OON COLLATZ CONJECTURE
ERHAN TEZCAN
Abstract.
The Collatz Conjecture can be stated as: using the reducedCollatz function C ( n ) = (3 n + 1) / x where 2 x is the largest power of2 that divides 3 n + 1, any odd integer n will eventually reach 1 in j iterations such that C j ( n ) = 1. In this paper we use reduced Collatzfunction and reverse reduced Collatz function. We present odd numbersas sum of fractions, which we call “fractional sum notation” and itsgeneralized form “intermediate fractional sum notation”, which we useto present a formula to obtain numbers with greater Collatz sequencelengths. We give a formula to obtain numbers with sequence length 2.We show that if trajectory of n is looping and there is an odd number m such that C j ( m ) = 1, n must be in form 3 j × k + 1 , k ∈ N where C j ( n ) = n . We use Intermediate fractional sum notation to show asimpler proof that there are no loops with length 2 other than trivialcycle looping twice. We then work with reverse reduced Collatz function,and present a modified version of it which enables us to determine theresult in modulo 6. We present a procedure to generate a Collatz graphusing reverse reduced Collatz functions. Keywords:
Number Theory, 3x+1 Problem, Collatz Conjecture,Unsolved Problem, Sequence, Cycle, Loop, Reverse, Reduced Collatz
MSC: Introduction
The Collatz Conjecture is a conjecture, also known as Ulam Conjecture,Syracuse Problem, Hasse’s Algorithm and possibly the most descriptively3 x + 1 Problem among many other names. The Collatz function F : N −→ N is defined by F ( n ) = (cid:26) n/ x ≡ n + 1 if x ≡ F i ( n ) denote the result of i iterations of function F on x . It isconjectured that there exists a number of iterations j such that F j ( n ) = 1.Whether this is true or not is an unsolved problem [10], it still is underspotlight when it comes to simple to express yet hard to solve problems.A bibliography regarding Collatz Conjecture can be read in this excellentpaper by J. C. Lagarias [5]. It has been computed that every number up to2 do end up reaching 1 [7]. Notice how after F j ( n ) = 1 the next iterationis F j +1 ( n ) = 3(1) + 1 = 4, F j +2 ( n ) = 4 / F j +3 ( n ) = 2 / −→ −→ −→ −→ −→ −→ −→ . . . This loop is oftenreferred to as “trivial cycle” [5]. When we refer to j th iteration such that Date : March 12, 2019. a r X i v : . [ m a t h . G M ] O c t ERHAN TEZCAN F j ( n ) = 1 we prefer to assume that this iteration is the first time the resultyields 1. Corollary 1.1.
Let n, i, j ∈ N and F j ( n ) = 1 where F i ( n ) (cid:54) = 1 , ≤ i < j .We can say that for any m ∈ N such that m = F i ( n ) , i < j the iteration F k ( m ) = 1 where k = j − i . This corollary formally states that if a number n reaches 1 in j iterations,every number in that trajectory also reaches 1. Corollary 1.2.
Let n, i, j ∈ N and F j ( n ) = 1 where F i ( n ) (cid:54) = 1 , ≤ i < j .Then F j +1 (2 n ) = 1 for every n and F j +1 (( n − / where n − ≡ . This reverse operation will be a point of discussion in the laterstages of our paper. Definition 1.3.
There are two phrases we will be using most often: Col-latz trajectory and Collatz sequence. A Collatz trajectory is a sequence ofnumbers starting at n ∈ N and ends at m ∈ N where m = F i ( n ) , i ∈ N . ACollatz Sequence is a sequence of numbers starting from any number n ∈ N and ending at , so it is the special case of the Collatz trajectory. We can show the Collatz trajectory using numbers and arrows where eacharrow points to an iteration over F . n −→ F ( n ) −→ F ( n ) −→ . . . −→ F i − ( n ) −→ mn and m are as defined in definition 1.3. Similarly, a Collatz sequence canbe shown as n −→ F ( n ) −→ F ( n ) −→ . . . −→ F j − ( n ) −→ Example 1.4.
The Collatz sequence for n = 3 is shown as −→ −→ −→ −→ −→ −→ −→ Reformulation of the problem.
A widely used reformulation is toomit the even numbers from the trajectories using a different function. Ifthe conjecture is proved for odd numbers it is intrinsically proved for evennumbers, as also seen in corollary 1.2. To do this we use the reduced Collatzfunction C : O + −→ O + defined by C ( n ) = 3 n + 12 a (1.2)where 2 a | n + 1 for the largest value of a ∈ N and O + is the set of positiveodd integers. The notation m | x states “ m divides x ” or as a congruence x ≡ m ), for the opposite purpose m (cid:45) x states that “ m does notdivide x ”. Letter C is preferred with respect to R. E. Crandall [2] who mostnotably used this reformulation, plus it is the first letter of Collatz! We willdefine a reduced Collatz trajectory where we will be able to see the exponent a explicitly. Definition 1.5.
The reduced Collatz trajectory using numbers and arrowswhere each arrow points to an iteration over C . n a −→ C ( n ) a −→ C ( n ) a −→ . . . a i − −−−→ C i − ( n ) a i −→ m N COLLATZ CONJECTURE 3
The reduced Collatz sequence is again similar, n a −→ C ( n ) a −→ C ( n ) a −→ . . . a j − −−−→ C j − ( n ) a j −→ We say that n has a “reduced Collatz trajectory length” j for C j ( n ) = m inthe trajectory written above. The reduced Collatz sequence has it m = 1 , sofor that we prefer to say n has a reduced Collatz sequence length j . In short,if we use the word “sequence” it describes that the end of the trajectory is . Example 1.6.
The reduced Collatz sequence for n = 3 is shown as −→ −→ The reduced Collatz sequence length of is . Fractional Sum Notation
We will present a way to represent numbers that eventually result in 1iterating over C . Definition 2.1.
Let n ∈ O + , C j ( n ) = 1 . The reduced Collatz trajectory is n a −→ C ( n ) a −→ C ( n ) a −→ . . . a j − −−−→ C j − ( n ) a j −→ Then we can write n as n = 2 a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − (2.1) We call this notation as the fractional sum notation, though it has beennoted as an “inverse canonical form” by G. Helms [3] . Note that a number n with reduced Collatz sequence length j will have j + 1 terms as seen in theequation above. We should explain the terms on the sum of fractions in the equation (2.1).First of all, for every term, the exponent of its numerator is greater thanthe exponent of the numerator of the term to its right. This is clearly seenby looking at the sums at the exponents, the rightmost sum being 0 as well.The rightmost term is very important: if the exponent of its numerator isgreater than 0, the whole expression is an even number. In fact, we defined n to be an odd number, but an even number m = 2 x × n can be representedsimilarly m = 2 a + a + ... + a j + x j − a + a + ... + a j − + x j − a + a + ... + a j − + x j − − . . . − a + x − x If the rightmost term have exponent 0 then n is an odd number. The firstiteration C ( n ) with exponent a will do the operation (3 n + 1) / a . Whatwe get is C ( n ) = 3 n + 12 a = 2 a + ... + a j j − − a + ... + a j − j − − a + ... + a j − j − − . . . − In other words, this representation is an explicit expression that shows allthe information about iterations of n over C . The exponent of denominatorof the left-most term tells us how many 3 n + 1 operations would happen if n were to iterate over F . The exponents of numerator also show how manydivisions by 2 occur at a certain iteration. This notation therefore gives ERHAN TEZCAN us the ability to use the Collatz trajectory all together in a mathematicalexpression.
Example 2.2.
The reduced Collatz sequence for n = 3 is shown as −→ −→ . The fractional sum notation for is then − − Odd Numbers with reduced Collatz Sequence length 1.
Con-sider a number n where C ( n ) = 1, the reduced Collatz sequence being n a −→
1. The fractional sum notation is then n = 2 a − (2.2)This reduces to n = 2 a −
13 (2.3)Since n is an odd integer, equation (2.3) tells us that2 a − ≡ Remark 2.3.
