#A50 INTEGERS 14 (014) ON RATS SEQUENCES IN GENERAL BASES Johann Thiel Dept. of Mathematics, New York City College of Technology, Brooklyn, New York jthiel@citytech.cuny.edu Received: 6/11/13, Revised: 6/17/14, Accepted: 7/15/14, Pulished: 10/8/14 Astract Conway s RATS sequences are generated y repeating the following process: egin with a positive integer, reverse the order of the digits, add the two integers together, then sort the sum s digits in increasing order from left to right. We consider this process for integers written in other ases esides 10 and study the long-term ehavior of such sequences. In particular, we examine and show the existence of periodic and quasiperiodic (a special type of divergence) sequences for certain choices of ase. 1. Introduction Conway s RATS [1] game has very simple set of rules: egin with a positive integer n, Reverse the digits of n, Add the two integers together, and Then Sort the digits of the sum in increasing order from left to right (discarding any zeros). Repeating this process generates an infinite sequence, ut the game is said to end if any term in the generated sequence is ever repeated. Example 1.1. The sequence generated y the RATS process eginning with n = 888 is given y: 888 7! 1677 7! 3489 7! 1333 7! 44556 7! 111 7! 7! 444 7! 888 7!. (1.1) This particular example shows that starting with n = 888, the RATS game ends ecause the sequence is periodic. Not all sequences generated y the RATS process are periodic, or even eventually periodic. Conway himself noted that the sequence generated y the RATS process starting with n = 1 diverges. Furthermore, he conjectured that this sequence is somewhat unique when it comes to the RATS game. Conjecture (Conway, [1]). Every RATS sequence is either eventually periodic or eventually part of the sequence 1 7! 7! 4 7! 7! 1333334444 7! 55666667777 7! 13333334444 7! 556666667777 7!.
INTEGERS: 14 (014) One can easily see that the RATS game can e extended to other integer ases esides 10. In this article, we will expand on some results of Cooper and Kennedy [] and Cooper and Shattuck [7] on the ehavior of RATS sequences for general ases. (Note that we always assume a nonunary ase.) In particular, we will answer an open question posed y Cooper and Shattuck on the existence of divergent RATS sequences for certain ases.. Notation and Terminology We egin with some definitions that will e used regularly throughout the following sections. Note that, unless otherwise specified, all variales used represent nonnegative integers. Definition.1. For any positive integer n written in ase, let n := n e the digit formed y reversing the digits of n, and let n 0 := n 0 e the digit formed y ordering the nonzero digits of n in increasing order from left to right. Define the function R : N! N y R (n) = (n + n) 0. To denote iterates of R, we adopt superscript notation. For any nonnegative integer t, let R t(n) = R (R t 1 (n)) if t 1 and R 0 (n) = n. Example.. R 3(110) = R 3 ((110 + 110) 0 ) = R 3 ((110 + 011) 0 ) = R 3 ((1010) 0 ) = R 3 (11) = (11 + 11) 0 = (1100) 0 = 11 Definition.3. We call {R i(n)}1 i=0 the RATS sequence generated y n in ase. Definition.4. If R p (n) = n for some p > 0, we say that n and the RATS sequence generated y n in ase are periodic. If p is the least integer with this property, we say that the period is p. If R t (n) is periodic for some t 0, we say that n and the RATS sequence generated y n in ase are ultimately periodic 1. Note that periodic sequences are also ultimately periodic, ut not vice versa. This helps to simplify some statements, avoiding the need to say that a result applies to periodic and ultimately periodic sequences. 1 In [6], periodic sequences are referred to as cycles and ultimately periodic sequences are triutaries to a cycle.
