On Arithmetic Functions Related to Iterates of the Schemmel Totient Functions
aa r X i v : . [ m a t h . N T ] J un On Arithmetic Functions Related toIterates of the Schemmel TotientFunctions
Colin DefantDepartment of MathematicsUniversity of FloridaUnited Statescdefant@ufl.edu
Abstract
We begin by introducing an interesting class of functions, knownas the Schemmel totient functions, that generalizes the Euler totientfunction. For each Schemmel totient function L m , we define two newfunctions, denoted R m and H m , that arise from iterating L m . Roughlyspeaking, R m counts the number of iterations of L m needed to reacheither 0 or 1, and H m takes the value (either 0 or 1) that the iterationtrajectory eventually reaches. Our first major result is a proof that,for any positive integer m , the function H m is completely multiplica-tive. We then introduce an iterate summatory function, denoted D m ,and define the terms D m -deficient, D m -perfect, and D m -abundant.We proceed to prove several results related to these definitions, cul-minating in a proof that, for all positive even integers m , there areinfinitely many D m -abundant numbers. Many open problems arisefrom the introduction of these functions and terms, and we mentiona few of them, as well as some numerical results. Throughout this paper, N , N , and P will denote the set of positive integers,the set of nonnegative integers, and the set of prime numbers, respectively.1or any function f , we will write f (1) = f and f ( k +1) = f ◦ f ( k ) for all k ∈ N .The letter p will always denote a prime number. For any n ∈ N , υ p ( n )will denote the unique nonnegative integer k such that p k | n and p k +1 ∤ n .Finally, in the canonical prime factorization Q ri =1 p α i i of a positive integer, itis understood that, for all distinct i, j ∈ { , , . . . , r } , we have p i ∈ P , α i ∈ N ,and p i = p j .The well-known Euler φ function is defined to be the number of positiveintegers less than or equal to n that are relatively prime to n . For each m ∈ N ,the Schemmel totient function L m ( n ) is defined as the number of positiveintegers k ≤ n such that gcd( k + s, n ) = 1 for all s ∈ { , , . . . , m − } [2]. Inparticular, L = φ . For no reason other than a desire to avoid cumbersomenotation and the possibility of dealing with undefined objects such as L (2)2 (6),we will define L m (0) to be 0 for all positive integers m .For any integer n >
1, let p ( n ) be the smallest prime number that divides n . Schemmel [7] showed that, for any positive integer m , L m is multiplicative.Thus, L m (1) = 1. Furthermore, for n > L m ( n ) = , if p ( n ) ≤ m ; n Y p | n (cid:18) − mp (cid:19) , if p ( n ) > m. (1)Letting Q ri =1 p α i i be the canonical prime factorization of n , we may rewritethe above formula as L m ( n ) = , if p ( n ) ≤ m ; r Y i =1 p α i − i ( p i − m ) , if p ( n ) > m (2)for n > φ function needed to reach 1 [5]. In the following section,we generalize Pillai’s function via the Schemmel totient functions. Then, inthe third section, we generalize the concept of perfect totient numbers withthe introduction, for each positive integer m , of a function D m , which sumsthe first R m iterates of L m . 2 The Functions R m and H m We record the following propositions, which follow immediately from (2), forlater use.
Proposition 2.1.
For x, y, m ∈ N , if x | y , then L m ( x ) | L m ( y ) . Repeatedly applying Proposition 2.1, we find
Proposition 2.2.
For x, y, m, r ∈ N , if x | y , then L ( r ) m ( x ) | L ( r ) m ( y ) . Proposition 2.3.
For m, n ∈ N , if m is even and n is odd, then either L m ( n ) = 0 or L m ( n ) is odd. In addition, the following theorem is now quite easy to prove.
Theorem 2.1.