Notice that in equation (2.3) if we try to get at the right-hand side of the equation the left-hand side becomes C ( n ) . n + 12 a = C ( n ) = 1 Lemma 2.4. k +1 − ≡ , k ∈ N .Proof. For k = 0 we have 2 − ≡ k = n therefore 2 n +1 − ≡ k = n + 1 we have 2 n +1)+1 − ≡ n +2+1 − n +1 × −
1= 2 n +1 × (3 + 1) −
1= 3 × n +1 + 2 n +1 − × n +1 + 2 n +1 − ≡ (cid:3) Lemma 2.5. k +2 − ≡ , k ∈ N .Proof. For k = 0 we have 2 − ≡ k = n therefore 2 n +2 − ≡ k = n + 1 we have 2 n +1)+2 − ≡ n +2+2 − n +2 × −
1= 2 n +2 × (3 + 1) −
1= 3 × n +2 + 2 n +2 − × n +2 + 2 n +2 − ≡ (cid:3) N = N ∪ { } N COLLATZ CONJECTURE 5
Theorem 2.6. C ( n ) = 1 where n = 2 k +2 − , k ∈ N (2.4) In other words, positive odd integers in the form above have reduced Collatzsequence length 1.Proof.
According to equation (2.3) and lemmas 2.4 and 2.5 this is correct. (cid:3)
It is important to mention that R. P. Steiner [9] proved that the trivialcycle 1 −→ Odd Numbers with reduced Collatz Sequence length 2.
Con-sider a number n where C ( n ) = 1, the reduced Collatz sequence being n a −→ C ( n ) a −→
1. The fractional sum notation is then n = 2 a + a − a − (2.5)The values of a and a for the first 22 odd numbers with reduced Collatzsequence length 2 are given in table 1. We see that for even values of a the first a for each even value of a is 8, then 14 which is 8 + 6. Similarly,looking at odd values of a the first a for each odd value of a is 4, then 10which is 4 + 6. Theorem 2.7. C ( n ) = 1 where n = 2 a +6 b +2( − a − a − (2.6) and a , b ∈ N .Proof. Equation (2.6) tells us that the congruence2 a +6 b +2( − a − a − ≡ ) , a ∈ N , b ∈ N (2.7)must hold. To show this we will be considering two separate equations, onefor the even values of a and the other for odd values of a . This will enableus to use the lemmas 2.4 and 2.5 which we proved already. If a is even thenit is in the form 2 k + 2 , k ∈ N . The congruence (2.7) becomes2 k +2+6 b +2 − k +2 − ≡ )Looking at b = 1 we have 2 k +2+6+2 − k +2 − ≡ ). Thisexpression reduces to2 k +2+6+2 − k +2 − k +2 × (2 − −
3= 2 k +2 × −
3= 2 k +2 × × −
3= 3(2 k +2 × − k +2 × (84 + 1) − × × k +2 + 3(2 k +2 − × × k +2 + 3(2 k +2 − ERHAN TEZCAN
Table 1.
Table of the first 22 odd numbers with Collatzsequence length 2 n a a a Parity3 1 4 Odd13 3 4 Odd53 5 4 Odd113 2 8 Even213 7 4 Odd227 1 10 Odd453 4 8 Even853 9 4 Odd909 3 10 Odd1813 6 8 Even3413 11 4 Odd3637 5 10 Odd7253 8 16 Even7281 2 14 Even13653 13 4 Odd14549 7 10 Odd14563 1 16 Odd29013 10 8 Even29125 4 14 Even54613 15 4 Odd58197 9 10 Odd58253 3 16 OddThe first term is divisible by 3 . As for the second term, lemma 2.5 showsus 2 k +2 − ≡ k +2 − ≡ ). We assumethat the congruence holds for b = n therefore 2 k +2+6 n +2 − k +2 − ≡ ). Looking at b = n + 1 we have 2 k +2+6( n +1)+2 − k +2 − ≡ N COLLATZ CONJECTURE 7 (mod 3 ). This expression reduces to2 k +2+6( n +1)+2 − k +2 − k +2 (2 n +6+2 − −
3= 2 k +2 (2 n +2 × − −
3= 2 k +2 (2 n +2 × (2 − − −
3= 2 k +2 (2 n +2 ×
63 + 2 n +2 − −
3= 2 k +2 (2 n +2 × ×
7) + 2 k +2 × (2 n +2 − −
3= 2 k +2 (2 n +2 × ×
7) + 2 k +2+6 n +2 − k +2 − , rest of the terms are exactly the same as b = n case which by induction proves that the congruence (2.7) holds foreven values of a . If a is odd then it is in the form 2 k + 1 , k ∈ N . Thecongruence (2.7) becomes2 k +1+6 b − − k +1 − ≡ )Looking at b = 1 we have 2 k +1+6 − − k +1 − ≡ ). Thisexpression reduces to2 k +1+6 − − k +1 − k +1 × (2 − −
3= 2 k +1 × −
3= 2 k +1 × (3 × −
3= 3 × (2 k +1 × (5) − × (2 k × (10) − × (2 k × (9 + 1) − × × k + 3 × (2 k − . As for the second term, if k = 0 then3(2 −
1) = 0, if k > k − ≡ k − ≡ ). We assume that the congruenceholds for b = n therefore 2 k +1+6 n − − k +1 − ≡ ). Looking at b = n + 1 we have 2 k +1+6( n +1) − − k +1 − ≡ ). This expressionreduces to2 k +1+6( n +1) − − k +1 − k +1 × (2 n − − −
3= 2 k +1 × (2 n − × − −
3= 2 k +1 × (2 n − × (2 + 1 − − −
3= 2 k +1 × (2 n − × (2 −
1) + 2 n − − −
3= 2 k +1 × n − ×
63 + 2 k +1 × (2 n − − −
3= 2 k +1 × n − × × k +1+6 n − − k +1 − , rest of the termsare exactly the same as b = n case which by induction proves that thecongruence (2.7) holds for odd values of a . The proofs for even and oddvalues of a therefore prove this theorem. (cid:3) ERHAN TEZCAN
Remark 2.8.
A pattern similar to what we briefly showed in paragraphbelow equation 2.5 is subtly noticed for odd numbers with Collatz sequencelength 3 too. Just like we did in table 1, sorting the numbers in ascendingorder and look at the exponents a , a and a , one can see a similar yet moreconvoluted pattern. We were unable to generalize this. Using fractional sum notation to obtain numbers withhigher sequence lengths
In this section we will see how we can use the fractional sum notationto obtain an odd number with higher sequence length. Consider a num-ber n with the reduced Collatz sequence n a −→ C ( n ) a −→ C ( n ) . . . a j − −−−→ C j − ( n ) a j −→ j . We will present a formula to obtain an odd num-ber m with the reduced Collatz sequence m a −→ C ( m ) a −→ C ( m ) . . . a j − −−−→ C j − ( m ) a j −→ C j ( m ) a j +1 −−−→ j + 1. This is different than obtaininga number by going reverse, as shown in corollary 1.2. Consider the same n before but now suppose n = 3 k + 1, in other words C ( k ) = n . Let theexponent be a k for that iteration, then the reduced Collatz sequence for k is k a k −→ n a −→ C ( n ) a −→ C ( n ) . . . a j − −−−→ C j − ( n ) a j −→
1. Notice how the newexponent is at the end of the trajectory for m and at the beginning of thetrajectory for k . Theorem 3.1.
