On generalizing happy numbers to fractional base number systems
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1 On generalizing hay numbers to fractional base number systems Enriue Treviño, Mikita Zhylinski October 17, 018 Abstract Let n be a ositive integer and S (n) be the sum of the suares of its digits. It is well known that if you iterate S enough times you reach a cycle. When the cycle is {1}, then n is called hay. The notion of hay numbers has been generalized to different bases, different owers and even negative bases. In this article we consider generalizations to fractional number bases. Let S e, (n) be the sum of the e-th owers of the digits of n base /. We rove that iterating S e, we show that the number of times the iteration is reuired to reach a cycle can be arbitrarily large. reaches a cycle and, if = or e = 1, 1 Introduction Let n be a ositive integer and S (n) be the sum of the suares of its digits. It is well known (for a comlete roof look at [6]) that if you iterate S enough times you either reach 1 or you reach the cycle When the iteration reaches 1, we say n is hay. A natural generalization is to relace sum of suares of digits, with sums of e-owers for integers e 1. Let S e (n) be the sum of e-th owers of n. If we iterate extensively and reach 1, we say n is an e-ower hay number. The study of which cycles can be reached for 1 e 10 has been done by Grundman and Teele in [3]. Another generalization considered in [3] is to change the base of the number system. Let S e,b (n) be the sum of the e-th owers of the digits of n in base b. The cycles that can be reached by iterating S e,b has been calculated for e {, 3} and bases b 10. The techniues in [3] can easily find the cycles for other choices of e and b. Another generalization is to allow the base b to be a negative number. It turns out that for a ositive integer n, there is a uniue set of digits 0 a i b 1 such that n = r a ib i. Grundman and Harris, in [5], find the cycles reached for b 10 and e =. The authors also study when can you have consecutive elements in an arithmetic rogression be b-hay (it means that the iterations of S,b reach 1 eventually), generalizing work of El-Sedy and Siksek [] and the work of Grundman and Teele [4]. In [1], Bland et al. addressed a series of uestions regarding a generalization of hay numbers to fractional bases. Any number n has a uniue reresentation modulo / when > and gcd(, ) = 1. Namely, there exist 0 a i < such that n = ( ) i a i. 1
2 For our notation, we will say n = a r a r 1 a 1 a 0. Let S e, (n) be the sum of the e-th owers of the digits of n when written in fractional base /, i.e., S e, (n) = a e i. In [1], the authors studied the case when e = and roved that there are no hay numbers greater than 1 for any fractional base. They mainly study the fractional base 3/, finding the ossible cycles that S, 3 can reach. They end the aer with several uestions. The three we will focus on in this aer are the following: 1. Can the cycles for different e-th owers be found for / = 3/?. Can the cycles for different / be found if we restrict e =? 3. Are there ositive integers n of arbitrary height? For the last uestion, we need to exlain what the height of a number is. The height is the number of iterations of S e, that are reuired to reach a cycle. In the case of ositive integer bases, that there are numbers with arbitrary height is relatively simle to rove because you can find an n such that S e,b (n) = k for any ositive integer k by having n be a number with k 1 s in its base b exansion. For examle, let a 1 = 10, then a 1 has height 1 since S (10) = 1. Let a = } 11 {{ 1}. 10 Since S (a ) = 10, then a has height. Let a n = } 11 {{ 1}. a n 1 Then a n has height n. This is a simle rocess to create a seuence of numbers with larger and larger heights by attaching the aroriate number of 1 s to a number. The roblem with fractional bases is that not every choice of digits leads to an integer. For examle 11 3 is not an integer, since Z. We answer the three uestions with a coule of theorems. The first theorem answers two of the uestions. Theorem 1. Let > be ositive integers with gcd(, ) = 1, and let e be a ositive integer. Then, for every ositive integer n, the reeated iterations of the function S e, on n will eventually reach a cycle. In articular, the ossible cycles reached for 1 e 1, / = 3/ can be found in Table, answering the first uestion. Also, the ossible cycles reached for e {, 3, 4} and / {5/, 5/3, 5/4, 7/} are in Table 3, answering the second uestion. The second theorem answers the third uestion for a secial class of fractional bases that includes 3/, and for all fractional bases when e = 1. Theorem. Let > be ositive integers with gcd(, ) = 1, and let e and H be ositive integers. If = or e = 1, then there exists an integer n such that the height of n is H. In section, we will resent useful background on fractional base number systems. In section 3, we rove Theorem 1. Finally, in section 4, we rove Theorem.
