An Investigation of the Collatz Conjecture
aa r X i v : . [ m a t h . G M ] J a n AN INVESTIGATION OF THE COLLATZ CONJECTURE
JOHN G. KOELZER † , ROCKHURST UNIVERSITYDANIEL J. WELLINGJanuary 7, 2019 Abstract:
This paper explores special conditions on the starting value of a Collatz sequence whichimply that the Collatz conjecture is true. This is the result of the collaboration of a retired mathematicsprofessor (Koelzer) and a retired physics professor (Welling). † email: [email protected]. AN INVESTIGATION OF THE COLLATZ CONJECTURE
The Collatz conjecture was formulated by L. Collatz in 1937. The conjecture concerns the definition ofa sequence of positive integers by means of a simple algorithm. The conjecture states that no matterwhat the starting value is, the sequence eventually equals 1. For more information on the history of theCollatz problem see [1] and [2].
The Collatz Conjecture:
Let n be a positive integer. If n is even, divide it by 2 to get n/
2. If n is odd, multiply it by 3, add 1 and divide the result by 2 to obtain the number (3 n + 1) /
2. Repeatthe process indefinitely. The conjecture is that no matter what number you start with, you will alwaysreach 1. Our definition follows that of Terras in [3]. This is equivalent to the original Collatz algorithmbut it results in a somewhat shorter sequence. We will assume that our starting value is an odd integer;otherwise we can simply divide by 2 until the number is odd.
Example:
The Collatz sequence for n = 19 is: 19 → → → → → → → → → → → → → →
1. (14 steps)In this paper we will present some results concerning the Collatz conjecture. We will assume that thestarting integer N in a Collatz sequence is an odd integer, since, if N is even, it can be repeatedly bedivided by 2 until the resulting integer is odd.The following theorem expresses the relation between an odd integer in a Collatz sequence and thenext even integer in the sequence. Theorem 1:
Let N be an odd integer in a Collatz sequence and define E to be N + 1. The successiveodd integers in the sequence starting with N are3 (cid:18) E (cid:19) − , (cid:18) E (cid:19) − , (cid:18) E (cid:19) − , · · · , k − (cid:18) E k − (cid:19) − , where k is the highest power of 2 contained as a factor in E . The integer 3 k (cid:18) E k (cid:19) − N . Proof:
Let m be an integer such that 1 ≤ m ≤ k − . Then E m is an even number and hence3 m (cid:18) E m (cid:19) − N . Case 1: m = 1. The next term in the Collatz sequence after N is3 N + 12 = 32 N + 12 = 32 ( E −
1) + 12 = 32 E − m = 1. Case 2: ≤ m ≤ k − . Since 3 m − (cid:18) E m − (cid:19) − (cid:18) m − (cid:18) E m − (cid:19) − (cid:19) + 12 , which simplifies algebraically to 3 m (cid:18) E m (cid:19) − N INVESTIGATION OF THE COLLATZ CONJECTURE 3 in this case.Finally, since k is the highest power of 2 contained as a factor in E , E k will be an odd integer.This means that 3 k (cid:18) E k (cid:19) − N .Given an odd integer N , let u be the number of steps to the next even integer, say M , and let d bethe number of steps from M to the next odd integer N ′ . By Theorem 1 u is the largest power of 2 in N + 1 and, since M is even, d is the largest power of 2 in M . The relation between N and N ′ is givenby the formula N ′ = (cid:2) ( ) u ( N − (cid:3) ( ) d Here is a simple corollary to illustrate the theorem:
Corollary 1:
Suppose k is a positive integer. Let N = 10 k − k − Proof:
Using the notation from Theorem 1, E = 10 k and the largest power of 2 in E is 2 k . So by theTheorem, the first even integer following 10 k − k ( E k ) − k − Theorem 2:
Let k , l and m be positive integers with m odd and let N = (cid:0) (cid:1) k (cid:0) l m + 1 (cid:1) −
1. If N isan odd integer then N reaches m in l + k steps. Proof:
Let E be the even integer N + 1. Note that E can be written as 2 k (cid:0) l m + 1 (cid:1) k . We claim that k is the highest power of 2 contained as a factor in E . Suppose p is the highest power; Since 2 k is a factorof E , we must have k ≤ p .Furthermore, we can write E p as 2 k (cid:18) l m + 13 k (cid:19) p . Since 2 l m + 1 and 3 k are odd, their quotient is oddand so 2 p must divide 2 k . This implies p ≤ k and we conclude k = p .By Theorem 1, N is followed by k − k (cid:18) E k (cid:19) −
1. This even number can be rewritten as (cid:18) (cid:19) k (cid:20) k (2 l m + 1)3 k (cid:21) −
1, which algebraically reducesto 2 l m . This means that the next l terms of the Collatz sequence are all products of m multiplied bypowers of 2 terminating at m and the total number of steps from N to m is l + k . Example: k = 3 , l = 2 , d = 47 → N = (8 / ·
47 + 1) − → → → → →
47. (5 steps)
Remarks:
To apply Theorem 1, it is required to find integers k , l and m that make N odd. Also, itwould be nice to show that there is an infinite number of values of k , l and m satisfying the hypothesesof the theorem, but that is not essential to the proof.Here is a special case of Theorem 2: AN INVESTIGATION OF THE COLLATZ CONJECTURE
Corollary 2:
Let k and l be positive integers and let N = (cid:0) (cid:1) k (cid:0) l + 1 (cid:1) −
1. If N is an odd integerthen N reaches 1 in l + k steps. Proof:
Let m = 1 in Theorem 1. Examples: k = 1 , l = 5 → N = (cid:0) + 1 (cid:1) − . The Collatz sequence for 21 is: 21 → → → → → → . (6 steps) k = 2 , l = 9 → N = (cid:0) (cid:1) (cid:0) + 1 (cid:1) − . The Collatz sequence for 227 is: 227 → → → → → → → → → → →
1. (11 steps) k = 3 , l = 9 → N = (cid:0) (cid:1) (cid:0) + 1 (cid:1) − . The Collatz sequence for 151 is: 151 → → → → → → → · · · → → → . (12 steps)We prove a result that will be useful later in the paper. Lemma 1:
Let n be a positive integer. Then 2 (3 n − ) ≡ − n . Proof:
We will use mathematical induction [4].
Case 1: If n = 1, 2 (3 ) = 2 ≡ − Case 2:
Assume the lemma is true for n = k ; i.e., 2 (3 k − ) ≡ − k . We may write 2 (3 k − ) = 3 k q − q . To show the equation is true for n = k + 1, we cube both sides of this equation usingthe binomial theorem: 2 k = (2 k − ) = (cid:0) k q − (cid:1) = (3 k q ) − k q ) + 3(3 k q ) − q k − q k + 3(3 k q ) − q k − k +1 − q k k +1 + 3 k +1 q − (cid:0) q k − − q k + q (cid:1) k +1 − k − ( −
1) is a multiple of 3 k +1 and the formula is verified for n = k + 1. Example:
For n = 4, 2 (3 n − ) = 2 = 134 , , ≡ ≡ − Theorem 3:
Suppose k is a positive integer and l = 3 k − · r , where r is an odd integer. If N = (cid:0) (cid:1) k (cid:0) l + 1 (cid:1) − N terminates at 1 after l + k steps. Proof:
By invoking Corollary 1, it suffices to show that N is an odd integer under the given hypotheses.By Lemma 2, 2 (3 k − ) ≡ − k .2( k − ) ≡ − k ⇒ ( k − m ≡ ( − m mod 3 k ⇒ l + 1 ≡ ( − m + 1 ≡ k , since m is an odd integer.Because 2 l + 1 is a multiple of 3 k , it follows that N is an odd integer. N INVESTIGATION OF THE COLLATZ CONJECTURE 5
Example: k = 2 , l = 3 · → N = (cid:0) (cid:1) (cid:0) + 1 (cid:1) − . The Collatz sequence for 14563 is:14563 → → → → → → · · · → → → → Corollary 3: If q is an integer and N = 4 q −
13 , then the Collatz sequence starting with N reaches 1in 2 q steps. Proof:
Let k = 1 in Theorem 2 and simplify algebraically. Example: q = 4 → N = 4 −
13 = 85.The Collatz sequence for 85 is: 85 → → → → → → → →
1. (8 steps).