INTEGERS: 14 (014) 3 Example.5. The sequence (1.1) shows that 888 is periodic with period 8. For convenience, we will adopt the exponential notation used y Guy [5] when necessary. In exponential notation, the ase denotes the digit and the exponent denotes the numer of times that the digit appears; e.g., 118110005 will e denoted as 1 81 0 3 5. For ases larger than 10, we place parentheses around a digit larger than 9. For example, in ase, 1(11) 3 (14) is the five digit numer with digits 1, 11, 11, 11, 14. Definition.6. If n = 1 m1 (k 1) m k 1 k m k (k + 1) m k+1 ( 1) m 1 and R qt (n) = 1m1 (k 1) m k 1 k mk+t (k + 1) m k+1 ( 1) m 1, with m i 0 for 1 apple i apple 1, for some q > 0 and all t 0, we say that n and the RATS sequence generated y n in ase are quasiperiodic. If q is the least integer with this property, we say that the quasiperiod is q. The digit k > 0 which increases in count after q iterations is called the growing digit. If R t (n) is quasiperiodic for some t 0, we say that n and the RATS sequence generated y n in ase are ultimately quasiperiodic. Example.7. From Conway s conjecture we see that 13 5 4 4 and 5 6 5 7 4 are quasiperiodic elements in ase 10 with quasiperiod and growing digits 3 and 6, respectively. Definition.8. Let P, called the period set in ase, e the set of all p for which there are periodic elements with period p in ase. From the first example (1.1), the sequence 888 7! 1677 7! shows that 888 is periodic with period 8, so 8 P 10. In fact, Cooper and Kennedy were ale to completely characterize the period set P 10 (see [] and [3]). Definition.9. Let Q, called the quasiperiod set in ase, e the set of all q for which there are quasiperiodic elements with quasiperiod q in ase. Conway s conjecture gives that Q 10. Unlike the case for P 10, we do not have a complete description of Q 10. However, if Conway s conjecture is true, then it follows that Q 10 = {}. In its full strength, this remains open, ut one can show that Q 10 can only contain even numers. One way to make further progress towards Conway s conjecture would e to show that, indeed, Q 10 = {}. This is, however, not enough to prove the conjecture, since the possiility exists that there are elements that are neither ultimately periodic nor ultimately quasiperiodic, or that there are multiple disjoint sequences with quasiperiod. The di culty in showing that certain quasiperiods are not in Q 10 is that they require dealing with a tremendous numer of cases. Even showing that there are no quasiperiodic RATS sequences, with quasiperiod, that do not intersect with Conway s sequence seems di cult. In [7], quasiperiodic sequences are called divergent and the quasiperiod is called the length.
INTEGERS: 14 (014) 4 In the susequent sections, we will study the sets P and Q for certain choices of. In Section 3, we will show that there is an infinite family of ases such that P = 1. In Section 4, we show how to construct a ase such that Q contains any desired finite set of positive integers. Lastly, in Section 5, we investigate the ehavior of the growing digit of a quasiperiodic integer. 3. Existence of Periodic Sequences From the work of Gentges [4] and Cooper and Shattuck [7], it follows that there are ases such that P = 1, namely = 3 and = 10. In this section, we show that there is an infinite family of ases for which the same result is true. By a sum written in the form we mean the following sum: 3 3 3 4 + 4 3 3 3 33344 + 44333 This will help visualize what the reverse-add portion of the RATS process produces. For example, if the aove sums were in ase 10, they would give the sum 5 3 3 4 + 5 3 3 4 = 5 6 6 5. Theorem 3.1. For m > 1 and = 3 m we have: (i) If m is odd, then P contains all su (ii) If m is even, then P contains all su, we have P = 1. More precisely, ciently large even integers. ciently large integers. Proof. We will first show that, for any even k > 0, 1 k ( m 1) ( m 1) k 1 is periodic with period m + 1 + k. This proves that when m is odd, P contains all su ciently large even numers, and when m is even, P contains all su ciently large odd numers. In particular, P = 1. We egin y iterating the RATS process m + 1 times with the staring element 1 k ( m 1) ( m 1) k 1. Since k ( m 1) = k 1 + k ( m ) + 1, the reverse-add portion of the RATS process produces 1 k 1 1 k ( m )+1 ( m 1) k 1 + ( m 1) k 1 1 k ( m )+1 1 k 1 ( m ) k 1 k ( m )+1 ( m ) k 1
INTEGERS: 14 (014) 5 (Note that no carries occur since the ase = 3 m sorting step, we otain satisfies > m.) After the R (1 k ( m 1) ( m 1) k 1 ) = k ( m )+1 ( m ) (k 1). Since k ( m ) + 1 = ( k 1) + k ( m 4) + 3, we get Therefore, (k 1) k ( m 4)+3 ( m ) (k 1) + ( m ) (k 1) k ( m 4)+3 (k 1) ( m + ) (k 1) 4 k ( m 4)+3 ( m + ) (k 1) R ( k ( m )+1 ( m ) (k 1) ) = 4 k ( m 4)+3 ( m + ) 4(k 1). Again, since k ( m 4) + 3 = 4( k 1) + k ( m 8) + 7, we get Therefore, 4 4(k 1) 4 k ( m 8)+7 ( m + ) 4(k 1) + ( m + ) 4(k 1) 4 k ( m 8)+7 4 4(k 1) ( m + 6) 4(k 1) 8 k ( m 8)+7 ( m + 6) 4(k 1) R (4 k ( m 4)+3 ( m + ) 4(k 1) ) = 8 k ( m 8)+7 ( m + 6) 8(k 1). By continuing this process, we see that if 0 < j apple m, then the jth term in the RATS sequence generated y 1 k ( m 1) ( m 1) k 1 is given y ( j ) k ( m j )+ j 1 ( m + j+1 ) j ( k 1). (3.1) In particular, y (3.1), the mth term in the RATS sequence is ( m ) m 1 ( m+1 ) m ( k 1). The mth term is the first instance in which the larger digit has the higher exponent. So now the reverse-add portion of the RATS process has the following form: ( m ) m 1 ( m+1 ) m ( k )+1 ( m+1 ) m 1 + ( m+1 ) m 1 ( m+1 ) m ( k )+1 ( m ) m 1 1 m ( m 1) m ( k )+1 1 m 0 (Note that carries occur at every step in the sum.) Therefore, R (( m ) m 1 ( m+1 ) m ( k 1) ) = 1 m+1 ( m 1) m ( k )+1.