3 Fractional base number systems As mentioned in the introduction, for any / with gcd(, ) = 1 and >, for every ositive integer n, there exist fractional digits a 0, a 1,..., a r satisfying 0 a i < such that n = ( ) i a i. We will use the following notation to denote that a i are the fractional digits of n. n = a k a k 1 a k... a a 1 a 0 3. For examle base 3/ uses numbers 0, 1, as digits. Table 1 is a table of numbers in base 3/: n n in base 3/ n n in base 3/ Table 1: The first 1 non-negative integers in the 3/ base number system It is easy to find n given its exansion in base /, but going the other way around is a little harder. Suose we have the number n and we want to find its fractional digits base /. Let n = a k a k 1 a 1 a 0. Then ( ) n a 0 = a k a k 1 a 1 The left side is an integer, so the right side is also an integer. Since gcd(, ) = 1, then a k a k 1 a 1, and so (n a 0 ). Therefore n a 0 mod. There is a uniue a 0 in {0, 1,,, 1} that is congruent to n modulo. But we also have ( ) a k a k 1 a 1 = (n a 0 ). We reeat the rocess and we can say that ( ) n (n a 0 ) a 1 mod. Therefore, we can find a 1. We can reeat this rocess until we reach 0 and find all of the digits of n. We can summarize the algorithm to translate numbers into the fractional base as follows: 1. n 0 = n (mod ). (. n = (n n 0 ) ). 3
4 3. Reeat stes 1 and, until n is zero. As an examle, suose we want to find the digits of 1 in base 3/. First we have 1 0 mod 3, so a 0 = 0. Then we calculate (1 0) = 8. We find 8 mod 3, so a 3 1 =. Then we find (8 ) = 4 and 4 1 mod 3, so a 3 = 1. Then we find (4 1)(/3) = and mod 3, so a 3 =. Since the next ste yields 0, we ve found that 1 = The cycles formed when iterating S e, 3 Numbers n > 1 in fractional base number systems cannot be hay. Indeed suose that n is hay, then S m e, (n) = 1 for some minimal number m. But then k = S m 1 (n) must satisfy e, that the sum of the e-th owers of its digits is 1. Therefore the fractional base exansion of k is But that means k = (/)r for some integer r. This number is not an integer unless r = 0, which would imly k = 1, but we assumed k > 1. While hainess is imossible, we can still search for what cycles the iterations can land on. For us to be able to rove that the determination of cycles is comlete, we need to first rove the following lemma. Lemma 1. Let / satisfy > and gcd(, ) = 1, and let e be a ositive integer. Then, there exists an n such that S e, (n ) n, and S e, (n) < n for all n > n. The values of n for different values of e and / = 3/ can be found in the last column of Table. The values of n for e {, 3, 4} and / {5/, 5/3, 5/4, 7/} are in Table 3. Proof. Let n be a ositive integer. Then so r log (n). But then n = a r a r 1 a 1 a 0 S e, (n) = a e i < Since e is a constant, then eventually = ( a i ) i ( ) r a r ( ) r, e = (r + 1) e (log (n) + 1)e. n > (log (n) + 1)e > S e, (n). (1) Indeed, one can confirm with L Hoital s rule that n/(c log(n)) as n. Therefore, there is a maximum n such that n < S e, (n ). To calculate the recise value of n, we use a comuter to find an N for which (1) is satisfied. Then we evaluate S e, (n) for all n N and record which one is the largest satisfying that n < S e, (n). Proof of Theorem 1. Let S(n) = S e, (n). Let n be as in Lemma 1. Now, for each n n, comute n, S(n), S(S(n)),... until it cycles. The rocess terminates because S(n) < n for 4
5 e Cycles n 1 [1], [] [1], [5, 8, 9] 8 3 [1], [9], [10], [17, 18] 3 4 [1], [51], [5] 77 5 [1], [131], [98, 99] [1], [197, 60, 387, 33, 63, 450], [34, 131, 59] [1], [771, 516, 643, 518] [1], [1539, 775, 184], [187, 1794, 1796, 05], [103], [1033] 73 9 [1], [566], [565] [1], [1047] [1], [1434, 16388, 14344], [14341], [14340] [1], [8678], [8677] 6933 Table : Cycles reached when iterating S e, 3, and the value of n for different values of e. all n > n. For n > n, eventually S k (n) n. Therefore, the cycle n reaches is one that was already comuted. Therefore, we need only find the cycles for n n. The outcome of erforming these calculations for different values of e and / = 3/ is recorded in Table. The outcome of erforming these calculations on e {, 3, 4} with / {5/, 5/3, 5/4, 7/} is recorded in 3. 4 Arbitrary Heights in fractional base number systems The key to our roof of Theorem is showing that there is n such that S e, (n) = k for every large enough k. The following lemma handles the case when =. Lemma. Let e 1 and > be an odd ositive integer. For every integer k e, there exists n, such that S e, (n) = k. Proof. We will rove it by induction. To show that it is ossible for k = e, consider the number. is, therefore S e, () = e. Now suose there exists an even m such that S e, (m) = k for some k e. Let m = b c where b 1 and c is odd. Write m in base / as Then m = a r a r 1 a 1 a 0. ( ) b m + 1 = ar a r 1 a 1 a }{{} 1, b 1 where there are b 1 zero digits. Since m = b c, then (/) b m + 1 is even. Furthermore, since it has the same digits as before with b 1 zeroes added and one 1 added, then the sum of the e-th owers of the digits is k + 1. The following lemma handles the e = 1 case. 5