Let n, m ∈ N where n has reduced Collatz sequence length j and m has reduced Collatz sequence length j + 1 . m = 2 a + a + ... + a j j +1 × (2 a j +1 − ) + n (3.1) where a j +1 = 2 b j + 2 , b ∈ N .Proof. As defined in definition 2.1 we have n = 2 a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − (3.2)By the same definition m is m = 2 a + a + ... + a j +1 j +1 − a + a + ... + a j j +1 − a + a + ... + a j − j − . . . − a − (3.3)Notice that the exponents a , a , . . . , a j are mutual. Now suppose x and y are x = 2 a + a + ... + a j j y = − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − Clearly n = x + y . m = 2 a + a + ... + a j +1 j +1 − x y Adding x − x to the right side gives m = 2 a + a + ... + a j +1 j +1 − x − x + x + y N COLLATZ CONJECTURE 9
Since n = x + y m = 2 a + a + ... + a j +1 j +1 − x n Writing x explicitly we have m = 2 a + a + ... + a j +1 j +1 − × a a ... + aj j nm = 2 a + a + ... + a j +1 j +1 − a + a + ... + a j +2 j +1 + nm = 2 a + a + ... + a j j +1 × (2 a j +1 − ) + n which is the equation we have used for the theorem. Now we have to showthat a j +1 = 2 b j + 2 , b ∈ N . By definition, both n and m are integerstherefore the first term on the right-hand side of equation 3.1 must be aninteger. Since 3 j +1 (cid:45) a + a + ... + a j it is required that 3 j +1 | a j +1 − . Thisgives the congruence 2 a j +1 − ≡ j +1 )Plugging in a j +1 = 2 b j + 2 yields2 b j +2 − ≡ j +1 )We can parenthesize with 2 which gives2 (2 b j − ≡ j +1 )Since 3 j +1 (cid:45) our congruence is now2 b j − ≡ j +1 ) (3.4)We will use proof by induction to prove this congruence. First, we look at b = 0 which gives 2 − ≡ j +1 ) which is correct. We also look at b = 1 which gives 2 × j − ≡ j +1 ) (3.5)We will come to this in a moment. Next, we assume that b = k is truetherefore 2 × k × j − ≡ j +1 )Finally we look at b = k + 1 which gives2 × ( k +1) × j − ≡ j +1 )We can see that the expression 2 × ( k +1) × j − × k × j × × × j − × k × j − × k × j to this gives2 × k × j × × × j − × k × j − × k × j ≡ j +1 )Parenthesizing with 2 × k × j gives2 × k × j (2 × × j −
1) + 2 × k × j − ≡ j +1 )Here the second and third terms together is the same expression we had for b = k . For the first term, we have 2 × × j − b = 1 (equation (3.5)). This tells us that we can prove the congruence (3.4) by proving it is correct for b = 1. In otherwords, we will prove by induction that the congruence2 × j − ≡ j +1 ) (3.6)holds for every j . Starting with j = 1 we have2 × − ≡ )which gives 63 ≡ j = k × k − ≡ k +1 )Assuming this is correct we look at j = k + 12 × k +1 − ≡ k +1+1 )We can write 2 × k +1 as 2 × k × × k . Adding 2 × k − × k to this gives2 × k × × k − × k − × k ≡ k +1+1 )Parenthesizing with 2 × k gives2 × k × (2 × k −
1) + 2 × k − ≡ k +1+1 )Note that 2 × k = 2 × k × × k × k × (2 × k × × k −
1) + 2 × k − ≡ k +1+1 ) (3.7)To show the next step clearly, we will denote 2 × k − A , so A = 2 × k − A + 1)(( A + 1) −
1) + A ≡ k +1+1 )( A + 1)( A + 2 A + 1 −
1) + A ≡ k +1+1 )( A + 1)( A + 2 A ) + A ≡ k +1+1 ) A + 2 A + A + 2 A + A ≡ k +1+1 ) A + 3 A + 3 A ≡ k +1+1 )Opening A back up and writing the congruence shows us that(2 × k − + 3(2 × k − + 3(2 × k − ≡ k +1+1 ) (3.8)Now from congruence (3) we have 2 × k − ≡ k +1 ). From this wecan definitely say that the congruence for the first term (2 × k − ≡ k +1+1 ) holds, similarly the congruence for the second term 3(2 × k − ≡ k +1+1 ) holds and the last congruence 3(2 × k − ≡ k +1+1 ) holds. Therefore it is proven that the congruence (3.4) holdsfor all b ∈ N and j ∈ N . (cid:3) Sadly, this formula does not give every odd number m with reduced Col-latz sequence length j + 1 when we have an odd number n with reducedCollatz sequence length j . To demonstrate, we can look at the formulafor numbers with reduced Collatz sequence length 1 which we have shownin theorem 2.6. Let n ∈ O + with reduced Collatz sequence length 1 and m ∈ O + with reduced Collatz sequence length 2. Theorem 3.1 states that m = 2 a × (2 a − ) + n N COLLATZ CONJECTURE 11
Theorem 2.6 stated that n = 2 k +2 − So we have m = 2 k +2 × (2 a − ) + 2 k +2 − Theorem 3.1 states that a = 2 b + 2 so we have m = 2 k +2 × (2 b +2 − ) + 2 k +2 − m = 2 k +6 b +4 − k +4 + 2 k +2 − m = 2 k +6 b +4 − k +2 − (3.9)However, theorem 2.7 states that m = 2 a +6 b +2( − a − a − (3.10)Now that we know a = 2 k + 2 we get m = 2 k +6 b +4 − k +2 − (3.11)Equations (3.9) and (3.11) are same. The problem is, we have shown intheorem 2.6 that the equation (3.10) is also correct. Equation (3.10) yieldsmore numbers with reduced Collatz sequence length 2. Why does this hap-pen? It’s because equation (3.1), when used with a number n with reducedCollatz sequence length j to obtain a number m with reduced Collatz se-quence length j + 1, is constrained to the fact that n is an odd integer. Inthe case of reduced Collatz sequence length 2, we were constrained to thefact that n with reduced Collatz sequence length 1 was also valid. Havingproved that the exponent must be an even number for reduced Collatz se-quence length 1, we constrain ourselves to use even values for a when weuse (3.1). Theorem 2.6 clearly shows that it can be odd too.3.1. Trivial Cycle.
One pretty result of equation (3.1) is when a j +1 = 2. m = 2 a + a + ... + a j j +1 × (2 − ) + n (3.12)This yields m = 0 + n . The trajectory of m is then m a −→ C ( m ) a −→ C ( m ) . . . a j − −−−→ C j − ( m ) a j −→ C j ( m ) −→
1. Looking at a trajectory such as k −→ k + 1) / = 1 so k = 1. We also know m = n . If a j +1 = 2 it just means that the trajectory ended at 1 already and is looping.An interesting result of using the trivial cycle is that given a number n with reduced Collatz sequence length j we can find numbers with greatersequence lengths, not just j + 1 but any length j + k, k ∈ N . Corollary 3.2.
Let n, m, k ∈ N where n has reduced Collatz sequence length j and m has reduced Collatz sequence length j + k . m = 2 a + a + ... + a j +2( k − j + k × (2 a j + k − ) + n (3.13) where a j + k = 2 b j + k − + 2 , b ∈ N and a j +1 = a j +2 = . . . = a j + k − = 2 sofor k − iterations we set the new exponent and therefore obtain a newformula using the same base n . The reduced Collatz sequences can be shownas n a −→ C ( n ) a −→ . . . a j −→ m a −→ C ( m ) a −→ . . . a j −→ C j ( m ) −→ C j +1 ( m ) −→ . . . −→ C j + k − ( m ) (cid:124) (cid:123)(cid:122) (cid:125) k − times a j + k −−−→ k = 1 in equation (3.13) basically yields equation(3.1). Example 3.3.
Let us consider the number 5. We will demonstrate bothequations (3.13) and (3.1) with 5 as our base number ( n = 5 ). First letus look at the reduced Collatz sequence of 5 which is −→ . It has reducedCollatz sequence length . Equation (3.1) becomes m = 2 × (2 b +2 − ) + 5 (3.14) If b = 0 we get 5 with the sequence −→ −→ , which has looped over forone iteration and the new exponent is therefore − − If b = 1 we get m = 2 × (2 − ) + 5 We find m = 453 and the theory states that this number should have reducedCollatz sequence length . Indeed it has, with the sequence −→ −→ .
453 = 2 − − Finally we will show how we can obtain a number with reduced Collatz se-quence length by using 5, which has reduced Collatz sequence length .Corollary 3.2 stated that we can obtain length j + k with k − iterationswith exponent . We will do exactly that. In this example j = 1 and wewant j + k = 3 so k = 2 . Plugging these in equation (3.13) we get m = 2 − × (2 b − +2 − ) + 5 m = 2 × (2 b +2 − ) + 5 We will just give b = 1 which results in m = 2 × (2 − ) + 5 This yields m = 2485509 . It’s reduced Collatz sequence is −→ −→ −→ and the fractional sum notation is − − −
1N COLLATZ CONJECTURE 13 Intermediate Fractional Sum Notation
So far we have only used fractional sum notation for Collatz trajectoriesthat end with 1. In this section we will introduce the Intermediate FractionalSum Notation which does not have that requirement. Any Collatz trajectoryis applicable.