INTEGERS: 14 (014) 6 Again, continuing this process in a similar fashion, since m ( k )+1 = m+1 + m ( k 4) + 3, we get 1 m+1 ( m 1) m ( k 4)+3 ( m 1) m+1 + ( m 1) m+1 ( m 1) m ( k 4)+3 1 m+1 ( m ) m+1 ( m+1 ) m ( k 4)+3 ( m ) m+1. Therefore, R (1 m+1 ( m 1) m ( k )+1 ) = ( m ) m+ 4 ( m+1 ) m ( k 4)+3. Continuing this process, we see that if m + 1 < j apple k, then the jth term in the RATS sequence generated y 1 k ( m 1) ( m 1) k 1 is given y (A) j ( m 1) (B) m ( k j )+ j 1, (3.) where A = 1 and B = m 1 if j is even, and A = m and B = m+1 if j is odd. Taking j = k in (3.), we get that, since k is even, the (m + 1 + k)th term in the RATS sequence is 1 k ( m 1) ( m 1) k 1, our starting element. Thus this sequence is periodic with period m + 1 + k, as claimed. This proves that P = 1 and part (i) of the theorem. To otain part (ii), it is enough to show that for all even k > 0, 1 k ( m+1 )( m+1 +1) ( m 1) k ( m+1 1) is periodic with period (m+1)+k. When m is even, this will give that P contains all su ciently large even numers. This can e done using the same method shown aove. We omit the details. Notice that since 10 = 3, the theorem applies to ase = 10 with m =, and gives that P 10 contains all su ciently large integers. In this case, as a slightly stronger result, Cooper and Kennedy [], [3] showed that the period set is as large as possile, that is, P 10 = {p : p }. At this time it is not known if there are infinitely many ases such that their period sets contain all periods p p 0 for some fixed p 0 > 1. Another point of interest is that ase 3 is not in the family descried y Theorem 3.1, yet it too has the property that P 3 = 1. Clearly we have not identified all such ases for which the period set is infinite. In fact, the family {3 m : m > 1}, covered y Theorem 3.1, has asymptotic density zero. A reasonale follow-up to Theorem 3.1 would e to examine the possiility of the existence of a family of ases, with nonzero asymptotic density, having infinite period sets.
INTEGERS: 14 (014) 7 4. Existence of Quasiperiodic Sequences The sequence from Conway s original RATS conjecture shows that Q 10, the quasiperiod set in ase 10, contains the element q =. In particular, the set Q 10 is not empty. In this section we will construct infinite families of ases for which Q 6= ;. We egin with a theorem of Cooper, Shattuck, and Gentges. Theorem 4.1 (Cooper and Shattuck, [7], Gentges, [4]). For any ase satisfying 1 (mod 18) or 10 (mod 18), with 10, we have Q, i.e., there exists a quasiperiodic RATS sequence of period. Sketch of Proof. In the case = 18k + 1 for k > 0, the element 13 3 4 4 5 1 6 16 7 48 8 64 (6k) m (6k + 1) (3 64k 3)/36 is quasiperiodic with quasiperiod if m is su ciently large. In the case = 18k + 10 for k 0, the element 13 3 4 4 5 1 6 16 7 48 8 64 (6k + 1) (3 64k )/4 (6k + ) 64k (6k + 3) m (6k + 4) (44 64k 8)/9 is quasiperiodic with quasiperiod if m is su ciently large. Theorem 4.1 shows that ases containing at least one quasiperiodic sequence make up a positive proportion of all integers. The following corollary makes this statement explicit. Corollary 4.. Let Then 1/9. = lim inf N!1 #{ apple N : Q 6= ;}. (4.1) N Along the lines of Theorem 4.1, Cooper and Shattuck [7] were also ale to conclude that for ases = ( q 1) +1, with q a prime or a ase- Fermat pseudoprime, we have q Q. In the next theorem, we expand on their result in two directions. First, we will remove the (pseudo)primality constraint on the quasiperiod q and second, we will exhiit an infinite family of ases for which q is a quasiperiod. Theorem 4.3. Let q >. If 1 (mod ( q 1) ) with > 1, then q Q. Proof. Suppose that = ( q 1) k + 1, where k 1. Let v = ( q 1)k. We will construct a quasiperiodic integer in ase with quasiperiod q and growing digit v. Consider an integer of the form 1 m1 m... (v 1) mv 1 v M+mv (v + 1) P v mi, (4.)
