Ternary Modified Collatz Sequences And Jacobsthal Numbers

Transcription

1 Journal of Integer Sequences, Vol. 19 (016), Article Ternary Modified Collatz Sequences And Jacobsthal Numbers Ji Young Choi Department of Mathematics Shippensburg University of Pennsylvania 1871 Old Main Drive Shippensburg, PA 1757 USA jychoi@ship.edu Abstract We show how to apply the Collatz function and the modified Collatz function to the ternary representation of a positive integer, and we present the ternary modified Collatz sequence starting with a multiple of N for an arbitrary large integer N. Each ternary string in the sequence is shown to have a repeating string, and the number of occurrences of each digit in each repeating string is identified as a Jacobsthal number, or one more or one less than a Jacobsthal number. 1 Introduction Definition 1. [1] For any positive integer m, the Collatz function f on m is defined as follows: { m, if m is even; f(m) := (1) m+1, if m is odd. For any positive integer m, consider a sequence satisfying: a 0 = m and a n = f(a n 1 ) for n > 1. Collatz [1] asked if there exists n such that a n = 1, for any positive integer m = a 0. The Collatz problem is still open, but it is known [] that for every positive integer m < there exists n > 0 such that f n (m) = 1. If there is an integer m such 1

2 that f n (m) 1 for any integer n, we can assume m is a large odd integer. If m is even, we can divide m by a power of to find a large odd integer. To find the Collatz sequence starting with a large odd integer m, we calculate f(m) = m + 1 and f (m) = m+1. To make these calculations simpler, we would like to modify the Collatz function, since f (m) is always m+1 for any odd integer m. Definition. [] For any positive integer m, the modified Collatz function g on m is defined as follows: { f(m) = m g(m) :=, if m is even; () f (m) = m+1, if m is odd. To make the calculation m+1 for a large integer m easier, we use the ternary representation of integers. In this paper, we show how to apply the Collatz function and the modified Collatz function to the ternary representation of integers. Then we study the ternary representation of the modified Collatz sequence starting with N for an arbitrary large integer N. As the ternary representation of N is the string with a digit 1 and N 0 digits, the ternary representation of g n ( N ) has a repeating string of digits, for significantly many nonnegative integers n. The number of occurrences of each digit in each repeating string will be counted and identified. Then we consider a multiple of N for an arbitrary large integer N. Section clarifies notation for this paper. Sections and 4 show how to apply the Collatz function and the modified Collatz function to the ternary representation of integers. In Section 5, a repeating string of digits in the ternary modified Collatz sequence starting with the string is studied, and the number of occurrences of each digit in each repeating string is counted. Section 6 shows that the number of 1 digits counted in Section 5 is identified as a Jacobsthal number, and suggests an extension of Jacobsthal numbers. Section 7 generalizes Section 5. Notation Every ternary representation of an integer is a finite string in {0,1,}, which is the set of all finite strings consisting of digits 0,1,. The set {0,1,} also includes the empty string, which contains no digits, denoted by ǫ [5]. The notation for the number of digits in a string is as follows: Definition. [5] For any finite string x and a digit a, x is defined to be the number of digits in x, and x a is defined to be the number of occurrences of digit a s in x. Lemma 4. For any string x in {0,1,}, x = x 0 + x 1 + x.

3 For example, = 8, =, =, = and 8 = ++. The following operation shows how to create a new string from given ones [5, 7]: Definition 5. For any strings x and y and any positive integer n, the concatenation of x and y, denoted by xy, is the string obtained by joining x and y end-to-end, and x n denotes the concatenation of n x s. That is, if x = a 1 a a x and y = b 1 b b y for some a i,b i {0,1,}, xy = a 1 a a x b 1 b b y, and x n = xx x (n times). For a convention, x 0 is defined to be ǫ. Then concatenation is associative, and the length of the concatenation of two strings is the sum of each string length. Lemma 6. For any ternary strings x, y, and z, xyz = (xy)z = x(yz). Lemma 7. For any strings x and y and a nonnegative integer n, xy = x + y and x n = n x. For example, = 1000, (10) = , and 1 = 1 (10) = = + = 5 and (10) = 10 = = 5. For a finite string x, a substring y is called a head of x if there exists a string z such that x = yz, and a substring z is called a tail of x if there exists a string y such that x = yz [4]. For example, 10 and 10 are heads of 1000, and 00 and 0 are tails of Notice that a head or a tail of a string is not necessarily unique. Since the ternary representation of an integer is a string in {0,1,}, we call the ternary representation of an integer as a ternary string throughout this paper. When we have to distinguish an integer and its ternary string, we use the following notation. Notation 8. For any integer m with base -representation x, we write m = [x] or (m) = x. For example, 5 = [1] and (5) = 1. Then ([x] ) = x for any ternary string x and [(m) ] = m for any integer m. Throughout the paper we use the convention that m is an integer and x is its ternary representation. When we apply the functions f and g we often phrase this in terms of how f and g transform x to another ternary representation, and do not mention m. Notation 9. For a ternary string x, the ternary representation of f([x] ) and g([x] ) are denoted by f(x) and g(x), respectively. That is, f(x) = (f([x] )) and g(x) = (g([x] )). Then the ternary string g(x) is identified as follows: Lemma 10. For a ternary string x, { f(x), if x is even; g(x) = f(f(x)), if x is odd.