For any prime number p and positive integer x , L p − ( px ) = ( L p − ( x ) , if p ∤ x ; pL p − ( x ) , if p | x. Proof. If p ∤ x , it follows from the multiplicativity of L p − that L p − ( px ) = L p − ( p ) L p − ( x ) = L p − ( x ). If p | x , then we have L p − ( px ) = L p − (cid:18) p · p υ p ( x ) · xp υ p ( x ) (cid:19) = L p − (cid:0) p υ p ( x )+1 (cid:1) L p − (cid:18) xp υ p ( x ) (cid:19) = p υ p ( x ) L p − (cid:18) xp υ p ( x ) (cid:19) = pL p − (cid:0) p υ p ( x ) (cid:1) L p − (cid:18) xp υ p ( x ) (cid:19) = pL p − ( x ) . Notice that, for any positive integers m and n with n >
1, we have L m ( n ) < n and L m ( n ) ∈ N . It is easy to see that, by starting with apositive integer n and iterating the function L m a finite number of times, wemust eventually reach either 0 or 1. More precisely, there exists a positiveinteger k such that L ( k ) m ( n ) ∈ { , } . This leads us to the following definitions.3 efinition 2.1. For all m, n ∈ N , let R m ( n ) denote the least positive integer k such that L ( k ) m ( n ) ∈ { , } . Furthermore, we define the function H m by H m ( n ) = L ( R m ( n )) m ( n ) . Though the functions H m only take values 0 and 1, they prove to besurprisingly interesting. For example, we can show that, for each positiveinteger m , H m is a completely multiplicative function. First, however, wewill need some definitions and preliminary results. Definition 2.2.
For m ∈ N , we define the following sets: P m = { p ∈ P : H m ( p ) = 1 } Q m = { q ∈ P : H m ( q ) = 0 } S m = { n ∈ N : q ∤ n ∀ q ∈ Q m } We define T m to be the unique set of positive integers defined by the followingcriteria: • ∈ T m . • If p is prime, then p ∈ T m if and only if p − m ∈ T m . • If x is composite, then x ∈ T m if and only if there exist x , x ∈ T m such that x , x > x x = x . Lemma 2.1.
Let k, m ∈ N . If all the prime divisors of k are in T m , thenall the positive divisors of k (including k ) are in T m . Conversely, if k ∈ T m ,then every positive divisor of k is an element of T m .Proof. First, suppose that all the prime divisors of k are in T m , and let d be a positive divisor of k . Then all the prime divisors of d are in T m . Let d = Q ri =1 w α i i be the canonical prime factorization of d . As w ∈ T m , thethird defining criterion of T m tells us that w ∈ T m . Then, by the same token, w ∈ T m . Eventually, we find that w α ∈ T m . As w α , w ∈ T m , we have w α w ∈ T m . Repeatedly using the third criterion, we can keep multiplyingby primes until we find that d ∈ T m . This completes the first part of the4roof. Now we will prove that if k ∈ T m , then every positive divisor of k is an element of T m . The proof is trivial if k is prime, so suppose k iscomposite. We will induct on Ω( k ), the number of prime divisors (countingmultiplicities) of k . If Ω( k ) = 2, then, by the third defining criterion of T m ,the prime divisors of k must be elements of T m . Therefore, if Ω( k ) = 2, weare done. Now, suppose the result holds whenever Ω( k ) ≤ h , where h > k ) = h + 1. By the thirddefining criterion of T m , we can write k = k k , where 1 < k , k < k and k , k ∈ T m . By the induction hypothesis, all of the positive divisors of k and all of the positive divisors of k are in T m . Therefore, all of the primedivisors of k are in T m . By the first part of the proof, we conclude that allof the positive divisors of k are in T m . Theorem 2.2. If m is a positive integer, then S m = T m .Proof. Fix m ∈ N . Let u be a positive integer such that, for all k ∈{ , , . . . , u − } , either k ∈ S m and k ∈ T m or k S m and k T m . Wewill show that u ∈ S m if and only if u ∈ T m . First, we must show that if k ∈ { , , . . . , u − } , then k ∈ S m if and only if L m ( k ) ∈ S m . Suppose,for the sake of finding a contradiction, that L m ( k ) ∈ S m and k S m . As k S m , we have that k > k T m . Lemma 2.1 then guaranteesthat there exists a prime q such that q | k and q T m . As q T m , thesecond defining criterion of T m implies that q − m T m . We know that q > m because, otherwise, p ( k ) ≤ q ≤ m , implying that L m ( k ) = 0 S m .Therefore, q − m ∈ { , , . . . , u − } and q − m T m . By the inductionhypothesis, q − m S m . Therefore, there exists some q ∈ Q m such that q | q − m . Because q | k , Proposition 2.1 implies that L m ( q ) | L m ( k ). Thus, q | q − m = L m ( q ) | L m ( k ), which implies that L m ( k ) S m . This is a contra-diction. Now suppose, so that we may again search for a contradiction, that L m ( k ) S m and k ∈ S m . L m ( k ) S m implies that k >