INTEGERS: 14 (014) 8 where M > 0, m i 0 for i = 1,..., v, and where the digit in oldface denotes the proposed growing digit. We will derive a system of equations and conditions that the exponents m i must satisfy in order for the integer (4.) to have quasiperiod q. We then will show that this system has a solution provided M is large enough. We see that the reverse-add portion of the RATS process produces. 1 m1 m... v M (v + 1) mv... (v + 1) m1 + (v + 1) m1 (v + 1) m... v M v mv... 1 m1 (v + ) m1 (v + 3) m... (v) M (v + 1) mv... (v + ) m1 (Note that no carries occur in the sum and every digit of the sum is nonzero.) Sorting the result gives the second term in the RATS sequence, (v + ) m1 (v + 3) m... (v 1) mv (v) M+mv 1 (v + 1) mv. (4.3) In order for the growing digit v of (4.3) to (approximately) line up against itself in the next sum, we need P v m i m v. Let M 1 e such that and assume M 1 m v = M 1 + m i (4.4) 0 for now. We repeat the aove process and get. (v + ) m1... (v) M+mv 1 M1 (v + 1) M1... (v + 1) m1 + (v + 1) m1... (v) M+mv 1 M1 (v) M1... (v + ) m1 (3v + 3) m1... (4v) M+mv 1 M1 (4v + 1) M1... (3v + 3) m1 Sorting the result gives the third term in the RATS sequence, (3v + 3) 4m1 (3v + 4) 4m... (4v 1) 4mv 3 (4v) M+mv 1+4mv M1 (4v + 1) M1. (4.5) In order for the growing digit 4v of (4.5) to line up against itself in the next sum, we need 4 P v 3 m i M 1. Let M e such that M 1 = M + 4 P v 3 m i and assume M 0 for now. Repeating this process a total of q 1 times gives that the qth term in the RATS sequences is (( q 1 1)v + q) q 1 m 1 (( q 1 1)v + q + 1) q 1 m...... ( q 1 v 1) q 1 m q 1 ( q 1 v) M+P q 1 i m v i P q Mi ( q 1 v + 1) Mq, (4.6) where the M j s are defined to e such that M j = M j+1 + j+1 v (j+) X m j (4.7)
INTEGERS: 14 (014) 9 for j = 1,..., q and are assumed to e nonnegative. We repeat the reverse-add portion of the RATS process one more time and get (( q 1 1)v + q) q 1 m 1... ( q 1 v) M 0... ( q 1 v + 1) q 1 m 1 + ( q 1 v + 1) q 1 m 1... ( q 1 v) M 0... (( q 1 1)v + q) q 1 m 1 1 (q + 1) q 1 m 1... v M 00... (q + 1) q 1 m 1 1 q where M 0 = M + P q 1 (i m v i M i ) and M 00 = M + P q 1 (i m v i M i ). (This is the first instance where carries occur; in fact, they occur at every position in the sum, and the result has all nonzero digits.) Sorting the result gives the (q + 1)th term in the RATS sequence, 1q(q + 1) q m 1 1 (q + ) q m... v M+mv 1+ +q m v q (M 1+ +M q 1) (v + 1) Mq 1. (4.8) We seek to choose the m i s such that the (q + 1)th term is identical to the first term of the RATS sequence, except that the exponent of v in (4.8) should e one larger than the exponent of v in (4.), the starting element. Comparing exponents in (4.) and (4.8) for all digits except v leads to the following set of equations: m 1 = 1, m = 0, m 3 = 0,. m q 1 = 0, m q = 1, m q+1 = q m 1 1 = q 1, m q+ = q m = 0, m q+3 = q m 3 = 0, (4.9) and. m q 1 = q m q 1 = 0, m q = q m q = q, m i = q m i q for q < i apple v 1, M q 1 = m i. (4.10) Note that these conditions automatically imply that the exponent of v in (4.8) must e one larger than that in (4.). This is due to the fact that the total numer
INTEGERS: 14 (014) 10 of digits is preserved y the RATS process in every step leading from (4.) to (4.8) except for the final step, where carries result in one additional digit. We see that (4.9) immediately gives nonnegative integer values of m i for 1 apple i apple v 1. What remains is to show that m v and the quantities M j defined y (4.7) and (4.4) are nonnegative integers. Furthermore, we need to ensure that the exponents of the growing digit in (4.) (4.6) and (4.8) are all nonnegative, in order for these expressions to make sense. For m v, comining (4.10), (4.4) and (4.7) for j = 1,..., q, we get that m v = M 1 + m i, 4m v = M 1 + 4 m i, 3 4m v = M + 4 m i + 4 m i, 3 8m v = M + 8 m i + 8 m i,. Xv q q m v = M q 1 + q m i + + q m i, q m v = q Xv m i + q m i + + q m i. In particular, we get that m v = 1 q 1L(q, v), where v q apple q L(q, v) = X m i + + 1 m i + m i. (4.11) Assume that m v is an integer. (We will estalish elow that this is indeed the case.) It immediately follows from (4.4) that M 1 is also an integer and that M 1 = m v = q 1 q+1 = q 1 m i Xv q apple q v m i + + X m i + q 1 1 m i + Xv q apple q m i 3 m i + + m i 1 m i + m i.
