Compositions with theEuler and Carmichael FunctionsWilliam D. BanksDepartment of MathematicsUniversity of MissouriColumbia, MO 65211, USAbbanks@math.missouri.eduFlorian LucaInstituto de Matema´ticasUniversidad Nacional Auto´noma de Me´xicoC.P. 58089, Morelia, Michoaca´n, Me´xicofluca@matmor.unam.mxFilip SaidakDepartment of MathematicsWake Forest UniversityWinston-Salem, NC 27109, USAsaidakf@wfu.eduPantelimon Sta˘nica˘Mathematics DepartmentAuburn University MontgomeryMontgomery, AL 36124, USApstanica@mail.aum.eduOctober 8, 20051AbstractLet ϕ and λ be the Euler and Carmichael functions, respectively.In this paper, we establish lower and upper bounds for the number ofpositive integers n ≤ x such that ϕ(λ(n)) = λ(ϕ(n)). We also studythe normal order of the function ϕ(λ(n))/λ(ϕ(n)).1 IntroductionThe Euler ϕ-function (first introduced in [20] of 1760) and the Carmichaelλ-function (first introduced in [9] of 1910) are two of the most interesting,useful and versatile arithmetic functions that have ever been studied. Fora positive integer n, the value ϕ(n) of the Euler function is defined to bethe number of natural numbers less than or equal to n and coprime to n.Equivalently,ϕ(n) = #(Z/nZ)∗ =∏pα ‖npα−1(p− 1).For a positive integer n, the value λ(n) of the Carmichael function is definedto be the maximal order of any element in the multiplicative group (Z/nZ)∗.More explicitly, for a prime power pα,λ(pα) ={pα−1(p− 1), if p ≥ 3 or α ≤ 2;2α−2, if p = 2 and α ≥ 3;and for an arbitrary integer n ≥ 2 with prime factorization n = pα11 . . . pαkk ,one hasλ(n) = lcm[λ(pα11 ), . . . , λ(pαkk )].Also, λ(1) = 1. In contrast to ϕ, the function λ is not multiplicative.Over the years, the Euler and Carmichael functions have been exten-sively researched in the literature, and quite a large number of results havebeen obtained concerning the rate of growth and the arithmetical propertiesof these functions. To gain insight into the nature of these and other arith-metic functions, many subsequent investigations considered compositions (oriterations) of such functions. Let us mention a few relevant examples thathave motivated our present investigations.(1) In 1929, Pillai [41] was the first to study properties of the iterates{ϕ(k) : k ≥ 1} of the Euler function, where ϕ(1) = ϕ, and ϕ(k) = ϕ ◦ ϕ(k−1)2for k ≥ 2. Pillai showed that if W (n) = k is the least integer k for whichϕ(k)(n) = 1, then [log 3nlog 3]≤ W (n) ≤[log 2nlog 2].Later, Shapiro [44], Mills [37], Erdo˝s [14] and Erdo˝s and Hall [16] investigatedrelated questions. Extending some of Shapiro’s work [44], the problem offinding integers n with the property that ϕ(k)(n) | n was first considered byHausman [31] in 1982, and these results were later generalized by Halter-Koch and Steindl [28] and by Siva Rama Prasad and Fonseca [46].(2) For a positive integer n, let Ω(n) and ω(n) denote the number ofprime factors of n counted with and without multiplicity, respectively, andlet ∆(n) = Ω(n)− ω(n).Generalizing the fundamental theorem of Tura´n [48] from 1934, in 1984,Erdo˝s and Pomerance [18] and Murty and Murty [39] proved, independently,that both compositions Ω(ϕ(n)) and ω(ϕ(n)) have normal order 12(log2 n)2,and they also gave analogues of the Erdo˝s-Kac theorem (see [17]) for thesefunctions.1In 1999, Bassily, Ka´tai and Wijsmuller [7] gave proofs of similar theoremsfor ϕ(k). In particular, they showed that, as x→∞,∆(ϕ(k)(n)) = (1 + o(1))1k!