6 / e = e = 3 e = 4 5 / [16, 6, 5, 4], [3, 4, 9]; [65], [163, 190, 73, 118, 64], [81], [80], [66], [17]; [371, 76, 75, 74], [355, 130, 113], [195, 353]; n = 39 n = 39 n = / 3 [34, 50], [5], [6], [59], [3], [11], [10]; [100, 38, 64, 10, 46], [101, 39], [17, 107, 73, 135], [16], [193], [190, 166, 18], [199, 37]; [77, 804, 454, 788, 950, 658, 934, 116, 108, 10, 868, 936, 390], [107, 1137, 115], [11, 994], [199], [101], [100]; n = 59 n = 84 n = / 4 [66, 55], [50], [58, 75, 49, 56, 67], [74, 83], [51]; [311, 51, 47, 31, 371], [361], [417], [ 374], [360], [314], [44, 418, 436, 7, 38, 364]; [1786, 1880, 1403, 1594, 1659, 011, 075, 1579, 057, 1947, 1688, 19, 1641, 1676, 1946, 1673, 1851, 159, 1419, 1974, 058, 01, 090]; n = 74 n = 464 n = / [5, 5], [97]; n = 97 [341, 591, 376, 143, 187, 16, 35, 5, 80, 44, 469, 63, 18, 44, 141, 161, 197, 73, 307, 467, 377, 34, 18, 91], [35], [88, 343, 9, 16, 7], [36], [189], [190], [468]; n = 615 [914, 065, 1953, 1538, 819, 690, 10, 1507, 1491, 610, 1856, 1348, 1666, 59, 1808, 659, 3136, 184], [1634, 1731, 994], [371, 34, 1313], [130, 354, 89, 1938, 365, 930, 1474, 1570], [451, 195, 177, 1554, 179, 513, 034, 530]; n = 5417 Table 3: Cycles reached when iterating S e, 3, and the value of n for different values of e and /. Lemma 3. Let / > 1 be written in lowest terms. For every integer k S 1, (), there exists n, such that S 1, (n) = k. Proof. We will rove this by strong induction. Using the exansion for we get that the statement is true for k = S 1, () = k 0. Now suose there is an integer m that is a multile of such that S 1, (m) = k for some k k 0, and that for each s [k 0, k], there is an integer m s such that S 1, (m s) = s. Write m as m = α b for some α 1 and b relatively rime to. Suose m = a t a t 1 a 0. Then l = ( ) α m = a t a }{{}. α We know l 0 mod. Let r be the smallest ositive integer such that l + r 0 mod. Then 1 r 1 < 1. But then l + r = a t a 0 0 } {{ 0} r, α because r < 1. This imlies that the digital sum base / of the numbers l + 1, l +,..., l + r are k + 1, k +,..., k + r, resectively. Now l + r is a multile of with 6
7 S 1, (l + r) = k + r k + 1, and we have that for all k 0 s k + r, there exists m s such that S 1, (m s) = s. Using these two Lemmas, we can now resent the roof of Theorem. Proof of Theorem. Let n be as defined in Lemma 1 and let k 0 = S 1, (). Since the cycles that are reached by iterations of S e, are finite and there s finitely many of them, there is a largest integer K with height 0. Let n be an integer greater than M = max{n, e, k 0, K}. Since n > K, then n has some height h > 0. Then, S e, (n) has height h 1, S e, (n) has height h, and so on. Therefore, every height i h can be reached. Since n > e, by Lemma, if =, then there exists t such that S e, (t) = n. Since n > k 0, by Lemma 3, if e = 1, then there exists t such that S 1, (t) = n. Therefore, in either case ( = or e = 1), there exists an integer t such that S e, (t) = n. But t > n, which imlies that n = S e, (t) < t. Therefore t > n and t has height h + 1. The roof follows by induction. References [1] Andre Bland, Zoe Cramer, Phili de Castro, Desiree Domini, Tom Edgar, Devon Johnson, Steven Klee, Joseh Koblitz, and Ranjani Sundaresan, Hainess is integral but not rational, Math Horiz. 5 (017), no. 1, MR [] Esam El-Sedy and Samir Siksek, On hay numbers, Rocky Mountain J. Math. 30 (000), no., MR [3] H. G. Grundman and E. A. Teele, Generalized hay numbers, Fibonacci Quart. 39 (001), no. 5, MR [4], Seuences of consecutive hay numbers, Rocky Mountain J. Math. 37 (007), no. 6, MR [5] Helen G. Grundman and Pamela E. Harris, Seuences of consecutive hay numbers in negative bases, The Fibonacci Quarterly 56, no. 3. [6] Ross Honsberger, Ingenuity in mathematics, New Mathematical Library, vol. 3, Random House, Inc., New York, MR
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