INTEGERS: 14 (014) 11 It follows that M 1 0, since q+1 q 1 0 and m i 0 for all 1 apple i apple v 1. Similarly, using (4.7) with j = 1, since M 1 is an integer, it follows that M is also a nonnegative integer. Continuing in this way, we see that the numers M j defined y (4.7) and (4.4) are indeed nonnegative integers. Taking M > 0 su ciently large, which we are free to choose, we get that every growing digit exponent in (4.) (4.6) and (4.8) is a nonnegative integer. Therefore, (4.) is a quasiperiodic element with quasiperiod q. To complete the proof, we will show that m v = 1 q 1L(q, v) is an integer, or equivalently, that L(q, v) 0 (mod q 1). (4.1) The remainder of the proof is roken down into cases, depending on the residue of v modulo q. We will only provide the proof for the case v 0 (mod q), as the other cases are quite similar. Suppose that v 0 (mod q). Then v = sq for some s 1. By (4.11), Xv q L(q, v) = apple q = (s 1)q X apple q Using (4.9) and (4.10), we see that sq X1 m i + + sq X m i + + 1 m i + sq X1 m i + m i (4.13) m i. m i = 1 + 1 + ( q 1) + q + q ( q 1) + q + + (s )q + (s )q ( q 1) = 1 + q + q (s 1)q + + Xs 1 = qj. j=0 For the remaining sums in the definition of L(q, v), notice that m i = 0 for v q+1 < i < v, so the sums inside the parenthesis of (4.13) are identical, with the exception of the first sum, P (s 1)q m i, which is missing the term (s )q ( q 1). Thus, we get that Xs 1 L(q, v) = (q q 1) j=0 qj (s )q ( q 1) X s 1 + Xs 1 = ( q (q 1) + 1) qj (s 1)q ( q 1). j=0 j=0 qj
INTEGERS: 14 (014) 1 Reducing modulo q 1, it follows that Xs 1 (1(q 1) + 1) 1 (mod q 1) j=0 sq (mod q 1) v (mod q 1) ( q 1)k (mod q 1) 0 (mod q 1). Therefore, (4.1) holds, as desired. Similar computations for the remaining cases exhaust all possiilities for the residue of v modulo q, which gives that m v is always a nonnegative integer. Theorem 4.3 shows that the asymptotic density of ases with nonempty quasiperiod sets is higher than the lower ound given y Corollary 4.. 1 Corollary 4.4. With as in (4.1), we have 9 + 1 1 1 0.19. 49 9 Proof. By Theorem 4.1, any ase 1, 10 (mod 18) is such that Q. By Theorem 4.3 for q = 3, any ase 1 (mod 49) is such that 3 Q. The set of ases that are either congruent to 1, 10 (mod 18) or 1 (mod 49) has asymptotic density (1/9) + (1/49)(1 1/9) y the Chinese remainder theorem. It follows immediately that (1/9) + (1/49)(1 1/9). Theorems 4.1 and 4.3 also imply that there exist ases such that Q contains any prescried finite set of distinct positive integers > 1. Corollary 4.5. For any set A = {q 1, q,..., q k } of distinct positive integers > 1, there exists a ase such that A Q. Proof. Let = 18( q1 1) ( q 1) ( q k 1) + 1. Then 1 (mod 18) and 1 (mod ( qi 1) ) for each i. The first condition implies that Q y Theorem 4.1 and the second condition implies that q i Q for any q i >, y Theorem 4.3. Corollary 4.5 shows that there are ases with aritrarily large quasiperiod sets. However, it is still unknown if there are ases with distinct quasiperiodic RATS sequences with the same quasiperiod. It is conjectured that this is never the case. Conjecture (Cooper and Shattuck, [7]). No ase has two or more distinct quasiperiodic RATS sequences with the same quasiperiod.