(log2 x)k log4 xholds for almost all n ≤ x, andlimx→∞1x#{n ≤ x : ∆(ϕ(k)(n))− s(x)√log2 x log4 x< z}= Φ(z) =1√2π∫ z−∞e−t2/2dt,where s(x) = (log4 x+ c+ o(1)) log2 x. In other words, the quantity∆(ϕ(k)(n))− s(x)√log2 x log4 xis normally distributed.Earlier, in 1990, Erdo˝s, Granville, Pomerance and Spiro [15] proved that,under the Elliott-Halberstam conjecture, the normal order of Pillai’s function1logk denotes the k-th iterate of the natural logarithm.3W (n) is Θ logn for some constant Θ, and for every positive integer k, the nor-mal order of ϕ(k)(n)/ϕ(k+1)(n) is eγk log3 n, where γ is the Euler-Mascheroniconstant.(3) In 1944, Alaoglu and Erdo˝s [2] considered the compositions ϕ◦σ andσ ◦ ϕ, where σ is the sum of divisors function, and they proved thatlim infn→∞ϕ(σ(n))n= 0 and lim supn→∞σ(ϕ(n))n=∞.Conversely, in 1964, Makowski and Schinzel [35] proved thatlim supn→∞ϕ(σ(n))n=∞ and lim infn→∞σ(ϕ(n))n≤ 12+1234 − 4 ,and in 1989, Pomerance [43] showed thatlim infn→∞σ(ϕ(n))/n > 0.In 1992, Golomb [25] observed that the value of σ(ϕ(n))− ϕ(σ(n)) takesboth positive and negative values infinitely often and asked for the proportionof each. De Koninck and Luca [11] have shown that this function is positivefor almost all values of n.(4) Recently, Martin and Pomerance [36] have studied iterates of theCarmichael function and have shown that the normal order of the functionlog (n/λ(λ(n))) is (log2 n)2 log3 n. In other words,λ(λ(n)) = n exp(−(1 + o(1))(log2 n)2 log3 n) (1)as n→∞ through a set of integers of asymptotic density one.For a variety of other results with a similar flavor, we refer the readerto [3, 4, 5, 6, 8, 10, 12, 13, 21, 22, 23, 24, 26, 33, 34, 40, 42, 45, 49] and thereferences contained therein.In this paper, we initiate the study of the composite functions ϕ ◦ λ andλ ◦ ϕ by establishing lower and upper bounds for the counting function ofthe setA(x) = {n ≤ x : ϕ(λ(n)) = λ(ϕ(n))}.4Our main results are the following:Theorem 1. There exist positive constants C and x0 such that the followingbound holds for all x ≥ x0:#A(x) ≥ exp(Clog xlog log log x).Theorem 2. The inequality#A(x) ≤ x(log x)3/2+o(1)holds as x→∞.Remark. If we denote A = {n : ϕ(λ(n)) = λ(ϕ(n))}, then Theorem 2implies that ∑n∈A1n<∞.It is natural to conjecture that the estimate#A(x) = x(log x)c+o(1)(2)holds for some positive constant c. Since the Sophie-German primes (i.e.,primes p for which q = (p − 1)/2 is also prime) are all contained in A, theHardy-Littlewood conjectures (see [29]) would suggest that c ≤ 2. Taking intoaccount the very special structure of the integers in A, it is also natural toexpect the Sophie-German primes to form a subset of A of positive relativeasymptotic density, and we therefore conjecture that (2) holds with c = 2.Although our focus in this paper is primarily on the set A of positiveintegers for which the values of ϕ◦λ and λ◦ϕ coincide, we have also been ledto consider the related question: Which value is larger, ϕ(λ(n)) or λ(ϕ(n)),for a “typical” integer n? In the last section, we study the normal order ofthe function ϕ(λ(n))/λ(ϕ(n)); our result, which relies heavily on (1), is thefollowing:5Theorem 3. The estimateϕ(λ(n))λ(ϕ(n))= exp((1 + o(1))(log logn)2 log log logn)holds