Definition 4.1.
Let n, m ∈ O + , C j ( n ) = m . The reduced Collatz trajectoryis n a −→ C ( n ) a −→ . . . a j − −−−→ C j − ( n ) a j −→ m Then we can write n as n = m × a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − (4.1)Essentially, this notation and the fractional sum notation are same. Giv-ing m = 1 turns this into a fractional sum notation. Example 4.2.
The reduced Collatz sequence of is −→ −→ −→ −→ −→ So, the Collatz trajectory of up to is −→ −→ −→ Writing Intermediate Fractional Sum Notation for n = 7 and m = C ( n ) =13 gives us × − − − If there exists a number that does not reach 1 at some iteration, then itis either looping or diverging. So one may find that there is a number witha looping trajectory or a diverging trajectory to prove that this conjectureis false. The intermediate fractional sum notation is a tool we can use tostudy such trajectories.4.1.
Studying looping trajectories with intermediate fractional sumnotation.
Suppose there is a number n ∈ O + such that C j ( n ) = n . Thetrajectory can be shown as n a −→ C ( n ) a −→ . . . a j − −−−→ C j − ( n ) a j −→ n This number is in a looping trajectory of length j . In fact, every number inthis loop is looping (as the words suggest). The intermediate fractional sumnotation for n is n = n × a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − (4.2) Theorem 4.3.
Let n ∈ O + and C j ( n ) = n with the trajectory n a −→ C ( n ) a −→ . . . a n − −−−→ C j − ( n ) a j −→ n . If there exists a number m ∈ O + such that C j ( m ) = 1 and the trajectory is m a −→ C ( m ) a −→ . . . a j − −−−→ C j − ( m ) a j −→ then n is in the form j × k + 1 , k ∈ N . Proof.
We start by writing the intermediate fractional sum notation for n which is n = n × a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − (4.3)In the first term on the right-hand side of the equation, we write n + 1 − n , which gives us n = ( n +1 − × a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − n = ( n − × a + ... + a j j + 2 a + ... + a j j − a + ... + a j − j − a + ... + a j − j − − . . . − a − (4.4)If there is a number m ∈ O + with the trajectory m a −→ C ( m ) a −→ . . . a j − −−−→ C j − ( m ) a j −→ m is m = 2 a + a + ... + a j j − a + a + ... + a j − j − a + a + ... + a j − j − − . . . − a − Then equation (4.4) becomes n = ( n − × a + a + ... + a j j + m (4.5)We define n and m to be odd numbers, therefore the first term on the right-hand side must be an integer. Since 3 j (cid:45) a + a + ... + a j we must have 3 j | n − n − ≡ j ) (4.6)Looking at this congruence we can say that n = 3 j × k + 1 , k ∈ N . Thereis one more thing to consider, n is an odd number. For odd values of k theexpression 3 j × k + 1 yields even numbers because (odd × odd + odd) resultsin an even number. Since n must be odd we can not have odd values for k .For even values of k the expression 3 j × k + 1 yields odd numbers. Rewritingthe expression with 2 k instead of k gives n = 3 j × k + 1 , k ∈ N . (cid:3) Corollary 4.4.
Under the same conditions mentioned in theorem 4.3, since n = 3 j × k + 1 we can say n ≡ . Modulo 6 turns out to be quiteimportant, as we will see in section 5. So far we only have one known loop: 1 −→
1. Plugging n = 1 in equation(4.5) we get 1 = (1 − × + m which shows that m = 1, which is consistent with the condition stated intheorem 4.3. In fact, a looping trajectory1 −→ −→ . . . −→ −→ (cid:124) (cid:123)(cid:122) (cid:125) j iterations is consistent with the theorem. All exponents are 2 and if there are j itera-tions equation (4.5) becomes1 = (1 −
1) 2 j j + m N COLLATZ CONJECTURE 15 which still gives m = 1.4.1.1. Loop with length 2.
It has been proved by J. L. Simons [8] that thereare no loops with length 2 other than 1 −→ −→
1, the proof used upperbounds and lower bounds, following R. P. Steiner’s method [9]. We canuse intermediate fractional sum notation to quickly prove there are no loopswith length 2 other than 1 −→ −→ n a −→ C ( n ) a −→ n . The intermediate fractional sum notation is n = n × a + a − a − (4.7)This gives us the equation 9 n = n × a + a − a −
3. Leaving n alone, n = 2 a + 32 a + a − n = 2 a + 32 a (2 a + 1 − −
12 + 3 n = 2 a + 32 a + 3 + 2 a (2 a − − n is an integer, therefore numerator is greater than orequal to denominator. This means that 2 a + 3 ≥ a + 3 + 2 a (2 a − − ≥ a + a − a . Since n is a positive odd integer, inequation (4.8) at the denominator we get the inequality 2 a + a >
9. Thistells us a + a >
3. Also remember that a , a ∈ N . If a = 1 we get14 ≥ a , which is possible for a <
3, but then adding a to this gives1 + a < a + a >
3. If a = 2 we get16 ≥ a , which holds for a = 1 and a = 2 but 2 + 1 < a = 2. If a = 3 we get 20 ≥ a . Only valid valueis a = 1. If a = 4 we get 28 ≥ a . There are no valid values for a after this point. 2 pairs of ( a , a ) satisfy both inequalities: (2 ,
2) and (3 , ,
1) if we plug it in equation (4.8) we get n = 2 + 32 − n ∈ O + as 11 / ,
2) which gives n = 2 + 32 − n = 1. This shows that the only loop of length 2 is 1 −→ −→ Loop with greater lengths.
Can we do the same thing for a loop withlength 3? The trajectory would be n a −→ C ( n ) a −→ C ( n ) a −→ n . n = n × a + a + a − a + a − a − (4.9)Leaving n alone gives n = 2 a + a + 3 × a + 92 a + a + a −
27 (4.10)
This is equal to n = 2 a + a + 3 × a + 92 a + a + 3 × a + 9 + 2 a + a (2 a − − a ) − a + a +3 × a +9 ≥ a + a +3 × a +9+2 a + a (2 a − − a ) −
36 together with 2 a + a + a − >
0, but it seems thingsget too complicated at this point. Considering n a −→ C ( n ) a −→ . . . a j −→ n givesequation (4.2). Applying the same procedure to leave n alone yields n = j (cid:88) i =1 (cid:16) i − × (cid:80) j − ik =1 a k (cid:17) (cid:80) ji =1 a i − j (4.11)If there is an odd integer n with loop length j then it must satisfy thisequation. Not the most malleable formula, but it is there!4.2. Studying diverging trajectories with intermediate fractionalsum notation.
F.C. Motta et al. [6] worked on diverging trajectories intheir paper, they had a formula that gives an upper bound for a mono-tonically increasing trajectory. A monotonically increasing trajectory is atrajectory where on each iteration the result gets bigger, for example n −→ C ( n ) −→ C ( n ) −→ . . . We can use the intermediate fractional sum notation to find numbers whichmonotonically increase for j iterations, with the trajectory n −→ C ( n ) −→ . . . −→ C j ( n )The intermediate fractional sum notation is n = C j ( n ) × j j − j − j − j − j − − . . . − − (4.12) n = C j ( n ) × j j − × (cid:18) j − j − + 2 j − j − + . . . + 2 + 1 (cid:19) Inside the parentheses we have a finite geometric series .Writing the sumof j terms for that geometric series gives us n = C j ( n ) × j j − × (cid:32) − j j − (cid:33) Leaving C j ( n ) alone we get C j ( n ) = (cid:18) (cid:19) j ( n + 1) − The sum of first n terms in a geometric series is given by a × (1 − r n )1 − r where a is the first term and r is the common ratio, − < r <
1. In our geometric series a = 1 and r = N COLLATZ CONJECTURE 17
Example 4.5.
One can construct monotonically increasing trajectories us-ing equation (4.13) . Giving n = 2 j × k − , k ∈ N yields a monotonically in-creasing trajectory with j odd numbers. − , monotonically increasingtrajectory with numbers: −→ −→ .
15 = 2 − , monotonically increas-ing trajectory with numbers: −→ −→ −→ .
39 = 5 × − , mono-tonically increasing trajectory with numbers: −→ −→ .