INTEGERS: 14 (014) 13 The definition for quasiperiodic sequences can sometimes seem rather narrow. However, all computational evidence compiled so far has yielded that the ehavior of RATS sequences always fit the descriptions of ultimately periodic or ultimately quasiperiodic sequences. We conjecture that this is always the case. Conjecture. In any ase, every RATS sequence is ultimately periodic or ultimately quasiperiodic. 5. Behavior of Growing Digits In this section we introduce a finite map, f, that descries the ehavior of growing digits in a quasiperiodic sequence in ase. We estalish properties of this map and use these results to prove nonexistence results for quasiperiodic sequences in certain ases. We egin with Conway s sequence 7! 13 m 4 4 7! 5 6 m 7 4 7! 13 m+1 4 4 7! 5 6 m+1 7 4 7!. (5.1) as a motivating example. By examining this sequence, it is clear that 3 is the growing digit associated with the quasiperiodic element 13 m 4 4. When performing the reverse-add portion of the RATS process, we see that the digit 3 primarily lines up against itself in the sum. This is the reason that 6 = 3 + 3 is the growing digit of the next term in the sequence, 5 6 m 7. Performing the reverse-add portion of the RATS process, again, we see that the digit 6 primarily lines up against itself in the sum. Due to a propagating carry, the most popular digit of the next iterate is again 3, since 3 6 + 6 + 1 (mod 10). In general, if k is the growing digit of some quasiperiodic element in ase, and k < /, then k is the growing digit of the next term in the RATS sequence. If k /, then a propagating carry makes k+1 the growing digit of the next term in the RATS sequence. Motivated y this idea, we make the following definition. Definition 5.1. Given a ase, let ( k if 0 < k < /, f (k) = k + 1 if / apple k <. For a nonnegative integer t, let f t (k) = f (f t 1 (k)) if t 1 and f 0 (k) = f (k). Proposition 5.. Suppose that is a ase with a quasiperiodic element n. If k is the growing digit of n, then f (k) is the growing digit of R (n). In particular, if n is quasiperiodic with quasiperiod q, then f q (k) = k.
INTEGERS: 14 (014) 14 Proof. The discussion aove Definition 5.1 shows that given an element n, if the digit k 6= ( 1)/ appears in n with a vastly higher count than all other digits comined, then f (k) gives the digit that will e the most popular in R (n). Applying this reasoning to the growing digit of a quasiperiodic element, we otain the desired result. The case where k = ( 1)/ is the most popular digit if n requires some additional care. In this case, the predominant digit of R (n) could e either 1 or 0, depending on the existence of carries to the right of the lock of digits k. However, since the RATS process discards zero digits, this would result in R (n) having fewer digits than n (for n large enough), a contradiction to the definition of a quasiperiodic element. Thus, if n is a quasiperiodic element with growing digit k = ( 1)/, then the growing digit of R (n) must e k = 1 = f (k). From the aove work, it follows that f q (k) = k y repeating the aove reasoning q times. Example 5.3. From (5.1), the growing digit of 13 m 4 4 is 3 and f 10(3) = f 10 (6) = 3. Definition 5.4. Given a ase and a digit 0 < k <, we say that k is a fixed point of f if there is a t > 0 such that f t (k) = k. If t is the least integer with this property, we say that the order of k is t. Corollary 5.5. The quasiperiod of a quasiperiodic element is a multiple of the order of the corresponding growing digit with respect to the map f. Proof. Suppose that k is the growing digit of a quasiperiodic element with quasiperiod q. By Proposition 5., f q (k) = k; that is, k is a fixed point of f. Suppose that the order of k is t. By the Euclidean algorithm, there are nonnegative integers s and r such that q = st + r with 0 apple r < t. Then k = f q st+r (k) = f (k) = f r st (f (k)) = f r (k). If r > 0, then we have a contradiction to the assumed order of k. Hence r = 0 and t q. Therefore, the quasiperiod of a quasiperiodic element is a multiple of the order of the corresponding digit. The aove results show that the growing digit of a quasiperiodic element is a fixed point of the corresponding map f. However, the converse is not true in general. The element 1 is always a fixed point (of order 1), ut is never a growing digit of a quasiperiodic element, as the following results show. Lemma 5.6. For any ase, k is a fixed point of order 1 if and only if k = 1. Proof. ()) If k is a fixed point of order 1, then either k = f (k) = k or k = f (k) = k + 1. The first case gives that k = 0, which is a contradiction, since k > 0, and the second case gives that k = 1. (() A simple calculation shows that, since > 1, 1 /, so f ( 1) = ( 1) + 1 = 1.