4 To apply the Collatz function to a ternary string, it is important to know whether a given ternary string represents an even or odd integer. To simplify the discussion, we say a ternary string is even or odd. Definition 11. For any ternary string x, x is said to be odd if [x] is odd, and x is said to be even if [x] is even. The Collatz function on ternary strings Every ternary string represents a sum of powers of. For example, [1011] = To determine the evenness or oddness of a sum, it does not matter how many even numbers are added. Instead, we have to count how many odd numbers are added. Since an even digit times a power of is even, and an odd digit times a power of is odd, we count odd digits (i.e., the 1 s) in a ternary string. For example, the string 1011 is odd, since = is odd. Observation 1. For a ternary string x, x 1 is odd, if and only if x is odd. For an odd ternary string x, the ternary string f(x) = ([x] +1). Since () = 10, the ternary string of the product [x] is the concatenation x0. Adding 1 to [x] yields replacing the tail digit 0 in the string x0 with the digit 1. Lemma 1. For any odd ternary string x, f(x) = x1. Proof. [x] +1 = [10] [x] +1 = [x0] +1 = [x1]. For example, f(1110) = = Now we consider even ternary strings. First, consider a ternary string x without any 1 s, i.e., x only consists of 0 s or s. Since x is even, the Collatz function f divides [x] by. Sinceeverydigitinthestringxisdivisibleby, f replaceseachoccurrenceofthedigitwith the digit 1, and keeps each digit 0 as it is. For example, the ternary f(000) = Lemma 14. For a ternary string x = a 1 a a x, if each a i is either 0 or, f(x) = a 1a a x, where a i = { 0, if ai = 0; 1, if a i =. ( ) Proof. Since x 1 = 0, x is even, so f(x) = [x]. Since a 1 a a i 1 1 = 0, a 1 a a i 1 is even for any i = 1,..., x. Hence, when [a 1 a a i 1 ] is divided by, the remainder is 0. Hence, the ith digit a i in f(x) is a i, which is 0 if a i = 0; 1 if a i =. Secondly, consider another even ternary string 1x1, where the string x does not contain any digit 1. 4

5 Lemma 15. For a ternary string x = a 1 a a x, if each a i is either 0 or, { 1, if f(1x1) = 0a 1a a x, where a ai = 0; i =, if a i =. ( ) ) Proof. Since 1x1 1 = iseven, 1x1iseven, sof(1x1) = [1x1]. Theheaddigitin( [1x1] is 0, because the head digit 1 in 1x1 is less than the divisor. Since 1a 1 a a i 1 1 = 1 is odd, 1a 1 a a i 1 is odd for any i = 1,..., x, so the remainder when [1a 1 a a i 1 ] is divided by is 1. Hence, the (i + 1)th digit a i in f(1x1) is the quotient when [1a i ] is divided by, i.e. a i = [1a i], which is 1 if a i = 0; if a i =. Similarly, since 1x is odd, the remainder when [1x] is divided by is 1. Hence, the last digit in f(1x1) is [11] =. Note 16. In this paper, for any string x with the head digit 1, when f transforms the head digit 1 to digit 0, we keep the new head digit 0, so that x = f(x). For example, f(10001) = 0111, so f(10001) = Now consider a ternary string yz, where y and z are even ternary strings. Lemma 17. For any even ternary strings y and z, f(yz) = f(y)f(z). () ( ) ( ) Proof. Since y and z are even, y 1 and z 1 are even, and f(y) = [y] and f(z) = [z]. ( ) Since yz 1 = y 1 + z 1 is even, yz is even, so f(yz) = [yz]. Since [yz] = [y0 z ] +[z], [yz] = [y0 z ] + [z] = [( ) ] [y] 0 z + [z]. Since z = f(z) and [z] = [f(z)], f(yz) = ([ f(y)0 f(z) ] +[f(z)] ) = ([f(y)f(z)] ) = f(y)f(z). Now consider an arbitrary even ternary string x. By Observation 1, x has even number of 1 s, so we can separate x into small strings, each of which can be a string of the type in either Lemma 14 or Lemma 15. That is, x = y 1 y y k, for some integer k, such that y i = either x i or 1x i 1, where x i 1 = 0 for all i. For example, x = = Applying Lemma 14 and 15, we find each ternary string f(y i ) for all i. Then we use the following theorem to put them together for f(x). Theorem 18. For any even ternary string y i for i = 1,,...,k, f(y 1 y y k ) = f(y 1 )f(y ) f(y k ). (4) 5