on a set of positive integers n of asymptotic density one.In particular, one sees that ϕ(λ(n)) is much larger than λ(ϕ(n)) for almostall positive integers n.Acknowledgements. The authors would like to thank the anonymous ref-eree for a careful reading of the manuscript and for useful suggestions. Thispaper was written during an enjoyable visit by F. L. and P. S. to the Univer-sity of Missouri–Columbia; these authors wish to express their thanks to thatinstitution for its hospitality and support. Research of W. B. was supportedin part by NSF grant DMS-0070628, that of F. L. by grants SEP-CONACYT37259-E and 37260-E, and that of P. S. by a grant from his institution.2 NotationThroughout the paper, we use the Landau symbols ‘o’ and ‘O’ and the Vino-gradov symbols ‘≪’ and ‘≫’ with the understanding that the implied con-stants are absolute; we recall that, for positive functions U and V , the nota-tions U ≪ V , V ≫ U , and U = O(V ) are each equivalent to the assertionthat the inequality U ≤ cV holds for some constant c > 0. As usual, P (n)denotes the largest prime factor of n > 1, and ω(n) denotes the number ofdistinct prime factors of n. Throughout the paper, the letters p, q, and ralways denote prime numbers. For a positive real number x, we use log x todenote maximum of 1 and the natural logarithm of x. For an integer k ≥ 2,logk x denotes the k-th iterate of the function log x. For a positive real num-ber x and a subset B of the positive integers, we write B(x) = B ∩ [1, x].Finally, we use c0, c1, . . . to represent positive constants that are absolute.For the convenience of the reader, we have included a brief index at theend of the paper which contains, in particular, a list of notation for our proofof Theorem 2 below.63 Coincidences between ϕ ◦ λ and λ ◦ ϕThroughout this section, we focus our study on the setA = {n ≥ 1 : ϕ(λ(n)) = λ(ϕ(n))}.Our goal is establish lower and upper bounds for the counting function#A(x) = #(A ∩ [1, x]), where x is a real parameter.3.1 Lower BoundTheorem 1. There exists a positive absolute constant c0 such that the fol-lowing bound holds for all x ≥ 2:#A(x) ≥ exp(c0log xlog3 x).Proof. Observe that, since #A(2) = 2, it suffices to establish the inequalityfor all sufficiently large values of x.For a positive integer n, let ϕ(0)(n) = n, and define ϕ(k)(n) inductively byϕ(k)(n) = ϕ(ϕ(k−1)(n))for all k ≥ 1. Let P(n) denote the set of odd primefactors of the integer∏k≥0 ϕ(k)(n). By a result of Pillai [41], the equalityϕ(k)(n) = 1 holds for some k ≤ K = ⌈(log n)/ log 2⌉; consequently,∏p∈P(n)p ≤ nK+1 ≤ exp (2(logn)2) . (3)We also have#P(n) ≤ (K + 1)maxℓ≤n{ω(ℓ)} ≪ (log n)2log2 n. (4)Now let n ≥ 3 be an odd square-free integer that is coprime to ϕ(k)(n)for all k ≥ 1, and putM = 2∏p∈P(n)p and N = 2nMλ(M).Note that P(n) ∪ {2} is precisely the set of primes that divide N . Denotingby vp(·) the standard p-adic valuation, it is easy to check thatvp(N) =2 + maxq∈P(n){v2(q − 1)}, if p = 2;1 + maxq∈P(n){vp(q − 1)}, if p ∈ P(n) and p ∤ n;2, if p | n.7Putαp ={maxq∈P(n){vp(q − 1)}, if p ∈ P(n) ∪ {2} and p ∤ n;1, if p | n.Observe that αp ≥ 1 for all primes p ∈ P(n) ∪ {2}. SinceN = 2α2+2∏p∈P(n)pαp+1,we haveλ(N) = lcm[λ(2α2+2), λ(pαp+1) : p ∈ P(n)]= lcm [2α2 , pαp(p− 1) : p ∈ P(n)] = 2α2∏p∈P(n)pαp.Therefore,ϕ(λ(N))= 2α2−1∏p∈P(n)pαp−1(p− 1) = 2α2+δ2−1∏p∈P(n)pαp+δp−1,where