INTEGERS: 14 (014) 15 Lemma 5.7. For any ase, if a quasiperiodic element has growing digit k, then k 6= 1. We leave the proof of Lemma 5.7 to the reader. Theorem 5.8. For any ase, we have 1 / Q. In other words, in any ase, there are no quasiperiodic elements with quasiperiod 1. Proof. Some simple computations show that all RATS sequences in ase are periodic, giving that Q = ;. So the assertion holds for =. Suppose > and that there is a quasiperiodic element n with quasiperiod 1 and growing digit k. By Corollary 5.5, k must e a fixed point of f of order 1. By Lemma 5.6, k must e 1, a contradiction to Lemma 5.7. Theorem 5.8 serves as an analogue of the fact that 1 / P (see [3, Theorem 1] for the case = 10) for any ase >. The result is slightly di erent in that ase does not play a special role. Continuing in this same vein, we investigate the existence (or nonexistence) of fixed points of f of a given order for general ases. Lemma 5.9. For any ase >, there is a fixed point of f of order if and only if 1 (mod 3). Proof. ()). Suppose that k is a fixed point of f of order. We consider four cases. Case 1. 0 < k, f (k) < /. Then k = f (k) = 4k, which implies that k = 0, a contradiction, since k > 0. So this case cannot happen. Case. 0 < k < / apple f (k) <. Then k = f (k) = (k) + 1 = 4k + 1. This equation is equivalent to 3k = 1, which implies that 1 (mod 3). Case 3. 0 < f (k) < / apple k <. Then k = f (k) = (k +1 ) = 4k +. This equation is equivalent to 3k = ( 1), which implies that 1 (mod 3). Case 4. / apple k, f (k) <. Then k = f (k) = (k+1 )+1 = 4k+3 3. This equation is equivalent to k = 1, a contradiction, since, y Lemma 5.7, k < 1. So this case cannot happen. Having exhausted all possiilities, we see that we must have that 1 (mod 3). (() Suppose that 1 (mod 3). Let k = ( 1)/3. Then, since 0 < k < /, f (k) = k = ( 1)/3. For 4 we have ( 1)/3 > /, so f (k) = 4( 1)/3 + 1 = ( 1)/3 = k. Therefore, k is a fixed point of f of order. Corollary 5.10. If Q, for some ase >, then 1 (mod 3). Proof. The result follows from Lemma 5.9 and Proposition 5..
INTEGERS: 14 (014) 16 Corollary 5.10 shows that the density of ases containing a quasiperiodic sequence with quasiperiod is at most 1/3. On the other hand, Theorem 4.1 shows that quasiperiodic sequences with quasiperiod exist in any ase satisfying 1 (mod 18) or 10 (mod 18). Hence this density is at least 1/9. The following lemma shows that, for any given order t, there is a such that f has a fixed point of order t. It should not e surprising that this is the case, as Theorems 4.1 and 4.3 show that quasiperiodic sequences can e constructed for any desired quasiperiod. What is interesting aout the lemma is that the ases in the statement do not match the form of those in Theorems 4.1 and 4.3. Lemma 5.11. If = t, then 1 is a fixed point of order t. Proof. The lemma follows quickly from noticing that, for = t, we have f s (1) = s for 1 apple s < t, and f t(1) = t + 1 = 1. Results like Corollary 5.10 and Lemma 5.11 can e used to narrow down ases that allow the existence of quasiperiodic RATS sequences of a specified quasiperiod. So if we are interested in finding quasiperiodic RATS sequences, y Proposition 5., it seems natural to search for ases that have many fixed points of f. Definition 5.1. For a ase, let S = {k : 0 < k <, f t (k) = k for some t}. In other words, S is the set of all fixed points of f. The following result gives an exact description of ases for which f has only one fixed point. Notice that y Lemma 5.6, 1 is always a fixed point of f, so we always have S 1. Theorem 5.13. For any ase >, we have S = 1 if and only if = t + 1 for some positive integer t. Proof. (() Assume that = t + 1 for some positive integer t. If t = 1, then = 1 + 1 = 3, and direct verification shows that in this case, k = is the only fixed point of f. Hence, S = 1. Suppose now that t. Let {1,,..., 1} = S 1 i=0 A i, where A i = {k : 0 < k <, i k}. In other words, we partition the set {1,,..., 1} according to the highest power of dividing each digit. Notice that all ut finitely many A i s are empty; in particular, since 1 = t, we have {1,,..., 1} = S t i=0 A i. The set A 0 represents all odd numers in {1,..., t }. Let m A 0. If m < /, then f (m) = m. If m /, then f (m) = m + 1 = m t. Since f (m) and 4 - f (m) in either case, we get that f (A 0 ) A 1. Now let m 0 A 1. Then m 0 = m 00 with - m 00. In particular, m 00 A 0. Since m 00 = m 0 / < /, f (m 00 ) = m 00 = m 0. So we get that A 1 f (A 0 ). Therefore, we have f (A 0 ) = A 1. Continuing in this fashion, we get that f (A i ) = A i+1 for 0 apple i < t. Notice that A t = { t }, so for any starting digit k, there is a t 0 such that f t0(k) = t. Since, y