6 Proof. By Lemma 6, y 1 y y k = y 1 (y y k ). Then, by Lemma 17, f(y 1 y y k ) = f(y 1 )f(y y k ). Continuing this, (4) is obtained. Example 19. The Collatz function f transforms the ternary string to the following ternary string: f( ) = f( ) = f(101)f(00)f(11)f(0)f(1001) = = The modified Collatz function on ternary strings We have found how to transform a ternary string x into the ternary string f(x) in Section. We use the same method to find the ternary string g(x) for any ternary string x, since ( ) f(x) = [x], if x is even; g(x) = ( ) f(x1) = [x1] (5), if x is odd, by Lemma 10 and Lemma 1. Lemma 0. For any ternary strings y and z, if y is even, g(yz) = g(y)g(z). (6) Proof. Sincey iseven, y 1 iseven. Hence, yz 1 = y 1 + z 1 iseven, iff z 1 iseven. Therefore, if z is even, yz is even, so g(yz) = f(yz). By Lemma 17, (6) is obtained. If z is odd, yz is odd and z1 is even. By (5) and Lemma 17, g(yz) = f(yz1) = f(y)f(z1) = g(y)g(z). Theorem 1. Let y i and z be ternary strings for i = 1,,...k. If every y i is even, g(y 1 y y k z) = g(y 1 )g(y ) g(y k )g(z). (7) Proof. Since every y i is even, y 1 y y k is even, so g(y i ) = f(y i ) for each i and g(y 1 y y k ) = f(y 1 y y k ). By Lemma 0, g(y 1 y y k z) = g(y 1 y y k )g(z) = f(y 1 y y k )g(z). Applying Theorem 18, (7) is obtained. Lemma. For a ternary string x = a 1 a a x, if each a i is either 0 or, (a) g(x) = a 1a a x, where a i = 0, if a i = 0; a i = 1, if a i = ; (b) g(1x1) = 0a 1a a x, where a i = 1, if a i = 0; a i =, if a i = ; 6

7 (c) g(1x) = 0a 1a a x, where a i = 1, if a i = 0; a i =, if a i =. Proof. Since x and 1x1 are even, g(x) = f(x) and g(1x1) = f(1x1). Hence, (a) and (b) are obtained by Lemmas 14 and 15. Since 1x is odd, g(1x) = f(1x1) by (5). Since 1x1 is even, f(1x1) = g(1x1), so we apply (b) to get (c). Strategy. For any ternary string x, the modified Collatz function g transforms x into the ternary string g(x) as follows: 1. break x: x = y 1 y y k y k+1, for some integer k, such that y i = either x i or 1x i 1 for all i k, and y k+1 = x k+1, 1x k+1 1, or 1x k+1, where x i 1 = 0 for all i = 1,,...,k,k +1;. find g(y i ): apply Lemma to find each ternary string g(y i );. concatenate all g(y i ) s: apply Theorem 1 to find the ternary string g(x). Example 4. The modified Collatz function transforms the ternary string to the following ternary string: g( ) = g( ) = g(101)g(00)g(11)g(10) = = In this paper, for any ternary string x with the head digit 1, when g transforms the head digit 1 to digit 0, we keep the new head digit 0 in g(x), so the length of the ternary string g(x) can be calculated as follows: Note 5. For any ternary strings x, g(x) = x, if x is even; and x +1, if odd. In order to apply the modified Collatz function g to a lengthy ternary string in Section 5 and 7, we need to study more on g. Theorem 6. For any ternary string x, let a i {0,1,} such that x = a 1 a a x. Then { a g(x) = 1 a a x, if x is even; a 1a a x, if x is odd, where 0, if a i = 0; a i = 0, if a i = 1; 1, if a i =, if a 1 a a i 1 1 is even; 1, if a i = 0; a i =, if a i = 1;, if a i =, if a 1 a a i 1 1 is odd. Proof. By(5), g divides either [x] or [x1] by. Since x = a 1 a i 1 a i a x and x1 = a 1 a i 1 a i a x 1, the ith digit in g(x) is { a a i i =, if the remainder is 0 when [a 1 a i 1 ] is divided by ; 1a i, if the remainder is 1 when [a 1 a i 1 ] is divided by, 7

