. g ( u ) (cid:54) = u for any u ∈ U p >
1. In addition, no u ∈ U p ,including the root, can give rise to two or more equal siblings since every siblingset is an infinite strictly ascending sequence. Hence, the only scenario where anon-trivial cycle exists in G rests on the possibility that distinct parents can yieldequal children belonging to separate sibling sets under g iteration. Assume thatthere exist such non-sibling nodes v n , v m ∈ V ( G ), where v n = v m for n, m ∈ N and, necessarily, n (cid:54) = m . Without loss of generality, let n < m , with v n and v m being at the same or different k in G . Suppose u i and u j , where u i (cid:54) = u j , arethe non-trivial parents of v n and v m , respectively. We consider two cases where u i ≡ u j (mod 3) or u i (cid:54)≡ u j (mod 3). Noting that G excludes the trivial cycle,we show that Lemma 5 (Uniqueness) . Every vertex in G is unique for all k ∈ N .Proof. For v n = v m , where u i (cid:54)≡ u j (mod 3), assume that u i ≡ [1] and u j ≡ [2] , with multiples µ and µ , respectively. If n < m , then µ > µ . From (8),we have that z n + 2 n µ = z m + 2 m − µ . Let m = n + d , where d >
0, such that n (cid:88) i =1 i − − n + d (cid:88) i =1 i − = 2 n + d ) − µ − n µ , − n + d (cid:88) i = n i − = 2 n (cid:0) d − µ − µ (cid:1) , − d (cid:88) i =1 i − = 2 d − µ − µ . Expressing for µ , we get µ = 2 d − µ + d (cid:88) i =1 i − , µ ∈ E . In the casewhere u i ≡ u j (mod 3), it suffices to test only for u i ≡ u j ≡ [1] . For distinction,let µ and ν , where µ > ν , be the multiples of u i and u j (mod 3), respectively,such that z n + 2 n µ = z m + 2 m ν . As above, we obtain n (cid:88) i =1 i − − n + d (cid:88) i =1 i − = 2 n + d ) ν − n µ , − n + d (cid:88) i = n i − = 2 n (cid:0) d ν − µ (cid:1) , − d (cid:88) i =1 i − = 2 d ν − µ , leading to µ = 2 d ν + d (cid:88) i =1 i − , which also contradicts the condition that µ ∈ E . Thus, both results imply thatno distinct non-trivial parents can generate equal children (non-siblings), whichvalidates the claim of the lemma that every vertex in G is unique.Lemma 5 has the immediate consequence that (cid:92) k ∈ N (cid:92) u ∈ U p H k ( u ) = (cid:92) k ∈ N Π k = ∅ . A non-trivial cycle has a sequence of vertices that does not include 1. Lemma5 implies that no non-trivial cycles exist in G . For any u ∈ V ( G ), where u ∈ U ,there exists q ∈ N and r, m ∈ N , such that u − r = m · q ⇐⇒ u ≡ r (mod q ) . A system of n linear congruences of the form u ≡ r i (mod q i ) , i ∈ { , , , . . . , n } is called a covering system , or simply a covering , of U if every u ∈ U belongsto at least one of the residue classes. The concept of covering systems was firstintroduced by Erd˝os in 1950 ([6]) to prove that a positive proportion of U cannotbe expressed as the sum of a prime number and a power of 2. The modulus q i in a covering system may be the same or distinct between residue classes.In particular, a covering of U is exact if every u belongs to exactly one of theresidue classes. We next show that Lemma 6 (Completeness) . V ( G ) = U . roof. From (8), every parent u ≡ [1] yields v = 1 + 4 µ . Since µ ∈ E , weget µ = E , if µ ≡ [0] , U + 1 , if µ ≡ [1] , E + 2 , if µ ≡ [2] . Similarly, if u ≡ [2] such that v = 1 + 2 µ , where µ ∈ U , we obtain µ = U , if µ ≡ [0] , E + 1 , if µ ≡ [1] , U + 2 , if µ ≡ [2] . By substituting for E and U , noting that u ≡ [0] is a leaf node, we derive u ≡ [1] : v = N , if µ ≡ [0] ,
17 + 24 N , if µ ≡ [1] , N , if µ ≡ [2] ,u ≡ [2] : v = N , if µ ≡ [0] , N , if µ ≡ [1] ,