1, and k ∈ S m im-plies (by the induction hypothesis) that k ∈ T m . By Lemma 2.1, all positivedivisors of k are elements of T m . Let k = Q ri =1 p α i i be the canonical primefactorization of k . Then, by (2), L m ( k ) = , if p ( k ) ≤ m ; r Y i =1 p α i − i ( p i − m ) , if p ( k ) > m. p ( k ) ≤ m , then H m ( p ( k )) = 0, which mean that p ( k ) ∈ Q m . As p ( k ) | k , wehave contradicted k ∈ S m . Therefore, p ( k ) > m , so L m ( k ) = Q ri =1 p α i − i ( p i − m ). For each i ∈ { , , . . . , r } , p i is a positive divisor of k , so p i ∈ T m . Thesecond criterion defining T m then implies that p i − m ∈ T m , so all positivedivisors (and, specifically, all prime divisors) of p i − m are in T m . Thisimplies that all prime divisors of L m ( k ) are elements of T m , so Lemma 2.1guarantees that L m ( k ) ∈ T m . However, we have shown that 0 < L m ( k ) < k ,so L m ( k ) ∈ { , , . . . , u − } . By the induction hypothesis, we have L m ( k ) ∈ S m , a contradiction. Thus, we have established that if k ∈ { , , . . . , u − } ,then k ∈ S m if and only if L m ( k ) ∈ S m .We are now ready to establish that u ∈ S m if and only if u ∈ T m .Suppose that u ∈ S m and u T m . We know that u > m because, oth-erwise, L m ( p ( u )) = H m ( p ( u )) = 0, implying that p ( u ) ∈ Q m and con-tradicting u ∈ S m . If u is prime, then u ∈ S m implies that u ∈ P m . Then H m ( u ) = H m ( u − m ) = L ( R m ( u − m )) m ( u − m ) = 1 ∈ S m . As L ( R m ( u − m )) m ( u − m ) = L m (cid:16) L ( R m ( u − m ) − m ( u − m ) (cid:17) ∈ S m (we assume here and in the rest of the proofthat R m ( u − m ) is large enough so that the notation L ( · ) m makes sense as wehave defined it, but the argument is valid in any case), it follows from the pre-ceding argument that L ( R m ( u − m ) − m ( u − m ) = L m (cid:16) L ( R m ( u − m ) − m ( u − m ) (cid:17) ∈ S m . Continuing this pattern, we eventually find that L m ( u − m ) ∈ S m ,so u − m ∈ S m . By the induction hypothesis, u − m ∈ T m . However,by the second criterion defining T m , the primality of u then implies that u ∈ T m , a contradiction. Thus, u must be composite. We assumed that u T m , so Lemma 2.1 guarantees the existence of a prime q T m suchthat q | u . As u is composite, q ∈ { , , . . . , u − } . The induction hypothesisthen implies that q S m , so q ∈ Q m . However, this contradicts u ∈ S m ,so we have shows that if u ∈ S m , then u ∈ T m . Suppose, on the otherhand, that u S m and u ∈ T m . Again, we begin by assuming u is prime.Then, because u ∈ T m , we must have u − m ∈ T m . Therefore, by the in-duction hypothesis and the fact that u − m ∈ { , , . . . , u − } , it followsthat u − m ∈ S m . Now, u S m , so we must have u ∈ Q m . Therefore, H m ( u ) = H m ( L m ( u )) = H m ( u − m ) = L ( R m ( u − m )) m ( u − m ) = 0 S m . How-ever, as L ( R m ( u − m )) m ( u − m ) = L m (cid:16) L ( R m ( u − m ) − m ( u − m ) (cid:17) S m , it followsthat L ( R m ( u − m ) − m ( u − m ) = L m (cid:16) L ( R m ( u − m ) − m ( u − m ) (cid:17) S m . Again, wecontinue this pattern until we eventually find that L m ( u − m ) S m , which6eans that u − m S m . This is a contradiction, and we conclude that u must be composite. From u ∈ T m and Lemma 2.1, we conclude that all ofthe prime divisors of u are elements of T m . Furthermore, as u is composite,all of the prime divisors of u are elements of { , , . . . , u − } . Then, by theinduction hypothesis, all of the prime divisors of u are in the set S m . Thisimplies that none of the prime divisors of u are in Q m , so u ∈ S m . This isa contradiction, and the induction step of the proof is finally complete. Allthat is left to check is the base case. However, the base case is trivial because1 ∈ S m and 1 ∈ T m .We may now use the sets S m and T m interchangeably. In addition, partof the above proof gives rise to the following corollary. Corollary 2.1.