8 for all i x. Applying Observation 1 and calculating a i and 1a i, a i can be obtained as desired. If x is odd, g(x) = x +1, so we have to find( one) more digit. Since the remainder is 1 when [x] is divided by, the last digit in g(x) = [x1] is [11] =. Theorem 7. For any odd ternary string x, let a i {0,1,} such that x = a 1 a a x. Then the ternary string g(xx) = a 1a a x a x +1 a x such that {0,1}, if a i = 0; {a i,a x +i} = {0,}, if a i = 1; (8) {1,}, if a i =. Proof. Let a x +i be the ( x +i)th digit in the string xx for any i x. Then a x +i = a i, so a 1 a a x +i 1 1 = a 1 a a x 1 + a x +1 a x + a x +i 1 1 = x 1 + a 1 a a i 1 1.Sincexis odd, x 1 is odd. Hence, a 1 a a i 1 1 is odd, iff a 1 a a x +i 1 1 is even. Since a i = a x +i, {a i,a x +i } collects both images of a i in Theorem 6. Corollary 8. For an odd ternary string x, the ternary string g(xx) satisfies the following: (a) g(xx) 0 = x 0 + x 1 ; (b) g(xx) 1 = x 0 + x ; (c) g(xx) = x 1 + x. Proof. By Theorem 7, g transforms one-half of the 0 digits and one-half of the 1 digits into the string xx as 0 s, so g(xx) 0 1 xx xx 1 = x 0 + x 1. Since there is no other way to get digit 0 s in g(xx), (a) is obtained. Similarly, (b) and (c) are proved. Corollary 9. For any odd ternary strings x, (a) g(xx) is odd, if and only if x is even; (b) g(xx) = x. Proof. By Corollary 8 (b), g(xx) 1 = x 0 + x = x x 1. Since x is odd, x 1 is odd by Observation 1. Hence, (a) is obtained. Since xx 1 = x 1 is even, xx is even. By Note 5, g(xx) = xx = x. Corollary 0. For any ternary strings x and z and for any positive integer k, where z 1 = xz, if k is odd, and z 1 = z, if k is even. g(x k z) = (g(xx)) k g(z 1 ), (9) Proof. Since k = k +p, where p = 1 for odd k; 0 for even k, xk = (xx) k x p. Since xx is even, apply Theorem 1 to provide (9). 8

9 Theorem 1. For any ternary strings y and z, ( ) [y] g(yz) = z, (10) for some ternary string z such that z = z, if yz is even; z +1, if odd. Proof. If y is even, the ( ternary ) string g(yz) = g(y)g(z) by Lemma 0, and yz is even, iff z is even. Since g(y) = [y], let z = g(z). Then z = z if yz is even; z +1 if odd. Assume that y is odd. If z is odd, yz is even. Then ( )[y] 1 and 1z are even, and [yz] = [([y] 1) 0 z ] +[1z]. Since 1z = z +1 and [1z] = 0z for some ternary string z such that z = z, ( ) ( ) ( ) [yz] [([y] 1) 0 z ] g(yz) = = +[0z [y] 1 ] = z. If z is even, yz is odd. Then [y] ) 1 and 1z1 are even and [yz1] = [([y] 1) 0 z1 ] +[1z1]. Since 1z1 = z + and = 0z for some ternary string z such that z = z by Theorem 6, ( ) [yz1] g(yz) = Since [y] 1 ( [1z1] = [y], (10) is obtained. ( ) [([y] 1) 0 z +1 ] = +[0z ] ( ) [y] 1 = z. 5 Ternary modified Collatz sequences with a repeating string A power of, N, for an arbitrary large positive integer N is complicated, but its ternary representation is simple: ( N ) = 1 0 N. Consider the ternary representation of the modified Collatz sequence starting with the string 1 0 N. For example, the first few numbers in the ternary modified Collatz sequence starting with are as follows: = ; g 1 ( ) = ; g ( ) = ; g ( ) = ; g 4 ( ) = Notice that there is a substring repeating in the ternary string g n ( ), when we ignore the head digit and a tail string. For example, the string 0 repeats 5 times and the string 00 repeats 1 times in the ternary string g ( ). 9