127 = 2 − ,monotonically increasing trajectory with numbers: −→ −→ −→ −→ −→ −→ and so on... Reverse Formulation of the problem
A widely used reformulation of the problem is using the reverse reducedCollatz function R : O + −→ O + defined by R ( n ) = 2 x × n −
13 (5.1)where n ∈ O + and x ∈ N . The problem so far was asking whether any oddnumber n reach 1 at some iteration C i ( n ). We can ask the same by sayingthat for every odd number n there is an iteration R i (1) = n . Regarding R function there are 3 lemmas, relating to modulo 6. Lemma 5.1. If n ≡ then x is an even number.Proof. We can write n = 6 a + 1 , a ∈ N and x = 2 b + 2 , b ∈ N . Equation(5.1) becomes R (6 a + 1) = (6 a + 1) × b +2 −
13 (5.2)For this function to be valid the congruence (6 a + 1) × b +2 − ≡ a × b +2 + 2 b +2 − ≡ | b +2 × a and 3 | b +2 − (cid:3) Lemma 5.2. If n ≡ there are no possible values for x .Proof. We can write n = 6 a + 3 , a ∈ N . Equation (5.1) becomes R (6 a + 3) = (6 a + 3) × x −
13 (5.3)For this function to be valid the congruence (6 a + 3) × x − ≡ | (6 a + 3) × x but 3 (cid:45) − x . (cid:3) Lemma 5.3. If n ≡ then x is an odd number.Proof. We can write n = 6 a + 5 , a ∈ N and x = 2 b + 1 , b ∈ N . Equation(5.1) becomes R (6 a + 5) = (6 a + 5) × b +1 −
13 (5.4)For this function to be valid the congruence (6 a + 5) × b +1 − ≡ a + 3) × b +1 + 2 × b +1 − ≡ | (6 a + 3) × b +1 and 3 | × b +1 − (cid:3) These 3 lemmas show us R ( n ) can only be used when n ≡ n ≡ Definition 5.4.
The function R : N × N −→ O R is defined as R ( a, b ) = (6 a + 1) × b +2 −
13 (5.5) O R ⊂ O + . This function acts as a reverse reduced Collatz iteration on anumber n = 6 a + 1 . Definition 5.5.
The function R : N × N −→ O R is defined as R ( a, b ) = (6 a + 5) × b +1 −
13 (5.6) O R ⊂ O + . This function acts as a reverse reduced Collatz iteration on anumber n = 6 a + 5 . We will soon show that O R ∩ O R = ∅ and O R ∪ O R = O + but to dothat we need to introduce one more function and prove several lemmas. Definition 5.6.
The function X : N × N −→ O X is defined as X ( a, b ) = 2 b +3 a + 2 b +4 −
13 (5.7) O X ⊂ O + . We will soon see why this function is important. Definition 5.7.
We need to provide one more useful notation for our proofs.The set of all possible values function R ( a, b ) can produce where a ∈ N and b ∈ { , , . . . , k } will be shown as O R ,k . The set of all possible valuesfunction R ( a, b ) can produce where a ∈ N and b ∈ { , , . . . , k } will beshown as O R ,k . The set of all possible values function X ( a, b ) can producewhere a ∈ N and b ∈ { , , . . . , k } will be shown as O X,k . If k is infinity, itbasically means b ∈ N . In that case we do not write infinity explicitly, forexample O R , ∞ is wrong, we use O R as defined before. Lemma 5.8. O R , ∪ O R , ∪ O X, = O + and O R , , O R , , O X, are disjointsets.Proof. Let us look at functions R ( a, , R ( a,
0) and X ( a, R ( a,
0) = (6 a + 1) × −
13 = 8 a + 1 R ( a,
0) = (6 a + 5) × −
13 = 4 a + 3 X ( a,
0) = 2 a + 2 −
13 = 8 a + 5Let ¯ r denote the set of positive odd integers congruent to r modulo 8, ¯ r = { n ∈ O + | n ≡ r (mod 8) } . It is evident that O R , = ¯1, O R , = ¯3 ∪ ¯7 and O X, = ¯5. We are able to say that ¯1 ∪ ¯3 ∪ ¯5 ∪ ¯7 = O + as well as see that¯1 , ¯3 , ¯5 , ¯7 are disjoint. (cid:3) Lemma 5.9. ( O X,k \ O X,k − ) = ( O R ,k +1 \ O R ,k ) ∪ ( O R ,k +1 \ O R ,k ) ∪ ( O X,k +1 \ O X,k ) N COLLATZ CONJECTURE 19
Proof.
Now this may appear to be a rather complicated lemma, but it is not,the notation is. We can elucidate it a bit: Consider an arbitrary positiveinteger k and a ∈ N , the lemma states that the set of all possible valuesproduced by X ( a, k ) is equal to the union of sets all possible values producedby R ( a, k + 1), R ( a, k + 1) and X ( a, k + 1). We just used the set differenceto achieve it. To show this, consider the cases of a in modulo 4: it canbe 0, 1, 2 or 3. We can say that the set of all possible values producedby X ( a, k ) is then equal to the union of sets all possible values producedby X (4 a, k ), X (2 a + 1 , k ) and X (4 a + 2 , k ). Let us look at the equation X (4 a, k ) = R ( a, k + 1).2 k +3 (4 a ) + 2 k +4 −
13 = (6 a + 1)2 k +1)+2 − × k +3 (2 a ) + 2 k +4 − a + 1)2 k +4 − a k +4 + 2 k +4 = (6 a + 1)2 k +4 (6 a + 1)2 k +4 = (6 a + 1)2 k +4 X (2 a + 1 , k ) = R ( a, k ).2 k +3 (2 a + 1) + 2 k +4 −
13 = (6 a + 5)2 k +1)+1 − × k +3 (2 a + 1) + 2 x +4 − a + 5)2 k +1)+1 − a k +3 + 3 × k +3 + 2 k +4 = 6 a k +3 + 5 × k +3 × k +3 + 2 × k +3 = 5 × k +3 × k +3 = 5 × k +3 X (4 a +2 , k ) = X ( a, k +1). 2 k +3 (4 a + 2) + 2 k +4 −
13 = 2 k +1)+3 a + 2 k +1)+4 − k +5 a + 2 k +4 + 2 k +4 −
13 = 2 k +5 a + 2 k +2+4 − × x +4 + 2 x +4 − x +6 − × x +4 = 4 × x +4 (cid:3) Lemma 5.10. ( O R ,k + x \ O R ,k ) ∪ ( O R ,k \ O R ,k − y ) = O R ,k + x \ O R ,k − y .This is also true for O R and O X .Proof. Consider ( O R ,k + x \ O R ,k ) ∪ ( O R ,k \ O R ,k − y ) and a ∈ N . O R ,k + x \ O R ,k is the set of all possible results produced by R ( a, k + x ) , R ( a, k + x − , . . . , R ( a, k + 1). Similarly O R ,k \ O R ,k − y is the set of all possible resultsproduced by R ( a, k ) , R ( a, k − , R ( a, k − , . . . , R ( a, k − y + 1). As aresult, ( O R ,k + x \ O R ,k ) ∪ ( O R ,k \ O R ,k − y ) is the set of all possible results produced by R ( a, k + x ) , R ( a, k + x − , . . . , R ( a, k +1) , R ( a, k ) , R ( a, k − , R ( a, k − , . . . , R ( a, k − y + 1), therefore we can see that( O R ,k + x \ O R ,k ) ∪ ( O R ,k \ O R ,k − y ) = O R ,k + x \ O R ,k − y (5.8)This applies to the functions R and X too, by definition. Therefore( O R ,k + x \ O R ,k ) ∪ ( O R ,k \ O R ,k − y ) = O R ,k + x \ O R ,k − y (5.9)and ( O X,k + x \ O X,k ) ∪ ( O X,k \ O X,k − y ) = O X,k + x \ O X,k − y (5.10) (cid:3) Theorem 5.11. O R ∩ O R = ∅ and O R ∪ O R = O + Proof.