Corollary 4: If r is an odd integer and N = 2 r +2 −
59 , then the Collatz sequence starting with N reaches 1 in 3 r + 2 steps. Proof:
Let k = 2 in Theorem 2 and simplify algebraically. Example: r = 3 → N = 2 −
59 = 227.The Collatz sequence for 227 is: 227 → → → → → → → → → → → Definitions:
Every odd positive integer N falls into one of three categories:(i) N = 6 r + 1 , r = 0 , , , ..., (call this type A ) Examples: 1, 7, 13,...(ii) N = 6 r + 3 , r = 0 , , , ..., (call this type B ) Examples: 3, 9, 15,...(iii) N = 6 r + 5 , r = 0 , , , ..., (call this type C ) Examples: 5, 11, 17,... Theorem 4:
Every odd integer in a Collatz sequence except possibly the first one is of type A or C.
Proof:
Let N be an odd integer of type B in a Collatz sequence. Then there is an integer r such that N = 6 r + 3. Suppose the number in the Collatz sequence preceding N in the sequence is the odd number N ′ . Then by the Collatz algorithm, 3 N ′ + 12 = N = 6 r + 3. We then have 3 N ′ + 1 = 12 r + 6. Since 3divides the right-hand side of this equation it must divide the left-hand side, which is impossible. Weconclude that N cannot be immediately preceded by an odd integer.Therefore we can assume that either N is at the start of the Collatz sequence or there is an integer k such that the even numbers 2 k N, k − N, · · · , N comprise the sequence before N . This implies thatthere is an odd number N ′ preceding 2 k N . This means that 3 N + 12 = 2 k N = 2 k (6 r + 3) = 2 k r + 1).So 3 N ′ + 1 = 3(2 k +1 )(2 r + 1). Since 3 divides the right hand side of this equation, it must divide theleft side, which is a contradiction and the theorem is proved. Remark:
This means that no Collatz sequence contains an odd multiple of three; i.e., Type B exceptpossibly at the start of the sequence.
AN INVESTIGATION OF THE COLLATZ CONJECTURE
Example:
The Collatz sequence for N = 9: 9(type B) → → → → → → → → → → → → → Conjecture:
Let N be an odd integer in a Collatz sequence and let N ′ = N +12 be the next term inthe sequence. If N ′ factors into 2 l m then m , the next odd integer in the sequence, is of type A if l isodd and is of type C if l is even. Examples: N = 15 → N ′ = 23 = 2 · l = 0 (even) and m = 23 (type C). N = 117 → N ′ = 176 = 2 · l = 4 (even) and m = 11 (type C). N = 49 → N ′ = 74 = 2 · l = 1 (odd) and m = 37 (type A). N = 133 → N ′ = 200 = 2 · l = 3 (odd) and m = 25 (type A). N = 341 → N ′ = 512 = 2 · l = 9 (odd) and m = 1 (type A). References [1] J. C. Lagarias. The 3x + 1 Problem and its Generalizations.
The American Mathematical Monthly , 92:3–23, 1985.[2] Eric W. Weisstein. Collatz Problem. from Mathworld–a Wolfram Web Resource. http://mathworld.wolfram.com/CollatzProblem.html .[3] R. Terras. A Stopping Time Problem on the Positive Integers.
Acta Arithmetica , 3(30):241–252, 1976.[4] Wikipedia Contributors. Mathematical Induction — Wikipedia, The Free Encyclopedia. https://en.wikipedia.org/w/index.php?title=Mathematical_induction&oldid=874142784https://en.wikipedia.org/w/index.php?title=Mathematical_induction&oldid=874142784