Let $f\colon\mathbb{N}\rightarrow\mathbb{N}_0$ be a multiplicative arithmetic
function such that for all primes $p$ and positive integers $\alpha$,
$f(p^{\alpha})<p^{\alpha}$ and $f(p)\vert f(p^{\alpha})$. Suppose also that any
prime that divides $f(p^{\alpha})$ also divides $pf(p)$. Define $f(0)=0$, and
let $H(n)=\displaystyle{\lim_{m\rightarrow\infty}f^m(n)}$, where $f^m$ denotes
the $m^{th}$ iterate of $f$. We prove that the function $H$ is completely
multiplicative.Comment: 5 pages, 0 figure

Defant, Colin

Miskolc Mathematical Notes

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Miskolc Mathematical Notes HU e-ISSN 1787-2413Vol. 16 (2015), No. 2, pp. 763–767 DOI: 10.18514/MMN.2015.1448A NOTE ABOUT ITERATED ARITHMETIC FUNCTIONSCOLIN DEFANTReceived 01 December, 2014Abstract. Let f WN!N0 be a multiplicative arithmetic function such that for all primes p andpositive integers ˛, f .p˛/ < p˛ and f .p/jf .p˛/. Suppose also that any prime that dividesf .p˛/ also divides pf .p/. Define f .0/D 0, and let H.n/D limm!1fm.n/, where f m denotesthe mth iterate of f . We prove that the function H is completely multiplicative.2010 Mathematics Subject Classification: 11A25; 11N64Keywords: arithmetic function, iteration, multiplicative, Euler totient, Schemmel totient1. INTRODUCTIONThe study of iterated arithmetic functions, especially functions related to the Eulertotient function ', has burgeoned over the past century. In 1943, H. Shapiro’s monu-mental work on a function C.n/, which counts the number of iterations of ' neededfor n to reach 2, paved the way for subsequent number-theoretic research [5]. In thispaper, we study a problem concerning the limiting behavior of iterations of functionsrelated to the Euler totient function.Throughout this paper, we letN,N0, and P denote the set of positive integers, theset of nonnegative integers, and the set of prime numbers, respectively. We will letf WN0!N0 be a multiplicative arithmetic function which has the following proper-ties for all primes p and positive integers ˛.I. f .p˛/ < p˛.II. f .p/jf .p˛/.III. If q is prime and qjf .p˛/, then qjpf .p/.IV. f .0/D 0.First, note that property IV does not effectively restrict the choice of f . Indeed, wemay let f be any multiplicative arithmetic function that satisfies properties I, II, andIII and then simply define f .0/D 0. One class of arithmetic functions which satisfyI, II, and III are the Schemmel totient functions. For each positive integer r , thec 2015 Miskolc University Press764 COLIN DEFANTSchemmel totient function Sr is a multiplicative arithmetic function which satisfiesSr.p˛/D(0; if p  r Ip˛ 1.p  r/; if p > rfor all primes p and positive integers ˛ [4]. These interesting generalizations of theEuler totient function have applications in the study of magic squares [3, page 184]and in the enumeration of cliques in certain graphs [1].Because f is multiplicative, properties I and II of f are equivalent to the followingproperties, which we will later reference.A. For all integers n > 1, f .n/ < n.B. If p is a prime divisor of a positive integer n, then f .p/jf .n/.Let f 0.n/ D n and f kC1.n/ D f .f k.n// for all nonnegative integers k and n.Observe that, for any n 2 N, f n.n/ 2 f0;1g. Furthermore, f n.n/ D limm!1fm.n/,so we will define H.n/D limm!1fm.n/. The author has shown that the functionH WN! f0;1g is completely multiplicative for the case in which f is a Schemmeltotient function [2]. Our purpose is to prove that H is completely multiplicative forany choice of a multiplicative arithmetic function f that satisfies properties