10 Definition. For an arbitrary large integer N and a positive integer n, the nth repeating string r n of the ternary string g n (1 0 N ) is defined to be the shortest string repeating times in g n (1 0 N ) such that for some ternary string t. g n (1 0 N ) = 0 (r n ) rn N t, For n = 1,,...,7, the nth repeating string r n in the ternary string g n (1 0 N ), for an arbitrary large integer N, is as follows: r 1 = 1; r = 0; r = 01; r 4 = 001; r 5 = ; r 6 = ; r 7 = Theorem. For any positive integer n, r n+1 = g(r n r n ), if r n is odd. Proof. By Definition, g n (1 0 N ) = 0 (r n ) rn N t for some ternary string t. Then, by Theorem 1 and Corollary 0, g n+1( 1 0 N) = g (0 (r n ) ) rn N t = g(0) (g(r n r n )) N / rn g(t 1 ), N where t 1 = r n t, if r n is odd; t, if even. Since N / N r n = r n g(r n r n ) is a string repeating times in the ternary string g n+1 (1 0 N ). N r n N r n, the ternary string Since r n is the shortest string repeating in g n (1 0 N ), r n+1 should be a multiple of r n. Since r n is odd, g(r n r n ) g(r n )g(r n ) by Theorem 7. Hence, r n+1 r n. Since g(r n r n ) = r n by Corollary 9 (b), the ternary string g(r n r n ) is the shortest string repeating times in the ternary string g n+1 (1 0 N ). Corollary 4. For every integer n, r n is odd and r n = n. N g(r nr n) Proof. Theproofisdonebymathematicalinduction. r = 01isoddand 01 =. Assume that r n is odd and r n = n. Then r n+1 = g(r n r n ) by Theorem. By Corollary 9, r n+1 is odd, since r n is even, and r n+1 = r n = (n+1). Remark 5. For an integer n > +log N, r n does not exist. Now let s count the number of 0 s, 1 s, and s in each r n for n. First, we simplify the notation. 10

11 n n 0 n 1 n n Table 1: n, 0 n, 1 n, and n Definition 6. For any positive integer n and any digit a {0,1,}, a n is defined to be the number of a s in the (n+)th repeating string r n+ in g n+ (1 0 N ), i.e., a n := r n+ a. For example, r n+ = n, 0 n, 1 n, and n, for n = 1,,...,5, are shown in Table 1. Lemma 7. For any positive integer n, (a) 0 n +1 n + n = n ; (b) 1 n is odd. Proof. Since r n+ = n by Corollary 4, (a) is obtained. Since r n+ is odd by Corollary 4, r n+ 1 is odd. Hence, (b) is obtained. Theorem 8. For any positive integer n, (a) 0 n+1 = 0 n +1 n ; (b) 1 n+1 = 0 n + n ; (c) n+1 = 1 n + n. Proof. Since n 1, r n+ is odd by Corollary 4, so r n+ = g(r n+ r n+ ) by Theorem. Then, by Corollary 8, r n+ 0 = r n+ 0 + r n+ 1 ; r n+ 1 = r n+ 0 + r n+ ; r n+ = r n+ 1 + r n+. By Definition 6, (a), (b), and (c) are obtained. Corollary 9. For any positive integer n, (a) n +0 n+1 = n ; (b) 1 n +1 n+1 = n ; (c) 0 n + n+1 = n. Proof. Apply each relation in Theorem 8 to Lemma 7 (a). Corollary 40. For any positive integer n, 11