Let k, m, n ∈ N . Then L ( k ) m ( n ) ∈ S m if and only if n ∈ S m .Proof. The proof follows from the argument in the above proof that L m ( n ) ∈ S m if and only if n ∈ S m whenever n ∈ { , , . . . , u − } . As we now knowthat we can make u as large as we need, it follows that L m ( n ) ∈ S m if andonly if n ∈ S m . Then L (2) m ( n ) ∈ S m if and only if L m ( n ) ∈ S m , L (3) m ( n ) ∈ S m if and only if L (2) m ( n ) ∈ S m , and, in general, L ( r +1) m ( n ) ∈ S m if and only if L ( r ) m ( n ) ∈ S m ( r ∈ N ). The desired result follows immediately. Corollary 2.2.
Let m, n ∈ N . Then H m ( n ) ∈ S m if and only if n ∈ S m .Proof. It is clear that H m ( n ) ∈ S m if and only if H m ( n ) = 1. Therefore, theproof follows immediately from setting k = R m ( n ) in Corollary 2.1.Notice that, for a given positive integer m , Corollary 2.2, along with The-orem 2.2 and the defining criteria of T m , provides a simple way to constructthe set of all positive integers x that satisfy H m ( x ) = 1. Corollary 2.2 alsoexpedites the proof of the following theorem. Theorem 2.3.
The function n H m ( n ) is completely multiplicative for all m ∈ N .Proof. Choose some m, x, y ∈ N . First, suppose H m ( x ) = 0. By Corollary2.2, x S m . Therefore, there exists q ∈ Q m such that q | x . This implies7hat q | xy , so xy S m . Thus, H m ( xy ) = 0. A similar argument shows that H m ( xy ) = 0 if H m ( y ) = 0. Now, suppose that H m ( x ) = H m ( y ) = 1. ThenCorollary 2.2 ensures that x, y ∈ S m . Therefore, xy ∈ S m , so H m ( xy ) = 1.As the function H m can only take values 0 and 1, the proof is complete.In concluding this section, we note that if m + 1 is composite, then it isimpossible for any integer greater than 1 to be in S m . Therefore, whenever m + 1 is composite, we have H m (1) = 1 and H m ( n ) = 0 for all integers n > A perfect totient number is defined [3] to be a positive integer n > n = R ( n ) X i =1 φ ( i ) ( n ) . In the following definitions, we generalize the concept of perfect totient num-bers. We also borrow some other traditional terminology related to perfectnumbers.
Definition 3.1.
Let m be a positive integer. We define the arithmetic func-tion D m by D m (1) = 0 and D m ( n ) = R m ( n ) X i =1 L ( i ) m ( n )for all integers n >
1. If D m ( n ) < n , we say that n is D m -deficient . If D m ( n ) = n , we say that n is D m -perfect . If D m ( n ) > n , we say that n is D m -abundant . Finally, in the case when D m ( n ) = 0, we say that n is D m -stagnant .We now present a series of theorems related to these definitions. Theorem 3.1. If m > is odd, then all positive integers are D m -deficient. roof. Let m > n be any positive integer. If n = 1 or p ( n ) ≤ m , then n is D m -stagnant. A fortiori , n is D m -deficient. If p ( n ) > m , then p ( n ) − m is even and p ( n ) − m | L m ( n ) (by Proposition 2.1).Thus, 2 | L m ( n ), which implies that L (2) m ( n ) = 0. Hence, D m ( n ) = L m ( n ) < n . Theorem 3.2.