INTEGERS: 14 (014) 17 Lemma 5.6, 1 = t is a fixed point of order one, there cannot e any other fixed points. This is equivalent to the statement that S = 1. ()) For the converse direction, we first note that if m is such that 0 < m <, then, y Definition 5.1, 8 m ><, m + 1 if m is even and is odd, f 1 ; if m is odd and is odd, (m) = (5.) m if m is even and is even, >: m + 1 if m is odd and is even. We will use this fact extensively in this portion of the proof. Assume now that S = 1. By Lemma 5.6, we know that 1 is a fixed point of f, so S = { 1}. By the pigeonhole principle, for every 0 < k < 1 there is a t > 0 such that f t(k) is a fixed point of f. By assumption, 1 is the only fixed point, so f t 1 (k) = 1. In particular, f ( 1) > 1, which, y (5.), implies 1 is even and c( 1) f 1 ( 1) = : 1 apple c apple, where c runs through integers in the given range. Thus must e of the form 1 = i0 0 with i 0 > 0 and - 0. If i 0 > 1, then using (5.) again, we get c( 1) f ( 1) = : 1 apple c apple 4. 4 Continuing in this way, we see that, for 1 apple i apple i 0, In particular, for 1 apple i apple i 0, f i ( 1) = c( 1) i : 1 apple c apple i. f i ( 1) = i. (5.3) Notice that for i < i 0, the elements in f i ( 1) are all even, ut in f i0 ( 1), half of the elements are odd. This implies, y (5.), that f (i0+1) ( 1) = f i0 ( 1) and hence f (i0+t) ( 1) = f i0 ( 1) for all t > 0. Hence {0 < k < : k odd} f i0 ( 1). Since, y (5.3), f i0 ( 1) contains i0 elements, and, y the aove oservation, half of these elements are odd, this implies that #{0 < k < : k odd} apple i0 1 and hence ( 1)/ = i0 1 0 apple i0 1. Thus 0 = 1 and = i0 0 +1 = i0 +1, so is of the desired form. In the case i 0 = 1, a similar argument shows that = 1 + 1 = 3, which is also of the desired form.
INTEGERS: 14 (014) 18 Corollary 5.14. For any ase = t + 1, where t 0, we have Q = ;. In other words, there are no quasiperiodic RATS sequences in ase = t + 1. Proof. Suppose that there is a quasiperiodic element n in ase = t + 1. By Proposition 5., the growing digit of n must e fixed point of f. Theorem 5.13 shows that if = t + 1, then the only fixed point of f is 1. It follows that 1 must e the growing digit of n, a contradiction to Lemma 5.7. Therefore, there are no quasiperiodic RATS sequences in ase. Corollary 5.14 shows that there is an infinite family of ases for which no quasiperiodic RATS sequences exist. Unfortunately, this is not enough to conclude that all sequences in such ases are ultimately periodic. For example, McMullen s conjecture [5], that all RATS sequences are ultimately periodic for ases 4 apple < 10, is still open in the cases = 5 = + 1 and = 9 = 3 + 1. References [1] John H. Conway, Play it again and again..., Quantum, 63:30 31, Novemer/Decemer 1990. [] Curtis Cooper and Roert E. Kennedy, Base 10 RATS Cycles and Aritrarily Long Base 10 RATS Cycles, Applications of Fionacci numers, Vol. 8 (Rochester, NY, 1998), pages 83 93. Kluwer Acad. Pul., Dordrecht, 1999. [3] Curtis Cooper and Roert E. Kennedy, On Conway s RATS, Mathematics in College (1998), 8 35. Availale at http://www.math-cs.ucmo.edu/~curtisc/articles/rats.pdf. [4] Karen Gentges, On Conway s RATS, Master s thesis, Central Missouri State University, 1998. [5] Richard K. Guy, Conway s RATS and other reversals, Amer. Math. Monthly, 96:45 48, 1989. [6] Richard K. Guy, Unsolved Prolems in Numer Theory, Prolem Books in Mathematics. Springer-Verlag, New York, third edition, 004. [7] Steven Shattuck and Curtis Cooper, Divergent RATS sequences, Fionacci Quart., 39():101 106, 001.