11 + 12 N , if µ ≡ [2] . (22)Observe that v ≡ [0] , i.e. a leaf node, when µ ≡ [2] in u ≡ [1] and µ ≡ [1] in u ≡ [2] ; hence, v ≡ [1] is the first parent node in H k ( u ) in such cases. Byapplying (9) to (22), we find that u ≡ [1] : H k ( u ) ≡ { , (5 , , } (mod 24) , if µ ≡ [0] , { , (21 , , } (mod 24) , if µ ≡ [1] , { , (13 , , } (mod 24) , if µ ≡ [2] ,u ≡ [2] : H k ( u ) ≡ { , (5 , , } (mod 12) , if µ ≡ [0] , { , (1 , , } (mod 12) , if µ ≡ [1] , { , (9 , , } (mod 12) , if µ ≡ [2] , (23)where ( r , r , r ) indicates a cycle of r i (mod 24) in each sequence.Partition V ( G ) into subsets V [1] and V [2] defined by V [1] = (cid:91) k ∈ N (cid:91) u ≡ [1] H k ( u ) , V [2] = (cid:91) k ∈ N (cid:91) u ≡ [2] H k ( u ) , (24)where u ∈ U p at k − τ is equal to v ∈ H ( τ ). Clearly, we havethat V [1] ∪ V [2] = V ( G ) . (25)Write A ≡ { r i } (mod q i ) to mean that every a ∈ A is congruent to exactly one r i (mod q i ) for i ∈ { , , , . . . , n } . We obtain from (23) that V [1] ≡ { , , } (mod 12) , V [2] ≡ { , , , , , } (mod 12) , (26)9hich reveals that no element in V [1] is congruent to { , , } (mod 12). Thisis easy to confirm since from (1) we have that f (3 + 12 N ) = 5 + 18 N = 2 + 3(1 + 6 N ) ,f (7 + 12 N ) = 11 + 18 N = 2 + 3(3 + 6 N ) ,f (11 + 12 N ) = 17 + 18 N = 2 + 3(5 + 6 N ) , which are all congruent to [2] , but V [1] only consists of vertices arising from f ( x ) = u ≡ [1] .Let V = { v ∈ V [1] } and V = { v ∈ V [2] } , which means that V r contains allinitial vertices in V [ r ] for r = 1 ,
2, such that V ⊂ V [1] and V ⊂ V [2] . It is trivialthat V ∩ V = ∅ , (27)since g ( u ) (cid:54) = g ( v ) for any u ≡ [1] and v ≡ [2] . To show this, let µ and µ bethe multiples of u and v , respectively. Then one obtains1 + 4 µ = 1 + 2 µ , µ = µ (28)which is false since µ ∈ E and µ ∈ U . In fact, one easily finds from (8) that V = { N } and V = { N } , which are mutually exclusive. Hence,(27) also implies Lemma 5 since every sibling set is an infinite strictly ascendingsequence that is fully determined by its initial element. It follows that V [1] and V [2] are mutually disjoint, i.e. V [1] ∩ V [2] = ∅ . (29)Since { , , , , , } (mod 12) exactly covers U and V [1] ∪ V [2] ≡ { , , , , , } (mod 12), with each vertex in G being unique, it follows that V ( G ) = U andthe proof is complete.It is obvious that G is weakly connected since if it is treated as an undirectedgraph, then every pair of distinct vertices u and v can be joined by an undirectedpath. For any u ∈ P , where P is the subset containing all parent nodes in G such that P ⊂ V ( G ), notice that v > u if u ≡ [1] and v < u if u ≡ [2] . Wesay that e = ( u, v ) is ascending if u ≡ [1] or descending if u ≡ [2] . Any path p ( u , u k ) = { u , u , u , . . . , v k } is strictly ascending or descending if u i ∈ V [1] or u i ∈ V , respectively, for all i = 0 , , , . . . , k − G . The end vertex u k in p ( u , u k ) can be any element in the sibling set arising from u k − , such that u i ∈ U p for all i = 1 , , , . . . , k − u k ∈ U . If u k is a leaf node, i.e. u k ≡ [0] ,then p ( u , u k ) cannot extend to any other vertices beyond k , in which case wesay that u k is a dead end and p ( u , u k ) is truncated at u k . This leads to the factthat every path from the root in G eventually terminates in a dead end. Hence,there can only be at most one vertex ≡ [0] in any p ( u , u k ) and if it exists, itmust be u k .To reinforce the proof of Lemma 6, we further demonstrate how V ( G ) exactlycovers