I, II, III,and IV. To help do so, we define the following sets.P D fp 2 P WH.p/D 1gQD fq 2 P WH.q/D 0gS D fn 2NWq − n8q 2QgWe define T to be the unique set of positive integers defined by the following criteria: 1 2 T . If p is prime, then p 2 T if and only if f .p/ 2 T . If x is composite, then x 2 T if and only if there exist x1;x2 2 T such thatx1;x2 > 1 and x1x2 D x.Note that T is a set of positive integers; in particular, 0 62 T . We may now establisha couple of lemmas that should make the proof of the desired theorem relativelypainless.Lemma 1. Let k 2N. If all the prime divisors of k are in T , then all the positivedivisors of k (including k) are in T . Conversely, if k 2 T , then every positive divisorof k is an element of T .Proof. First, suppose that all the prime divisors of k are in T , and let d be apositive divisor of k. Then all the prime divisors of d are in T . Let d DrYiD1p˛iibe the canonical prime factorization of d . As p1 2 T , the third defining criterion ofT tells us that p21 2 T . Then, by the same token, p31 2 T . Eventually, we find thatITERATED ARITHMETIC FUNCTIONS 765p˛11 2 T . As p˛11 ;p2 2 T , we have p˛11 p2 2 T . Repeatedly using the third criterion,we can keep multiplying by primes until we find that d 2 T . This completes the firstpart of the proof. Now we will prove that if k 2 T , then every positive divisor ofk is an element of T . The proof is trivial if k is prime, so suppose k is composite.We will induct on ˝.k/, the number of prime divisors (counting multiplicities) ofk. If ˝.k/ D 2, then, by the third defining criterion of T , the prime divisors of kmust be elements of T . Therefore, if ˝.k/ D 2, we are done. Now, suppose theresult holds whenever ˝.k/  h, where h > 1 is an integer. Consider the case inwhich ˝.k/D hC 1. By the third defining criterion of T , we can write k D k1k2,where 1 < k1;k2 < k and k1;k2 2 T . By the induction hypothesis, all of the positivedivisors of k1 and all of the positive divisors of k2 are in T . Therefore, all of theprime divisors of k are in T . By the first part of the proof, we conclude that all of thepositive divisors of k are in T . Lemma 2. The sets S and T are equal.Proof. First, note that 1 2 S \T . Let m > 1 be an integer such that, for all k 2f1;2; : : : ;m  1g, either k 2 S \T or k 62 S [T . We will show that m 2 S if andonly if m 2 T . First, we must show that if k 2 f1;2; : : : ;m  1g, then k 2 S if andonly if f .k/ 2 S . Suppose, by way of contradiction, that f .k/ 2 S and k 62 S . Ask 62 S , we have that k > 1 and k 62 T . Lemma 1 then guarantees that there existsa prime q such that qjk and q 62 T . As q 62 T , the second defining criterion of Timplies that f .q/ 62 T . As f .k/ 2 S , f .k/ ¤ 0. By property B of f , f .q/jf .k/,so f .q/ ¤ 0. Therefore, f .q/ 2 f1;2; : : : ;m  1g, and f .q/ 62 T . By the inductionhypothesis, f .q/ 62 S . Therefore, there exists some q0 2Q such that q0jf .q/. Thus,q0jf .q/jf .k/, which contradicts the assumption that f .k/ 2 S .Conversely, suppose that f .k/ 62 S and k 2 S . The fact that f .k/ 62 S implies thatk > 1, and the fact that k 2 S implies (by the induction hypothesis) that k 2 T . ByLemma 1, all prime divisors of k are elements of T . The second criterion defining Tthen implies that f .p/ 2 T for all prime divisors p of k. Using Lemma 1 again, weconclude that, for any prime divisor p of k, all prime divisors of f .p/ are in T . Byproperty III of f , all prime divisors of f .k/ are elements of T . Therefore, Lemma1 guarantees that f .k/ 2 T . From property A of f and the fact that 0 62 T , we