Lemma 5.9 shows us that when k = 0 we have( O X, \ O X, − ) = ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O X, \ O X, ) O X, − is not possible since b ≥ O X, − = ∅ . Therefore O X, = ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O X, \ O X, ) (5.11)For k = 1 we have( O X, \ O X, ) = ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O X, \ O X, )Writing ( O X, \ O X, ) in equation (5.11) O X, = ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O X, \ O X, )Continuing this process yields the expression below O X, =( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , ) ∪ ( O R , \ O R , )... ∪ ( O R ,k \ O R ,k − ) ∪ ( O R ,k \ O R ,k − ) ∪ ( O X,k \ O X,k − )Applying lemma 5.10 on this yields O X, = ( O R ,k \ O R , ) ∪ ( O R ,k \ O R , ) ∪ ( O X,k \ O X,k − )We can infer that O X, = ( O R \ O R , ) ∪ ( O R \ O R , )as k goes to infinity. Lemma 5.8 states that O R , ∪ O R , ∪ O X, = O + Now that we have shown O X, = ( O R \ O R , ) ∪ ( O R \ O R , ) we can useit in lemma 5.8 to have O R , ∪ O R , ∪ ( O R \ O R , ) ∪ ( O R \ O R , ) = O + Since ( O R \ O R , ) ∪ O R , = O R and ( O R \ O R , ) ∪ O R , = O R wefinally have O R ∪ O R = O + (cid:3) N COLLATZ CONJECTURE 21
This proof is a safety assurance that we can use the functions R and R to study reverse reformulation of the problem. It is important to understandwhat these functions mean. R ( a, b ) is a reverse iteration of an odd number n such that n ≡ b +2 andthen 1 is subtracted from this, finally being divided by 3. Let the final resultbe m . See that m = 2 b +2 × n − n alone, 3 m + 12 b +2 = n So actually C ( m ) = n and m = R ( n ). Similarly for R function we workwith n such that n ≡ m = 2 b +1 × n − n alone, 3 m + 12 b +1 = n Again, C ( m ) = n and m = R ( n ). The proof tells us that every odd numberis a result of reverse iteration of some odd number. This is a rather obviousstatement, but a required one for the function to be valid. Another thingthis proof shows is that every odd number is the result of a reverse iterationof exactly one odd number. Using two functions for cases of modulo 6 canalso be seen in K. Andersson’s work [1].5.1. Determining modulo 6 after a single reverse iteration.
In thissection we present a modified version of R and R , namely RR and RR .Before we continue we need to provide several lemmas. Lemma 5.12. x +1 ≡ where x ∈ N .Proof. At x = 0 we get 2 ≡ x = 1 we get 8 ≡ x = k the congruence2 k +1 ≡ x = k + 1 we have2 k +1)+1 ≡ k +1 − k +1 to this expression yields2 k +2+1 + 2 k +1 − k +1 ≡ k +1 yields2 k +1 + 2 k +1 (2 − ≡ k +1 + 2 k × ≡ | k × k +1 ≡ x = k therefore thislemma is proven by induction. (cid:3) Lemma 5.13. x +2 ≡ where x ∈ N . Proof. At x = 0 we get 4 ≡ x = 1 we get 16 ≡ x = k the congruence2 k +2 ≡ x = k + 1 we have2 k +1)+2 ≡ k +2 − k +2 to this expression yields2 k +2+2 + 2 k +2 − k +2 ≡ k +2 yields2 k +2 − k +2 (2 − ≡ k +2 + 2 k +1 × ≡ | k +1 × k +2 ≡ x = k thereforethis lemma is proven by induction. (cid:3) Lemma 5.14. x +2 − ≡ where x ∈ N .Proof. At x = 0 we get 1 ≡ x = 1 we get 85 ≡ x = k the congruence2 k +2 − ≡ x = k + 1 we have2 k +1)+2 − ≡ k +2 − k +2 to 2 k +1)+2 yields2 k +1)+2 + 2 k +2 − k +2 − ≡ k +2 yields2 k +2 (2 − k +2 − ≡ k +2 ×
42 + 2 k +2 − ≡ | k +2 ×
42 and k +2 − ≡ x = k thereforethis lemma is proven by induction. (cid:3) Lemma 5.15. x +4 − ≡ N COLLATZ CONJECTURE 23
Proof.
Notice that this is equal to4 × x +2 − ≡ x +2 −
13 + 2 x +2 ≡ x )+2 − ≡ x )+2 ≡ x +2 − +2 x +2 ≡ (cid:3) Lemma 5.16. x +6 − ≡ Proof.
Notice that this is equal to4 × x +4 − ≡ x +4 −
13 + 2 x +4 ≡ x +4 − ≡ x +1)+2 ≡ x +4 − + 2 x +4 ≡ (cid:3) Theorem 5.17.
Let a, b ∈ N and c ∈ { , , } . The function RR : N × N × { , , } −→ O R is defined as RR ( a, b, c ) = (6 a + 1) × b +( c − a mod 3)) mod 3)+2 −
13 (5.15) where RR ( a, b, c ) ≡ c (mod 6) .Proof. We will consider the cases of a in modulo 3 and c . a mod 3 can be0 , c ∈ { , , } . In total we have 9 pairs of( a, c ) which are (0 , , (0 , , (0 , , (1 , , (1 , , (1 , , (2 , , (2 , , (2 , RR ( a, b, c ) ≡ c (mod 6). Let us set c = 1 first to show RR ( a, b, ≡ RR ( a, b,
1) = (6 a + 1)2 b +(( a mod 3)) mod 3)+2 − a = 3 k, k ∈ N so a ≡ RR (3 k, b,
1) = (6(3 k ) + 1)2 b +((3 k mod 3)) mod 3)+2 − RR (3 k, b,
1) = (18 k + 1)2 b )+2 − RR (3 k, b,
1) = 6 k × b )+2 + 2 b +2 − | k × b )+2 and lemma 5.14 shows that b +2 − ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
1) = (6(3 k + 1) + 1)2 b +(((3 k +1) mod 3)) mod 3)+2 − RR (3 k + 1 , b,
1) = (18 k + 6 + 1)2 b +1)+2 − RR (3 k + 1 , b,
1) = 6 k × b +1)+2 + 2 × b +1)+2 + 2 b +1)+2 − RR (3 k + 1 , b,
1) = 6 k × b +1)+2 + 2 b +2)+1 + 2 b +4 − | k × b )+2 , lemma 5.12 shows that 2 b +2)+1 ≡ b +4 − ≡ RR (3 k + 1 , b, ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
1) = (6(3 k + 2) + 1)2 b +(((3 k +2) mod 3)) mod 3)+2 − RR (3 k + 2 , b,
1) = (18 k + 12 + 1)2 b +2)+2 − RR (3 k + 2 , b,
1) = 6 k × b +2)+2 + 4 × b +2)+2 + 2 b +2)+2 − RR (3 k + 2 , b,
1) = 6 k × b +2)+2 + 2 b +3)+2 + 2 b +6 − | k × b +2)+2 , lemma 5.13 shows that 2 b +3)+2 ≡ b +6 − ≡ RR (3 k + 2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ c = 3 to show RR ( a, b, ≡ RR ( a, b,
3) = (6 a + 1)2 b +(2+( a mod 3)) mod 3)+2 − a = 3 k, k ∈ N so a ≡ RR (3 k, b,
3) = (6(3 k ) + 1)2 b +(2+(3 k mod 3)) mod 3)+2 − RR (3 k, b,
3) = (18 k + 1)2 b +2)+2 − RR (3 k, b,
3) = 6 k × b +2)+2 + 2 b +6 − N COLLATZ CONJECTURE 25 | k × b +2)+2 and lemma 5.16 shows that b +6 − ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
3) = (6(3 k + 1) + 1)2 b +(2+((3 k +1) mod 3)) mod 3)+2 − RR (3 k + 1 , b,
3) = (18 k + 6 + 1)2 b )+2 − RR (3 k + 1 , b,
3) = 6 k × b )+2 + 2 × b )+2 + 2 b )+2 − RR (3 k + 1 , b,
3) = 6 k × b )+2 + 2 b +1)+1 + 2 b +2 − | k × b )+2 , lemma 5.12 shows that 2 b +1)+1 ≡ b +2 − ≡ RR (3 k + 1 , b, ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
3) = (6(3 k + 2) + 1)2 b +(2+((3 k +2) mod 3)) mod 3)+2 − RR (3 k + 2 , b,
