Construction of periods for 3x+1 problem Use of division algorithm by 2 in 3-base number system for construction of 3-adic numbers as periods of Collatz sequence Yagub N. Aliyev Dept. of Mathematics Qafqaz University Khirdalan AZ0101, Azerbaijan yaliyev@qu.edu.az Vugar A. Suleymanov (Masters Student) Dept. of Computer Science Qafqaz University Khirdalan AZ0101, Azerbaijan s.vugar1988@gmail.com Abstract—The division by 2 algorith for numbers written in 3- base number system is dicussed. This algorithm can be easily generalized to other odd base number systems. We used the algorithm for study of periods in sequence which appear in famous "3x+1 problem" or "Collatz problem". The idea of using 3-adic numbers naturally arises from our discussion. We proved an interesting fact about the nature of periods in digits of these 3- adic numbers. Index Terms—3x+1 problem, Collatz problem, p-adic numbers, 3-adic numbers, periods. I. INTRODUCTION Let us define the following sequence. Suppose x0 is a positive integer. For n>0 we define recursively xn=xn-1/2 if xn-1 is even number and xn=(3xn-1+1)/2 if xn-1 is odd number. For example if we start with x0=7 we obtain x1=11, x2=17, x3=26, x4=13, x5=20, x6=10, x7=5, x8=8, x9=4, x10=2, x11=1, x12=2, x13=1, …. As we see we obtained number 1 at 11 th step. After that step we obtain cyclic sequence 1,2,1,2,…. Various interesting names are used to describe this sequence. In the book which inspired us to write the current paper it is called as “devil's sequence” [1]. It is also called as “hailstone numbers” [2]. The following two problems are still unsolved. Problem 1. Prove that for arbitrary positive integer x0 there is a number n such that xn=1. Problem 2. If the sequence xn is periodic then its period is 1,2,1,2,…. Many professional and amateur mathematicians worked on this problem: Lothar Collatz, Bryan Thwaites. Therefore this problem is sometimes called as “Collatz conjecture”, “Syracuse problem”, “Hasse’s algorithm”, “Kakutani’s problem”, “Ulam’s problem”. But we prefer to use generally accepted name “3x+1 problem” [3-5]. It is interesting to note remarks of Paul Erdös about this problem “Mathematics is not yet ripe for such a question”. Richard Guy put this problem in his list: “Don’t try to solve these problems”. We studied some number theoretical problems related to this problem in paper [6]. In the current paper we discussed an interesting approach to Problem 2. We supposed that some period with given operations x/2 or (3x+1)/2 exists and then constructed the elements of the periodic sequence as 3-adic numbers. If these numbers have infinite number of nonzero digits then the period with given operations doesn’t exist. For this we must first discuss algorithm which we used for division by 2. II. 3-BASE NUMBER SYSTEM Suppose that number is written in 3-base number system. Therefore only digits 0,1,2 are used. It is easy to decide whether this number is even or odd. Just add the digits if this sum is even the number is also even, otherwise it is odd. We can also use simpler test: If the number of unit digits (1) is even then the number itself is also even, otherwise it is odd. For example the number 1102022100122021 is an odd number but 21221200200020 is an even number. If the number is even then it is possible to group the digits of this number so that: 1) Either there is no unit digits in the group or two unit digits 2) Unit digits appear only at the beginning and the end of group. So the even number 2122120102010012102 can be grouped as 2 1221 20 10201 00 121 02. This grouping is useful in application of the algorithm which we are now going to discuss. III. DIVISION BY 2 ALGORITHM We use two substitutions:          1 2 0 2 1 0 A ,          2 0 1 2 1 0 B . 1) We start from right with 1 st digit and apply A. 2) If n th digit (counted from right) is even then we apply to the next (n+1) th digit the same substitution which was applied to n th digit. 3) If n th digit (counted from right) is odd then we apply to the next (n+1) th digit the substitution which is different from what was applied to n th digit i.e. if A applied to n th digit then we apply B to (n+1) th digit and vice versa. Each time we write the obtained digits under the corresponding digits of given number. For example: xk = 12100120212