seethat f .k/ 2 f1;2; : : : ;m  1g. The induction hypothesis then implies that f .k/ 2 S ,which is a contradiction. Thus, we have established that if k 2 f1;2; : : : ;m 1g, thenk 2 S if and only if f .k/ 2 S .We are now ready to establish that m 2 S if and only if m 2 T . Assume, first, thatm is prime. By the second criterion defining T ,m2T if and only if f .m/2T . By theinduction hypothesis and property A of f , f .m/ 2 T if and only if f .m/ 2 S . Fromthe preceding argument, we see that f .m/ 2 S if and only if f 2.m/ 2 S . Similarly,f 2.m/ 2 S if and only if f 3.m/ 2 S . Continuing this pattern, we eventually find thatm 2 T if and only if f m.m/ 2 S . Observe that f m.m/DH.m/ and that 0 62 S and1 2 S . Hence,m 2 T if and only ifH.m/D 1. Becausem is prime,H.m/D 1 if and766 COLIN DEFANTonly if m 62Q. Finally, it follows from the definition of S that m 62Q if and only ifm 2 S . This completes the proof of the case in which m is prime.Assume, now, that m is composite. By Lemma 1, m 2 T if and only if all theprime divisors ofm are in T . Becausem is composite, all the prime divisors ofm areelements of f1;2; : : : ;m  1g. Therefore, by the induction hypothesis, all the primedivisors of m are in T if and only if all the prime divisors of m are in S . It should beclear from the definition of S that all the prime divisors of m are in S if and only ifm 2 S . Hence, m 2 T if and only if m 2 S . We may now use the sets S and T interchangeably. In addition, part of the aboveproof gives rise to the following corollary.Corollary 1. Let k;r 2N. Then f r.k/ 2 S if and only if k 2 S .Proof. The proof follows from the argument in the above proof that f .k/ 2 S ifand only if k 2 S whenever k 2 f1;2; : : : ;m 1g. As we now know that we can makem as large as we need, it follows that f .k/ 2 S if and only if k 2 S . Repeating thisargument, we see that f 2.k/ 2 S if and only if f .k/ 2 S . The proof then followsfrom repeated application of the same argument. Corollary 2. For any positive integer k, H.k/D 1 if and only if k 2 S .Proof. It is clear that H.k/ D 1 if and only if H.k/ 2 S . Therefore, the prooffollows immediately from setting r D k in Corollary 1. Notice that Corollary 2, Lemma 2, and the defining criteria of T provide a simplerecursive means of constructing the set of positive integers x that satisfy H.x/D 1.We also have the following theorem.Theorem 1. The function H WN! f0;1g is completely multiplicative.Proof. Corollary 2 tells us that H is the characteristic function of the set S ofpositive integers that are not divisible by primes inQ. If x;y 2N, then it is clear thatxy 2 S if and only if x 2 S and y 2 S . The proof follows immediately. REFERENCES[1] C. Defant, “Unitary Cayley graphs of Dedekind domain quotients,” Submitted, 2014.[2] C. Defant, “On arithmetic functions related to iterates of the Schemmel totient functions,” Journalof Integer Sequences, vol. 18, no. 2, p. 3, 2015.[3] J. Sa´ndor and B. Crstici, Handbook of number theory II. Springer Science & Business Media,2004, vol. 2.[4] V. Schemmel, “U¨ber relative Primzahlen,” Journal fu¨r die reine und angewandte Mathematik, pp.191–192, 1869.[5] H. Shapiro, “An arithmetic function arising from the  function,” American Mathematical Monthly,pp. 18–30, 1943, doi: 10.2307/2303988.ITERATED ARITHMETIC FUNCTIONS 767Author’s addressColin DefantUniversity of Florida, Department of Mathematics, 1400 Stadium Rd., 32611 Gainesville, FL,United StatesE-mail address: cdefant@ufl.edu

A Note about Iterated Arithmetic Functions

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