3) = (18 k + 12 + 1)2 b +1)+2 − RR (3 k + 2 , b,
3) = 6 k × b +1)+2 + 4 × b +1)+2 + 2 b +1)+2 − RR (3 k + 2 , b,
3) = 6 k × b +1)+2 + 2 b +2)+2 + 2 b +4 − | k × b +1)+2 , lemma 5.13 shows that 2 b +2)+2 ≡ b +4 − ≡ RR (3 k + 2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ c = 5 to show RR ( a, b, ≡ RR ( a, b,
5) = (6 a + 1) × b +(4+( a mod 3)) mod 3)+2 − a = 3 k, k ∈ N so a ≡ RR (3 k, b,
5) = (6(3 k ) + 1)2 b +(4+(3 k mod 3)) mod 3)+2 − RR (3 k, b,
5) = (18 k + 1)2 b +1)+2 − RR (3 k, b,
5) = 6 k × b +1)+2 + 2 b +4 − | k × b +1)+2 and lemma 5.15 shows that b +4 − ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
5) = (6(3 k + 1) + 1)2 b +(4+((3 k +1) mod 3)) mod 3)+2 − RR (3 k + 1 , b,
5) = (18 k + 6 + 1)2 b +2)+2 − RR (3 k + 1 , b,
5) = 6 k × b +2)+2 + 2 × b +2)+2 + 2 b +2)+2 − RR (3 k + 1 , b,
5) = 6 k × b +2)+2 + 2 b +3)+1 + 2 b +6 − | k × b +2)+2 , lemma 5.12 shows that 2 b +3)+1 ≡ b +6 − ≡ RR (3 k + 1 , b, ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
5) = (6(3 k + 2) + 1)2 b +(4+((3 k +2) mod 3)) mod 3)+2 − RR (3 k + 2 , b,
5) = (18 k + 12 + 1)2 b )+2 − RR (3 k + 2 , b,
5) = 6 k × b )+2 + 4 × b )+2 + 2 b )+2 − RR (3 k + 2 , b,
5) = 6 k × b )+2 + 2 b +1)+2 + 2 b +2 − | k × b )+2 , lemma 5.13 shows that 2 b +1)+2 ≡ b +2 − ≡ RR (3 k + 2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ RR ( a, b, ≡ RR ( a, b, ≡ RR ( a, b, c ) ≡ c (mod 6) where c ∈ { , , } . We shouldalso show that the range of function RR is O R . Comparing RR to R ,only difference is the exponent of 2. The exponent in RR is 2(3 b + ( c − a mod 3)) mod 3) + 2 and in R it is 2 b + 2. For R we know b ∈ N ,therefore we have to show 3 b + ( c − a mod 3)) mod 3 can produce allnumbers in N . This can be seen if we look at the proofs for all 9 pairs of( a, c ) above. As we are working with the pairs, we see that the exponentbecomes 3 b, b + 1 and 3 b + 2. Since b ∈ N it is clear 3 b, b + 1 and 3 b + 2will produce numbers in N . (cid:3) Theorem 5.18.
Let a, b ∈ N and c ∈ { , , } . The function RR : N × N × { , , } −→ O R is defined as RR ( a, b, c ) = (6 a + 5) × b +( c − ( a mod 3)) mod 3)+1 −
13 (5.16) where RR ( a, b, c ) ≡ c (mod 6) .Proof. The proof will be similar to that of theorem 5.17. Let us set c = 1first to show RR ( a, b, ≡ RR ( a, b,
1) = (6 a + 5)2 b +(1 − ( a mod 3)) mod 3)+1 − N COLLATZ CONJECTURE 27
Let a = 3 k, k ∈ N so a ≡ RR (3 k, b,
1) = (6(3 k ) + 5)2 b +(1 − (3 k mod 3)) mod 3)+1 − RR (3 k, b,
1) = (6(3 k ) + 5)2 b +1)+1 − RR (3 k, b,
1) = (18 k + 3 + 2)2 b +1)+1 − RR (3 k, b,
1) = 6 k × b +1)+1 + 2 b +1)+1 + 2 × b +3 − RR (3 k, b,
1) = 6 k × b +1)+1 + 2 b +1)+1 + 2 b +4 − | k × b +1)+1 , lemma 5.12 shows that 2 b +1)+1 ≡ b +4 − ≡ RR (3 k, b, ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
1) = (6(3 k + 1) + 5)2 b +(1 − ((3 k +1) mod 3)) mod 3)+1 − RR (3 k + 1 , b,
1) = (6(3 k + 1) + 5)2 b )+1 − RR (3 k + 1 , b,
1) = (18 k + 9 + 2)2 b )+1 − RR (3 k + 1 , b,
1) = 6 k × b )+1 + 3 × b )+1 + 2 × b +1 − RR (3 k + 1 , b,
1) = 6 k × b )+1 + 6 × b ) + 2 b +2 − | k × b )+1 + 6 × b ) and lemma 5.14 shows that b +2 − ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
1) = (6(3 k + 2) + 5)2 b +(1 − ((3 k +2) mod 3)) mod 3)+1 − RR (3 k + 2 , b,
1) = (6(3 k + 2) + 5)2 b +2)+1 − RR (3 k + 2 , b,
1) = (18 k + 12 + 3 + 2)2 b +2)+1 − RR (3 k + 2 , b,
1) = 6 k × b +2)+1 + 4 × b +2)+1 + 2 b +2)+1 + 2 × b +5 − RR (3 k + 2 , b,
1) = 6 k × b +2)+1 + 2 b +3)+1 + 2 b +2)+1 + 2 b +6 − | k × b +5 , lemma 5.12 shows that 2 b +3)+1 +2 b +2)+1 ≡ b +6 − ≡ RR (3 k + 2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ c = 3 to show RR ( a, b, ≡ (mod 6). RR ( a, b,
3) = (6 a + 5)2 b +(3 − ( a mod 3)) mod 3)+1 − a = 3 k, k ∈ N so a ≡ RR (3 k, b,
3) = (6(3 k ) + 5)2 b +(3 − (3 k mod 3)) mod 3)+1 − RR (3 k, b,
3) = (18 k + 3 + 2)2 b )+1 − RR (3 k, b,
3) = 6 k × b )+1 + 2 b )+1 + 2 × b +1 − RR (3 k, b,
3) = 6 k × b )+1 + 2 b )+1 + 2 b +2 − | k × b )+1 , lemma 5.12 shows that 2 b )+1 ≡ b +2 − ≡ RR (3 k, b, ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
3) = (6(3 k + 1) + 5)2 b +(3 − ((3 k +1) mod 3)) mod 3)+1 − RR (3 k + 1 , b,
3) = (18 k + 9 + 2)2 b +2)+1 − RR (3 k + 1 , b,
3) = 6 k × b +2)+1 + 3 × b +2)+1 + 2 × b +5 − RR (3 k + 1 , b,
3) = 6 k × b +2)+1 + 6 × b +2) + 2 b +6 − | k × b +2)+1 + 6 × b +2) and lemma 5.16 shows that b +6 − ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
3) = (6(3 k + 2) + 5)2 b +(3 − ((3 k +2) mod 3)) mod 3)+1 − RR (3 k + 2 , b,
3) = (18 k + 12 + 3 + 2)2 b +1)+1 − RR (3 k + 2 , b,
3) = 6 k × b +1)+1 + 4 × b +1)+1 + 2 b +1)+1 + 2 × b +3 − RR (3 k + 2 , b,
3) = 6 k × b +1)+1 + 2 b +2)+1 + 2 b +1)+1 + 2 b +4 − | k × b +1)+1 , lemma 5.12 shows that 2 b +2)+1 + 2 b +1)+1 ≡ b +4 − ≡ RR (3 k +2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ c = 5 to show RR ( a, b, ≡ RR ( a, b,
5) = (6 a + 5)2 b +(5 − ( a mod 3)) mod 3)+1 − N COLLATZ CONJECTURE 29
Let a = 3 k, k ∈ N so a ≡ RR (3 k, b,
5) = (6(3 k ) + 5)2 b +(5 − (3 k mod 3)) mod 3)+1 − RR (3 k, b,
5) = (18 k + 3 + 2)2 b +2)+1 − RR (3 k, b,
5) = 6 k × b +2)+1 + 2 b +2)+1 + 2 × b +5 − RR (3 k, b,
5) = 6 k × b +2)+1 + 2 b +2)+1 + 2 b +6 − | k × b +2)+1 , lemma 5.12 shows that 2 b +2)+1 ≡ b +6 − ≡ RR (3 k, b, ≡ RR (3 k, b, ≡ a = 3 k + 1 , k ∈ N so a ≡ RR (3 k + 1 , b,
5) = (6(3 k + 1) + 5)2 b +(5 − ((3 k +1) mod 3)) mod 3)+1 − RR (3 k + 1 , b,
5) = (6(3 k + 1) + 5)2 b +1)+1 − RR (3 k + 1 , b,
5) = (18 k + 9 + 2)2 b +1)+1 − RR (3 k + 1 , b,
5) = 6 k × b +1)+1 + 3 × b +1)+1 + 2 × b +3 RR (3 k + 1 , b,
5) = 6 k × b +1)+1 + 6 × b +1) + 2 b +4 | k × b +1)+1 + 6 × b +1) and lemma 5.15 shows b +4 ≡ RR (3 k + 1 , b, ≡ a = 3 k + 2 , k ∈ N so a ≡ RR (3 k + 2 , b,
5) = (6(3 k + 2) + 5)2 b +(5 − ((3 k +2) mod 3)) mod 3)+1 − RR (3 k + 2 , b,
5) = (18 k + 12 + 3 + 2)2 b )+1 − RR (3 k + 2 , b,
5) = 6 k × b )+1 + 4 × b )+1 + 2 b )+1 + 2 × b +1 − RR (3 k + 2 , b,