All positive even integers are D m -deficient for all positiveintegers m .Proof. The proof is trivial for m > D m -stagnant. For m = 1, we use the fact that all totientnumbers greater than 1 are even. Therefore, D ( n ) = φ ( n ) + φ (2) ( n ) + · · · + φ ( R ( n )) ( n ) ≤ n + 14 n + · · · + 12 R ( n ) n < n. Theorem 3.2 is nothing revolutionary, but we include it because it fitsnicely with the next theorems.For the next two theorems, which are not quite as trivial as the previoustwo, we require the following lemma.
Lemma 3.1. If k > is an odd integer, then at least one element of the set { k, L ( k ) , L (2)2 ( k ) } is divisible by .Proof. Let k > p , and suppose 3 ∤ k and 3 ∤ L ( k ). We know that p p − | L ( k ), so p ≡ p − ≡ p − p ′ such that p ′ ≡ | p ′ − L ( p ′ ) | L ( p −
2) = L (2)2 ( p ) | L (2)2 ( k ). Theorem 3.3.
For any integer m > , all positive multiples of are D m -deficient.Proof. If m ≥
3, then any positive multiple of 3 is clearly D m -stagnant.Therefore, we only need to check the case m = 2. Write K = { n ∈ N :9 | n, D ( n ) ≥ n } . Suppose K = ∅ and let n be the smallest element of K . If n = 3 α for some α ∈ N , then D ( n ) = 3 α − +3 α − + · · · +3+1 = n − < n .Therefore, n must have some prime divisor p = 3. From Theorem 3.1, p = 2. Also, by Proposition 2.2, L ( p ) | L ( n ) and L (2)2 ( p ) | L (2)2 ( n ). ByLemma 3.1, at least one of L ( p ) and L (2)2 ( p ) must be divisible by 3. Suppose3 | L ( p ) so that 3 | L ( n ). By the choice of n as the smallest element of K , we have D ( L ( n )) < L ( n ). This implies that D ( n ) = L ( n ) + D ( L ( n )) < L ( n ) < n because 3 | n . From this contradiction, weconclude that 3 | L (2)2 ( p ), so 3 | L (2)2 ( n ). Again, by the choice of n , we have D (cid:16) L (2)2 ( n ) (cid:17) < L (2)2 ( n ). However, this implies that D ( n ) = L ( n ) + L (2)2 ( n ) + D (cid:16) L (2)2 ( n ) (cid:17) < L ( n ) + 2 L (2)2 ( n ) < L ( n ) < n , which is acontradiction. It follows that K is empty. Theorem 3.4. If m > is a positive integer and m = 4 , then all positivemultiples of are D m -deficient.Proof. Let m be a positive integer other than 1 or 4, and let n be a multipleof 5. If m ≥
5, then n is D m -stagnant. If m = 3, then n is D m -deficientby Theorem 3.1. We therefore only need to check the case m = 2. Write n = 5 α k , where α, k ∈ N . We may assume that 2 ∤ k and 3 ∤ k because,otherwise, the desired result follows immediately from either Theorem 3.2 orTheorem 3.3. We now consider two cases.Case 1: α ≥
2. Write L ( k ) = 3 α α t , where t is a positive integer notdivisible by 2, 3, or 5 and α , α ∈ N (we use Proposition 2.3 to concludethat t is odd). Then L ( n ) = L (5 α ) L ( k ) = 3 α +1 α + α − t and L (2)2 ( n ) = L (3 α +1 ) L (5 α + α − ) L ( t ) = 3 α +1 α + α − L ( t ). As 3 | L ( n ), we can useTheorem 3.3 to write D ( n ) = L ( n ) + L (2)2 ( n ) + D (cid:16) L (2)2 ( n ) (cid:17) < L ( n ) +2 L (2)2 ( n ) = 3 α +1 α + α − t + 2 (3 α +1 α + α − L ( t )) ≤ α +1 α + α − t ) ≤ α k = 2125 n . This completes the proof of the case when α ≥ α = 1. In this case, n = 5 k , so L ( n ) = 3 L ( k ). We may assumethat k > n = 5 is trivial. First, suppose that 3 | L ( k ). Inthis case, L (2)2 ( k ) ≤ L ( k ), and, by Theorem 2.1, L (2)2 ( n ) = 3 L (2)2 ( k ). Then,10sing Theorem 3.3, we have D ( n ) = L ( n ) + L (2)2 ( n ) + D (cid:16) L (2)2 ( n ) (cid:17) =3 L ( k ) + 3 L (2)2 ( k ) + D (cid:16) L (2)2 ( k ) (cid:17) < L ( k ) + 6 L (2)2 ( k ) ≤ L ( k ) ≤ n − ∤ L ( k ). By Lemma 3.1 and our assumption that 3 ∤ k ,we have 3 | L (2)2 ( k ). Using Theorem 2.1 and Theorem 3.3 again, we have D ( n ) = L ( n ) + L (2)2 ( n ) + D (cid:16) L (2)2 ( n ) (cid:17) = 3 L ( k ) + L (2)2 ( k ) + D (cid:16) L (2)2 ( k ) (cid:17) < L ( k ) + 2 L (2)2 ( k ) ≤ k −