U . Consider a number line representing only the odd positive integers.10n every H k ( u ) = { v n } n ∈ N for u ≡ U p and k ∈ N , both (8) and (19) specify that v = u + µ = 1 + 4 µ if u ≡ [1] and v = ( u + µ ) / µ if u ≡ [2] ,such that u < v if u ≡ [1] and u > v if u ≡ [2] as claimed above. The root τ gives rise to H ( τ ) = Z = { , , , , . . . } , where e = ( τ, v ) forms the trivialcycle that is ignored in G . Mark H ( τ ) as a sequence on the odd number lineat k = 1. Also apply g on every v n ∈ U p > H ( u ) to generate infinitelymany sibling sets, whose vertices are also marked on the number line at k = 2.Repeat the same process for increasing k to infinity, as depicted in Figure 2 withonly the first few vertices shown for the initial sibling sets at k = 1 to 7.In this context, a gap refers to the number of odd positive integers betweentwo successive vertices in a sibling set. In H ( τ ), there exist exactly 2 n − − v n and v n +1 , not inclusive, for n ∈ N . Hence, Lemma 7 is trueif the subsequent g iterations uniquely generate all odd integers between verticesin H ( τ ). Ignoring the trivial cycle, the first sibling set H (5) = { , , , . . . } at k = 2 arises from v = 5 in H ( τ ). Notice that each vertex in H (5) occupiesthe midpoint between successive vertices in H ( τ ) (Figure 2). Since the initialvertex completely determines the spacing between successive vertices in a siblingset, it is crucial to examine how the subsequent initial vertices are generated tofill the gaps in H ( τ ). Observe that v at k > u , depending on whether u ≡ [1] or [2] , respectively, since v = u + µ if u ≡ [1] , v = u − ( µ + 1) if u ≡ [2] . (30)With µ ∈ E and µ ∈ U , it is assured that v is always odd in (30). Equatingthe expressions in (30) leads to (28), thus implying that no two initial verticesin distinct sibling sets are equal at any k > G . v n+1 − v n = 2 k =1 k =2 k =3 k =4 k =5 k =6 9 37 k =7 19 77 51 25 49 35 23 15 61 81 Figure 2: Gap-filling process by g iteration. Numbers of the same color indicate the first fewvertices (siblings) in the same sibling set, with the encircled number as the initial vertex, ateach depth k in G . By applying (30) on any u (cid:54)≡ [0] at k with respect to the odd number line toyield v ∈ U at k +1, one can generate the sequence of vertices in H k +1 ( u ) by anymethod given above. Since the gap between v n and v n +1 strictly increases with n in every sibling set at k ∈ N , repeatedly iterating (30) eventually produces11ll odd integers to fill such gaps. The gaps tend to be fully filled in progressionfrom left to right in H ( τ ), thereby completely covering the odd number line as k goes to infinity. Since v is a leaf node in H (5), further iterations on v = 13produce the vertices { , , , , } at k = 2 to 6, which all lie between 5 and21 in H ( τ ) (Figure 2). For the other odd integers in the same gap, 15 resultsfrom successive iterations on v = 53 to yield { , , , } at k = 2 to 5,while 19 is the initial vertex arising from v = 29 at k = 5. As 9 and 15 are leafnodes, both paths p ( τ,
9) and p ( τ,