we have factored ∏p∈P(n)(p− 1) = 2δ2∏p∈P(n)pδp .Here, we have used the fact that for each p ∈ P(n), the odd prime factors ofp− 1 also lie in P(n).On the other hand, we haveϕ(N) = 2α2+1∏p∈P(n)pαp(p− 1) = 2α2+δ2+1∏p∈P(n)pαp+δp .Since α2, δ2 ≥ 1, and αp ≥ 1 for all p ∈ P(n), it follows thatλ(ϕ(N)) = lcm[λ(2α2+δ2+1), λ(pαp+δp) : p ∈ P(n)]= lcm[2α2+δ2−1, pαp+δp−1(p− 1) : p ∈ P(n)]= 2α2+δ2−1∏p∈P(n)pαp+δp−1 = ϕ(λ(N)).Thus, we have shown that the integer N lies in the set A.8We now use the fact that there exists an absolute constant c1 > 0 suchthat for all y > 10, there exists a positive integer ℓy ≤ y2 with the propertythat ∑(p−1) | ℓy1 ≥ exp(c1log ylog2 y)(5)(see [1], for example). Let S be the set of odd primes p such that p− 1 is adivisor of ℓy, but p does not lie in P(ℓy). Using the inequalities (4) and (5),it follows that#S ≥∑(p−1) | ℓyp 6=21−#P(ℓy) ≥ s := exp(c2log ylog2 y)(6)holds with c2 = c1/2, provided that y is sufficiently large. Replacing S byone of its subsets, if necessary, we can assume that #S = ⌈s⌉. We now sett = ⌊√s⌋ and consider subsets T ⊂ S of cardinality t. The number of thesesubsets is(⌈s⌉t)≥(⌈s⌉ − tt)t= exp((0.5 + o(1))s1/2 log s) ≥ exp(c3 s1/2 log ylog2 y),where c3 = c2/3, provided that y is large enough. For each subset T , putnT =∏p∈T p. Sinceϕ(nT ) =∏p∈T(p− 1)∣∣∣ ℓty,it follows thatϕ(k)(nT )∣∣∣ϕ(k−1)(ℓty), k ≥ 1.As P(ℓty) = P(ℓy), we see that nT is coprime to ϕ(k)(nT ) for all k ≥ 1, and∏p∈P(nT )p divides nT∏p∈P(ℓy)p. (7)The construction given at the beginning of the proof now shows that ifMT = 2∏p∈P(nT )p,then the positive integerNT = 2nTMT λ(MT )9lies in the set A. Moreover, it is clear that distinct subsets T give rise todistinct elements of A (for if NT1 = NT2, then by comparing those parts ofNT1 and NT2 composed of primes in S, we obtain that n2T1 = n2T2 , and byunique factorization this leads to T1 = T2). To bound the size of NT , we firstuse (3) and (7) to estimateMT ≤ 2nT∏p∈P(ℓy)p ≤ nT exp(O((log y)2)).Since λ(MT ) ≤MT , we can use this bound for λ(MT ) as well. Also,nT =∏p∈Tp ≤ (y2 + 1)t = exp ((2 + o(1))s1/2 log y) .Therefore,NT ≪ nTMT λ(MT ) ≤ exp((6 + o(1))s1/2 log y +O((log y)2))Now, given a large real number x, let y be defined implicitly by the equationx = exp(7s1/2 log y), where s is defined as in (6). Then NT ≤ x holds for allsuch subsets T , provided that x is sufficiently large. Sincelog2 x = (0.5 + o(1)) log s≫log ylog2 y,it follows thatlog2 y ≤ (1 + o(1)) log3 x.Therefore, if x is large enough, then#A(x) ≥ exp(c3s1/2 log ylog2 y)≥ exp(c0log xlog3 x),where c0 = c3/8. This completes the proof.3.2 Upper BoundWe begin this subsection with a few technical lemmas that are used in ourproof of Theorem 2 below.The following result is a weakened and simplified version of a well-knownresult of Hildebrand [32] (see, for example, Chapter III.5 in the book byTenenbaum [47]):10Lemma 1. Uniformly for exp ((log2 x)2) ≤ y ≤ x, the cardinality Ψ(x, y) ofthe set of smooth numbersS(x, y) = {n ≤ x : P (n) ≤ y}is bounded byΨ(x, y) ≤ xu−u+o(u),where u = (log x)/(log y).Lemma 2. Uniformly for x ≥ y ≥ 2, the cardinality of the setF(x, y) = {n ∈ x : q | gcd(n, ϕ(n)) for some prime q > y}is bounded by#F(x, y)≪ x log2 xy log y.Proof. If n lies in F(x, y), then either there exists a prime q > y such thatq2 | n, or there exists a prime