12 (a) 0 n = n +1; (b) 0 n = 1 n, if n is odd; 1 n +1, if n is even; (c) n = 1 n 1, if n is odd; 1 n, if n is even. Proof. Applying Theorem 8 (a) and (c), 0 n n = (0 n 1 +1 n 1 ) (1 n 1 + n 1 ) = 0 n 1 n 1. Then 0 n n = 0 n 1 n 1 = = = 1, so (a) is obtained. Applying Theorem 8 (b) and (c); (a) and (b), we have 1 n n = (0 n 1 + n 1 ) (1 n 1 + n 1 ) = 0 n 1 1 n 1 ; 0 n 1 n = (0 n 1 +1 n 1 ) (0 n 1 + n 1 ) = 1 n 1 n 1. Combining them, 1 n n = 1 n n and 0 n 1 n = 0 n 1 n. Hence, if n is even, 1 n n = 1 = 0 and 0 n 1 n = 0 1 = 1. If n is odd, 1 n n = = 1 and 0 n 1 n = = 0. Therefore, (b) and (c) are obtained. Corollary 41. For any positive integer n, (a) 0 n 1 n n ; (b) 0 n is odd, if and only if n is odd; n is odd, if and only if n is even. Proof. By Corollary 40, (a) is obvious. By Lemma 7 (b) and Corollary 40 (b), 0 n = 1 n is odd, iff n is odd. By Corollary 40 (a), n is even, iff 0 n is odd. Corollary 4. For any positive integer n, 0 n+ 0 n = n = n+ n. (11) Proof. By Corollary 9 (a), n+1 = n+1 0 n+, so Corollary 9 (c) becomes 0 n +( n+1 0 n+ ) = n. Hence, n = 0 n+ 0 n. Then, by Corollary 40 (a), n+ n = (0 n+ 1) (0 n 1) = n. Corollary 4. For any positive integer n, (a) 0 n +0 n+1 = n +1; (b) n + n+1 = n 1. Proof. Applying Corollary 40 (a) to Corollary 9 (a) and (c), we have (0 n 1)+0 n+1 = n ; ( n +1)+ n+1 = n. 1

13 Finally, the explicit formulae for 0 n, 1 n, and n are as follows: Theorem 44. For any positive integer n, if n is odd, and if n is even, 0 n = n +1 0 n = n + ; 1 n = n +1 ; 1 n = n 1 ; n = n, ; n = n 1. Proof. By Corollary 40 (a) and (c), { (n +1)+( 0 n +1 n + n = n +1)+ n = n +, if n is odd; ( n +1)+ n + n = n +1, if n is even. Then, by Lemma 7 (a), n = n +, if n is odd, and n = n +1, if n is even. Solve for n, first. Then apply Corollary 40 (a) and (c) to find 0 n and 1 n. 6 Jacobsthal numbers and their extension The Jacobsthal numbers (J n ) n 0 (A001045) are the numbers satisfying the following recurrence relation and initial conditions: J n = J n 1 + J n ; J 0 = 0, J 1 = 1, (1) and the nth Jacobsthal number J n is 0,1,1,,5,11,1,4,85,171,41, for n = 0,1,...,10 [6]. Notice that J n = 1 n for n = 1,,...,10, which is true for any positive integer n. Theorem 45. For any positive integer n, 1 n = J n. Proof. By Theorem 8 (b), 1 n n = (0 n + n )+ 1 n = (0 n +1 n )+(1 n + n ) = 0 n 1 + n 1 = 1 n. Hence, 1 n satisfies the recurrence relation in (1). Since 1 1 = 1 = J 1 and 1 = 1 = J, 1 n = J n for any integer n 1. We also find the sequences A 1n (A005578) and A n (A000975) such that A 1n = 0 n and A,n 1 = n for any positive integer n: n ( n 1) A 1n := ; A n :=, for any nonnegative integer n [6]. For n = 0,1,...,10, A 1n is 1, 1,,, 6, 11,, 4, 86, 171, 4, and A n is 0, 1,, 5, 10, 1, 4, 85, 170, 41. Notice that the index n of J n, A 1n, and A n starts from 0, and the index n of 1 n, 0 n, and n starts from 1. However, using the explicit formulae in Theorem 44, 0 n, 1 n, and n are extended for any integer n. 1