5) = 6 k × b )+1 + 2 b +1)+1 + 2 b )+1 + 2 b +2 − | k × b )+1 , lemma 5.12 shows that 2 b +1)+1 +2 b )+1 ≡ b +2 − ≡ RR (3 k +2 , b, ≡ RR (3 k + 2 , b, ≡ RR ( a, b, ≡ RR ( a, b, ≡ RR ( a, b, ≡ RR ( a, b, c ) ≡ c (mod 6)where c ∈ { , , } . We should also show that the range of function RR is O R . Comparing RR to R the only difference is the exponent of 2.The exponent in RR is 2(3 b + ( c − ( a mod 3)) mod 3) + 1 and in R it is2 b + 1. For R we know b ∈ N , therefore we have to show 2(3 b + ( c − ( a mod 3)) mod 3) + 1 can produce all numbers in N . This can be seen ifwe look at the proofs for all 9 pairs of ( a, c ) above. As we are working withthe pairs, we see that the exponent becomes 3 b , 3 b + 1 and 3 b + 2. Since b ∈ N it is clear 3 b, b + 1 and 3 b + 2 will produce numbers in N . (cid:3) One-tree using reverse reduced Collatz function.
Consider agraph G = ( V, E ) where V = O + and E = { ( n, m ) : C ( n ) = m where n, m ∈ V } . This graph is most notably known as Collatz Graph [4]. If we omit theself-loop (1 ,
1) from the Collatz Graph, the conjecture can be reformulatedin this context: “Is Collatz Graph a tree with root vertex 1”? The afore-mentioned tree is also named One-tree in the context of Collatz conjecture,or (3 x + 1)-tree [5].5.2.1. Constructing the Collatz graph.
Let s = R ( d ). By definition of R weknow that s does not have a single value, since the exponent x in R canhave many values (see (5.1)). Moreover, s does not have a value when d ≡ s separately for d ≡ d ≡ R and R functions. We get s = R ( a, b ) , d = 6 a + 1 or s = R ( a, b ) , d = 6 a + 5. Suppose we calculate s and d by calculating R ( a, b ) or R ( a, b ). The order of calculation or thestarting point does not matter, what matters is that we cover all possibilitiesfor both functions where a, b ∈ N . Suppose that we made a calculation andthereby have s and d . Suppose we also have a graph G = ( V, E ). We have4 cases regarding s and d being an element of V :(1) s (cid:54)∈ V, d (cid:54)∈ V . Neither s or d is present in our graph. What we dois, add s and d to our graph and connect s to d . Our new graph is G (cid:48) = ( V ∪ { s, d } , E ∪ { ( s, d ) } ).(2) s (cid:54)∈ V, d ∈ V . This means that d already exists in the graph, sowe add s to the graph and connect it to d . Our new graph is G (cid:48) =( V ∪ { s } , E ∪ { ( s, d ) } ).(3) s ∈ V, d (cid:54)∈ V . This means that s already exists in the graph, sowe add d to the graph and connect it to s . Our new graph is G (cid:48) =( V ∪ { d } , E ∪ { ( s, d ) } ).(4) s ∈ V, d ∈ V . This means that both numbers are present in the graphbut they are not connected. Our new graph is G (cid:48) = ( V, E ∪ { ( s, d ) } )Suppose we use G (cid:48) instead of G after every calculation of s and d . Startingfrom an empty graph G = ( ∅ , ∅ ) and calculating s , d for all possibilities of R ( a, b ) and R ( a, b ) using the logic described above, we expect to get theCollatz graph. From theorem 5.11 we can infer that a case where s, d ∈ V and ( s, d ) ∈ E at some calculation is not possible, because the theoremindicates that s will always be unique, as for the graph construction in allfour cases we add the edge ( s, d ) to E , if s is unique then the edge ( s, d ) willbe added to E only once, for every unique occurrence of s , that is.A short example of the procedure is given in figure 1. There are 8 cal-culations, order being from left to right and top to bottom. First R (0 , s = 1, d = 1. This is the only case where s = d ,which is the trivial cycle 1 −→
1. We add the vertex 1 and make a self-loop.Next, we calculate R (0 ,
1) and get s = 5, d = 1. 1 exists in the vertices(in fact it is the only one so far) but 5 does not. This is case 2 as described N COLLATZ CONJECTURE 31
Figure 1.
An example step-by-step construction of CollatzTree, steps from left to right and top to bottom.above. We add 5 and connect it to 1. The calculations go on like this; it isconjectured that in the end we will have a tree with root 1 (and the self-loopat 1 of course), similar to the graph shown at the bottom right corner offigure 1. 6.
Concluding Remarks
We believe the fractional sum notations (see definitions 2.1 and 4.1) are auseful way of analyzing reduced Collatz trajectories. A continuation of theloops discussed in section 4.1 could perhaps shed more light on the subject.The issue regarding equation 3.1 shows that it can perhaps be improved toresolve the problem, which would let us study Collatz sequences even better.We have shown that the reverse reformulation can used to predeterminethe numbers in modulo 6 (see theorems 5.17 and 5.18). For example, one canavoid bumping into numbers n such that n ≡ For a trajectory n a −→ C ( n ) a −→ . . . a j −→ R ( n ) x −→ n a −→ C ( n ) a −→ . . . a j −→
1. We have shown how to use equation(3.1) and equation (3.13) in theorem 3.1 to get m a −→ C ( m ) a −→ . . . a j −→ C j ( m ) a j +1 −−−→
1. This may provide an additional perspective on generatingtrajectories.
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Mathematics of Computation , 32(144):1281–1292, October 1978.[3] G. Helms. Collatz-Intro - Some general remarks, 2004. Accessed 12 March 2019.[4] J. C. Lagarias. The 3x+1 and Its Generalizations.
American Mathematical Monthly ,92(1):3–23, January 1985.[5] J. C. Lagarias. The 3x+1 Problem: An Annotated Bibliography, II (2000-2009). arXivMathematics e-prints , page math/0608208, Aug 2006.[6] F. C. Motta, H. R. de Oliveira, and T. A. Catalan. An Analysis of the CollatzConjecture.[7] E. Roosendall. On the 3x+1 Problem, 2019. Accessed 12 March 2019.[8] J. L. Simons. A simple (inductive) proof for the non-existence of 2-cycles of the 3x+1problem.
Journal of Number Theory , 123(1):10 – 17, 2007.[9] R. P. Steiner. A Theorem on the Syracuse Problem. In
A Theorem on the SyracuseProblem , pages 553–559. 7th Manitoba Conference on Numerical Mathematics, 1977.[10] Ian Stewart.
The Great Mathematical Problems , chapter 17, page 284. Profile BooksLTD, 2014.
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