2) + 2( k −
4) = n −
14. This completes the proofof all cases.The last few theorems have dealt with D m -deficient numbers, so it isnatural to ask questions about D m -abundant numbers. We might wish toknow the positive integers m for which D m -abundant numbers even exist.How many D m -abundant numbers exist for a given m ? How large can wemake D m ( n ) − n ? Theorem 3.1 deals with these questions for the cases when m is odd and greater than 1. Also, a great deal of literature [3, 4] alreadyexists concerning the case m = 1. In the following theorem, we answer all ofthe preceding questions for the cases when m is a positive even integer. Theorem 3.5.
Let m be a positive even integer. For any positive A and δ ,there exist infinitely many primes p such that L m ( p α )+ L (2) m ( p α ) > p α + Ap α − δ for all positive integers α .Proof. Fix m , A , and δ to be positive real numbers, where m is an eveninteger. Let p , p , . . . , p r be all the primes that divide m , and let q , q , . . . , q t be all the primes that are less than m and do not divide m . For each j ∈{ , , . . . , t } , define σ j by σ j = ( , if m q j ); − , if m ≡ q j ) . Write M = Y p ≤ m p . By the Chinese remainder theorem, there exists a uniquesolution modulo M to the system of congruences defined by ( x ≡ p i ) if i ∈ { , , . . . , r } ; x ≡ σ j (mod q j ) if j ∈ { , , . . . , t } . (3)11t is easy to see that if x is a solution to (3), then x and x − m areeach relatively prime to every prime less than or equal to m . By Dirichlet’stheorem concerning the infinitude of primes in arithmetic progressions, theremust be infinitely many primes that satisfy the system (3). Let p be onesuch prime, and write β = Y p | p − m (cid:18) − mp (cid:19) . As p is relatively prime to all primes less than or equal to m , we have Y m
Ap − δ + 3 mp .With this choice of p , we may write β ( p − m ) > βp − βmp ≥ βp − mp > Ap − δ + mp . But β ( p − m ) = L m ( p − m ) because p ( p − m ) > m .Thus, ( p − m ) L m ( p − m ) > Ap − δ + mp . (4)Let α be an integer, and, for now, assume α ≥
2. Rearranging and mul-tiplying the inequality (4) by p α − , we have − mp α − + p α − L m ( p − m ) >mp α − L m ( p − m ) + Ap α − δ . After further algebraic manipulation, we find p α − ( p − m ) + p α − ( p − m ) L m ( p − m ) > p α + Ap α − δ . Noticing thatthe left-hand side of the preceding inequality is simply L m ( p α ) + L (2) m ( p α ),we have L m ( p α ) + L (2) m ( p α ) > p α + Ap α − δ . This is the desired result for α ≥
2. To show that the result holds when α = 1, it suffices to show that L m ( p )+ L (2) m ( p ) > L m ( p ) + L (2) m ( p ) p . This reduces to p − m + L m ( p − m ) >p − m + p − mp L m ( p − m ), which is obviously true.12 orollary 3.1. For any positive even integer m , there exist infinitely many D m -abundant numbers. We conclude this section with a remark about D m -perfect numbers. Using Mathematica , one may check that for m ∈ { , , } , the only D m -perfectnumber less than 100 ,
000 is 37 , D -perfect. Unfortunately, thisdata is too scarce to make any reasonable conjecture about the nature ordistribution of D m -perfect numbers for positive even integers m . In 1943, H. Shapiro investigated a function C , which counts the number ofiterations of the φ function needed to reach 2 [8]. Shapiro showed that thefunction C is additive, and he established bounds for its values. In this paper,we have not gone into much detail exploring the functions R m because theyprove, in general, to be either completely uninteresting or very difficult tohandle. For example, for any integer n > R ( n ) = ( , if n , , if n ≡ , . On the other hand, the function R does not seem to obey any nice patternor exhibit any sort of nice additive behavior. There seems to be some hopein analyzing the function R , so we make the following conjecture. Conjecture. If x > R ( x ) ≥ log (cid:0) x (cid:1) log 7 . We note that it is not difficult to prove, using Lemma 3.1 and a bitof case work, that R ( x ) ≤ x + 2)log 3 − x > x = 7). However, as Figure 1 shows, this is a very weak13igure 1: A plot of the first 300 ,