14 Definition 46. For any integer n, 0 n, 1 n, and n are defined as follows: 1 n := n ( 1) n ; if n is odd, 0 n := 1 n and n := 1 n 1, and if n is even, 0 n := 1 n +1 and n := 1 n. For example, 0 n, 1 n, and n, for n = 5, 4,...,4,5, are shown in Table. n n n n Table : 0 n, 1 n, and n, for n = 0,±1,±,±,±4,±5 Notice that for any negative integer n, 0 n is always positive, n is always negative, and 1 n alternates the sign. More over, every a n, for any digit a {0,1,} and any negative integer n, is a fraction, whose numerator in the simplest form is a Jacobsthal number. Theorem 47. For any positive integer n, 0 n = { Jn n, for odd n; J n+1 n, for even n; 1 n = ( 1) n+1j n n; n = Proof. By Definition 46 and Theorem 44, for any positive integer n, 1 n = n ( 1) n { J n+1 n, for odd n; Jn n, for even n. = 1 ( )n n ( 1) n = ( 1) n+1 = ( 1) n+11 n n n n. Since 1 n = J n for any positive integer n, 1 n is as desired. Then, by Definition 46, 0 n and n are calculated as desired. Since the explicit formulae in Theorem 44 hold for any integer n by Definition 46, almost all of the rules in Section 4 hold for any integer n. Lemma 48. For any integer n, (a) 0 n +1 n + n = n ; (b) 0 n+1 = 0 n +1 n ; 1 n+1 = 0 n + n ; n+1 = 1 n + n ; (c) n = 1 n +1 n+1 = n+ n = 0 n+ 0 n = 0 n+1 +0 n 1 = n + n+1 +1; (d) 0 n 1 n n. 14

15 Hence, even if 0 n, 1 n, or n is not an integer for each negative n, we may say that an extension for each J n, A 1n, and A n is found. Claim 49. J n (A001045), A 1n (A005578), and A n (A000975) can be extended as follows: for any integer n, J n := 1 n ; A 1n := 0 n ; A,n 1 := n. In particular, an extension of the Jacobsthal numbers is as follows: Definition 50. For any positive integer n, J n is defined as follows: J n := ( 1) n+1j n n. Then the extended Jacobsthal numbers satisfy the same recurrence relation in (1): J n 1 + J n = J n, for any integer n. For example, the extended Jacobsthal numbers J n, for n = 7, 6,...,6,7, are shown in Table. n J n Table : J n, for n = ±1,±,...,±7 7 More repeating strings For an arbitrary large integer N, the ternary modified Collatz sequence starting with the ternary string 1 0 N has a repeating string, and J n counts the number of occurrences of digit 1 in the shortest repeating string in the ternary string g n+ (1 0 N ). Now, we would like to find more repeating strings, which have J n occurrences of digit 1. We consider the ternary modified Collatz sequence starting with a ternary string y 0 N for some ternary string y, which represents a multiple of N. Observation 51. For an arbitrary large integer N and any odd ternary string y, there exist ternary strings h and t such that the ternary string g (y 0 N ) = h r N t, where the string 01, if [y] 1 (mod 8); 10, if [y] r = (mod 8); 1, if [y] 5 (mod 8); 1, if [y] 7 (mod 8). ( ) Proof. Since y is odd, the ternary string y 0 N is odd, so g(y 0 N ) = [y 0 N 1] by (5), and g ( y 0 N) ) = z for some string z such that z = z + 1 by Theorem 1. Since ( [y] 1 15