15) thus terminate in dead ends. The siblingssets associated with these initial vertices, e.g. as determined by (9), containinfinitely many u ∈ U p on which (30) also applies recursively. In fact, furtherrecursions on u ∈ U p up to infinity will generate all distinct odd numbers tofill the gaps in H ( τ ), thus eventually covering the entire odd number line. Inshort, this gap-filling process by g iteration (Figure 2) generates Π k , with itselements ordered in monotone increasing sequence on the odd number line, forall k ∈ N . Therefore, since every vertex in V ( G ) is unique by Lemma 5 and V ( G ) = U by Lemma 6, we deduce from the gap-filling process that V ( G ) is acomplete and exact covering of U .The gap-filling process assures that no odd number is not covered by V ( G ),as claimed by Lemma 6. Assume the contrary that there exists u k ∈ U , where u k (cid:54)∈ V ( G ) such that V ( G ) (cid:54) = U . It suffices to consider only u k ≡ [0] , since,just the same, u k yields H k +1 ( u k ), with infinitely many leaf nodes to whichevery directed path eventually terminates, if u k (cid:54)≡ [0] . This implies that it isimpossible to have any finite subset of U that is not in V ( G ). It is always truethat u k = z n + 2 n − c or u k = z n + 2 n c for c ∈ N , thus implying that thereexists u k − ∈ U p such that g n ( u k − ) = u k for n ∈ N . It follows that u k − is theparent of u k , where c = µ r given that u k − = 3 µ r + r for u k − ≡ [ r ] . Suppose u k − is also not in G , then there exists u k − (cid:54)≡ [0] that is the parent of u k − . Ingeneral, consider { u i ∈ U p : g ( u i ) = u i +1 , ∀ i = k − , k − , . . . , , , } for some k ∈ N , where v i (cid:54)∈ V ( G ) for all i . Then we state the fact that p ( u , u k − ) is apath where deg − ( u ) = 0 and deg − ( u i ) = 1 for i = 1 , , , . . . , k −
1, whereasdeg + ( u i ) = ℵ for all i = 0 , , , . . . , k −
1. It is impossible that u > u (cid:54)∈ V ( G ), which is a contradiction. Hence, the only likelihood is that u = 1,which corresponds to the root τ of G , such that p ( u , u k − ) is connected to G .Since every parent in G gives rise to one and only one sibling set H k ( u k − ),with infinitely many elements, then u k ∈ H k ( u k − ). Therefore, u k is in G andthere exists a directed path p ( τ, u k ) from the root τ to u k , thus contradictingour assumption that u k is disjoint from G .Call u a non-initial vertex if it is not the initial vertex in a sibling set. Anypath consisting entirely of non-initial vertices in G is always strictly ascending.If a path is a mix of ascending and descending edges, then it is called a hailstonepath . We refer to any proper subset of a path p as sub-path or segment s , suchthat s ⊆ p . Obviously, p is a subset of itself. It is a trivial fact that infinitelymany vertices in a sibling set, say, H k ( u ) share the same sub-path s ( u , u ) fromthe root u to the parent node u (cid:54)≡ [0] at k − G . This implies that thereexist infinitely many odd natural numbers with convergent Collatz trajectories12f equal length k ∈ N under f iteration given in (1). In fact, we now claim that Lemma 7 (Convergence) . Every odd natural number has a convergent trajectoryunder f iteration.Proof. As G is an arborescence, there is always a directed path from the root τ to any other vertex u ∈ V ( G ). Since V ( G ) = U by Lemma 6 and every path p ( τ, u ) is a convergent Collatz trajectory in reverse, then the claim of the lemmafollows.A trivial corollary to Lemma 7 is the fact that the Collatz trajectory of anyodd natural number, as defined by (2), can never diverge. Hence, with Lemmas5-7, we have shown why the Collatz conjecture is true. Acknowledgments
Critical remarks from an anonymous reviewer of an earlier version of themanuscript substantially improved the approach presented in this paper. Thesupport from KAUST Core Labs is deeply appreciated.
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