000 values of the function R , as well as someimportant curves. Note that the black streaks in the figure are, in actuality,several overlapping dots.upper bound (at least for relatively small x ). It is tempting to think,based on the figure, that R ( x ) ≤ x log 3 for all positive integers x .However, setting x = 480 , ,
203 yields a counterexample because 3 +log 480 , , ≈ . <
22 = R (480 , , R natu-rally leads to a question about the infinitude of twin primes, which hints atthe potential difficulty of the problem. Indeed, Harrington and Jones [1] havearrived at the same conclusion while studying the function C ( x ) := R ( x ) − C ( x ) + C ( y ) − C ( xy ) can be arbi-trarily large. To avoid the unpredictability of the values of the function C , Harrington and Jones have restricted the domain of C to the set D ofpositive integers k with the property that none of the numbers in the set { k, L ( k ) , L (2)2 ( k ) , . . . } has a prime factor that is congruent to 1 modulo 3.With this restriction of the domain of C , these two authors have establishedresults analogous to those that Shapiro gave for the function C mentionedearlier. In fact, we speculate that methods analogous to those that Harring-14on and Jones have used could easily generalize to allow for analogous resultsconcerning functions C m ( x ) := R m ( x ) − C m .We next remark that, in Theorem 3.4, the requirement that m = 4 isessential. For example, write p = 306 , p = 4 + p , p = 4 + p , p = 4 + p , and p = 4 + p . Then the number 5 p is a D -abundantmultiple of 5.Lastly, we have not spent much effort analyzing the “sizes” of the func-tions D m or searching for D m -perfect numbers. We might inquire about theaverage order or possible upper and lower bounds for D m for a general pos-itive even integer m . In addition, it is natural to ask if there even are any D m -perfect numbers other than 37 ,
147 for even positive integers m . Dedicated to my parents Marc and Susan, my brother Jack, and my sisterJuliette.Also dedicated to Mr. Jacob Ford, who wrote a program to find valuesof the functions L m , R m , and H m . Mr. Ford and my father also sparkedmy interest in computer programming, which I used to analyze the functions D m .Finally, I would like to thank the unknown referee for taking the time toread carefully through my work and for his or her valuable suggestions. References [1] J. Harrington and L. Jones, On the iteration of a function related toEuler’s φ -function. Integers (2010), 497–515.[2] J. Herzog and P. R. Smith, Lower bounds for a certain class of errorfunctions. Acta Arith. (1992), 289–305.153] D. Iannucci, M. Deng, and G. Cohen, On perfect totient numbers. J.Integer Seq. (2003), Article 03.4.5.[4] F. Luca, On the distribution of perfect totients. J. Integer Seq. (2006),Article 06.4.4.[5] S. S. Pillai, On a function connected with φ ( n ). Bull. Amer. Math. Soc. (1929), 837–841.[6] J. Rosser and L. Schoenfeld. Approximate formulas for some functions ofprime numbers. Illinois J. Math. (1962), 64–94.[7] V. Schemmel, ¨Uber relative Primzahlen, Journal f¨ur die reine und ange-wandte Mathematik , (1869), 191–192.[8] H. Shapiro, An arithmetic function arising from the φ function. Amer.Math. Monthly (1943), 18–30.[9] G. K. White, Iterations of generalized Euler functions. Pacific J. Math. (1962), 777–783.2010 Mathematics Subject Classification : Primary 11N64; Secondary 11B83.