16 [y0 N 1] = [([y] 1) 0 N+1 ] + [10 N 1] and Hence, If [y] 1 ( ) [10 N 1] ( [y] 1 is even, in g (y 0 N ) by Theorem 6. Hence, by Theorem 1, g ( y 0 N) ( ) [y] 1 = 4 = 01 N by Theorem 6, z = 1 N. g ( y 0 N) ( ) ) [y] 1 = 1 N = (11) N 1 (N N ). [y] 1 = [y] 1, and g transforms each repeating string 11 in g(y 0 N ) to 0 4 (0) N t 1, for some string t 1. Then g transforms each repeating string 0 in g (y 0 N ) to 01 in g (y 0 N ) if [y] 1 is even; 1 if odd, by Theorem 6. Hence, 01 and 1 repeat N times in 4 g (y 0 N ), if [y] 1 (mod 8) and [y] 5 (mod 8), respectively. If [y] 1 is odd, [y] 1 = [y], and g transforms each repeating string 11 in g(y 0 N ) to 4 0 in g (y 0 N ) by Theorem 6. Hence, by Theorem 1, g ( y 0 N) ( ) [y] = (0) N t, 4 for some string t. Then g transforms each repeating string 0 in g (y 0 N ) to 10 in g (y 0 N ) if [y] is even; 1 if odd, by Theorem 6. Hence, 10 and 1 repeat N times in 4 g (y 0 N ), if [y] (mod 8) and [y] 7 (mod 8), respectively. Notice that each r in g (y 0 N ) is an odd ternary string of length. Hence, we use two copiesofr ing (y 1 0 N )tofindarepeatingstringing 4 (y 0 N ), andcontinuetofindarepeating string in g n+ (y 0 N ). Then each repeating string in g n+ (y 0 N ) has J n or J n ±1 occurrences of digit 0, J n occurrences of digit 1, and J n or J n ±1 occurrences of digit. This works for any ternary string y, since we can reduce y to an odd string. Theorem 5. Let N be an arbitrary large integer and y be a ternary string. Then, for any positive integer n, there exists a ternary string r n such that r n = n, r n 1 = J n, either r n 0 = 0 n and r n = n or r n 0 = n and r n = 0 n, and the ternary string g n+iy+( y 0 N) = h (r n ) N n t for some ternary strings h and t, where i y is the nonnegative integer such that [y] / iy is an odd integer. Proof. If y is even, g iy (y 0 N ) = g iy (y) 0 N by Theorem 1 and 6. Since g iy (y) = ( [y] iy ) odd, g n+iy (y 0 N ) = g n (y 1 0 N ) for some odd string y 1. Hence, without loss of generality, we assume y is an odd string and i y = 0. is 16

17 The proof is done by mathematical induction on n. The base case, when n = 1, is shown in Observation 51: the string of length repeating N times in the ternary g (y 0 N ) are 10, 01, 1, and 1. Hence, 10 1 = 01 1 = 1 1 = 1 1 = 1 = J 1 ; 10 0 = 01 0 = 1 = 0 1 and 10 = 01 = 0 = 1 ; 1 0 = 1 0 = 0 = 1 and 1 = 1 = 1 = 0 1. Assume r n is the string of length n repeating N times in g n+ (y 0 N ) such that r n n = n, r n 1 = J n, either r n 0 = 0 n and r n = n or r n 0 = n and r n = 0 n. Since J n = 1 n is odd for any positive n by Lemma 7 (b), r n is odd. Then g(r n r n ) repeats N times r n in g n+ (y 0 N ) by Corollary 0, and g(r n r n ) = r n = n+1 by Corollary 9 (b). Hence, g(r n r n ) is the string of length n+1 repeating N times in g n+ (y 0 N ), i.e., g(r n+1 n r n ) = r n+1. Then, by Corollary 8, r n+1 0 = r n 0 + r n 1 ; r n+1 1 = r n 0 + r n ; r n+1 = r n 1 + r n. If r n 0 = 0 n and r n = n are assumed in the induction hypothesis, by Theorem 8, r n+1 0 = 0 n +1 n = 0 n+1 ; r n+1 1 = 0 n + n = 1 n+1 ; r n+1 = 1 n + n = n+1. If r n 0 = n and r n = 0 n are assumed in the induction hypothesis, by Theorem 8, r n+1 0 = n +1 n = n+1 ; r n+1 1 = n +0 n = 1 n+1 ; r n+1 = 1 n +0 n = 0 n+1. Since 1 n+1 = J n+1, r n+1 1 = J n+1. 8 Acknowledgments I would like to thank Mr. Andrew Mosedale for proofreading, although all remaining errors are my own. References [1] L. Collatz, On the motivation and origin of the (n + 1) problem, J. Qufu Normal University, Natural Science Edition (1986) [] J. C. Lagarias, The Ultimate Challenge: the x+1 Problem, American Mathematical Society, 010. [] R. Terras, A stopping time problem on the positive integers, Acta Arithmetica 0(1976) [4] G. E. Révész, Introduction to Formal Languages, McGraw-Hill Book Company, 198. [5] J. Shallit, A Second Course in Formal Languages and Automata Theory, Cambridge University Press,

18 [6] N. J. A. Sloane, Online Encyclopedia of Integer Sequences, [7] E. W. Weisstein, MathWorld-A Wolfram Web Resource, Mathematics Subject Classification: Primary 11A67; Secondary 11B75. Keywords: Collatz sequence, Collatz problem, Jacobsthal number, ternary representation, x+1 problem. (Concerned with sequences A000975, A001045, and A ) Received December 015; revised versions received January 7 016; August 1 016; September 016. Published in Journal of Integer Sequences, September Return to Journal of Integer Sequences home page. 18