q > y and a prime p ≡ 1 (mod q) such thatpq | n. In the first case, the number of such integers n ≤ x is bounded by∑q>y⌊xq2⌋≤ x∑q>y1q2≪ xy log y,and in the second case, the number of such integers n ≤ x is at most∑q>y∑py1q∑p≤xp≡1 (mod q)1p≪ x log2 x∑q>y1q2≪ x log2 xy log y,where we have used the uniform bound∑p≤xp≡1 (mod m)1p≪ log2 xϕ(m), (8)which follows from the inequality (3.1) of [15] (see also Lemma 1 of [7]). Theresult follows.11Lemma 3. If x ≥ 2 and w ≥ ⌊2e log2 x⌋+ 1, then the cardinality of the setG(x, w) = {n ∈ x : ω(n) > w}is bounded by#G(x, y) ≤ xlog x2−w(1+o(1)).Proof. By results of Hardy and Ramanujan [30], the number of positive in-tegers n ≤ x for which ω(n) = k is bounded above byxlog x· 1(k − 1)! · (log2 x+O(1))k−1 ≤ xlog x(e log2 x+O(1)k − 1)k−1.In the above inequality (and in many others to follow), we have used Stirling’sformula to conclude that k! ≥ (k/e)k holds for all positive integers k. Inparticular, if w ≥ ⌊2e log2 x⌋+ 1, then#G(x, w) ≤ xlog x∑k>w(0.5 + o(1))k =xlog x2−w(1+o(1)),which is the desired estimate.Theorem 2. The inequality#A(x) ≤ x(log x)3/2+o(1)holds as x→∞.Proof. Our strategy is to express A(x) as a union of boundedly many subsets,each of which has a cardinality bounded above by x(log x)−3/2+o(1).Let x be a large positive real number. The first five subsets that weconsider are the following:(i) The subsetA1(x) = {n ∈ A(x) : P (n) ≤ y1},wherey1 = exp(log x log3 x3 log2 x).12Since A1(x) ⊂ S(x, y1), Lemma 1 immediately implies that#A1(x) ≤ x(log x)3+o(1). (9)Note that if x is large, then A1(x) contains all powers of 2 which aresmaller than x.(ii) The subsetA2(x) = {n ≤ x : q | gcd(n, ϕ(n)) or q2 | ϕ(n) for some q > y2 prime}where y2 = (log x)2. Write A2(x) = A′2(x) ∪ A′′2(x), where A′2(x) con-sists of those n ∈ A2(x) such that q | gcd(n, ϕ(n)) for some primeq > y2, and A′′2(x) consists of the remaining n ∈ A2(x). Since A′2(x) isa subset of F(x, y2), Lemma 2 provides the estimate#A′2(x)≪x(log x)2. (10)Now suppose that n ∈ A′′2(x). Then there exist a prime q > y2 andeither a prime factor p of n with p ≡ 1 (mod q2), or two prime factorsp1 < p2 of n such that q | gcd(p1 − 1, p2 − 1). Fix q, and p in the firstcase, and p1 and p2 in the second case, respectively. Then, the numberof possible multiples n ≤ x of p is ⌊x/p⌋ ≤ x/p in the first case, whilethe number of possible multiples n ≤ x of p1p2 is ⌊x/p1p2⌋ ≤ x/p1p2 inthe second case. Therefore,#A′′2(x) ≤ x∑q>y2∑p≤xp≡1 (mod q2)1p+∑q>y2∑p1y2∑p≤xp≡1 (mod q2)1p+∑q>y212 ∑p≤xp≡1 (mod q)1p2≪ x(log2 x)2∑q>y21q2≪ x(log2 x)2y2=x(log2 x)2(log x)2, (11)13where we have used the estimate (8) twice. The estimates (10) and (11)immediately imply that#A2(x) ≤ x(log x)2+o(1). (12)(iii) The subsetA3(x) = {n ∈ A(x) : ω(n) > w1},wherec1 = 2e and w1 = ⌊c1 log2 x⌋+ 1.Since A3(x) ⊂ G(x, w1), Lemma 3 provides the upper bound#A3(x) ≤ xlog x2−w1(1+o(1)) =x(log x)1+2e log 2+o(1). (13)(iv) The subsetA4(x) = {n ∈ A(x) : 2w2 | n or 2w2 | (p− 1) for some prime p | n},wherec2 = 2/ log 2 and w2 = ⌊c2 log2 x⌋.Clearly,#A4(x) ≤⌊ x2w2⌋+∑p≤xp≡1 (mod 2w2 )⌊xp⌋≤ x2w2+ x∑p≤xp≡1 (mod 2w2 )1p≪ x log2 x2w2=x(log x)2+o(1), (14)where we have used (8) in the last step.For the remaining subsets of A(x), our estimates are presented as a series oftechnical lemmas. We begin with the following:14Lemma 4. LetA5(x) ={n ∈ A(x)\ (∪4j=1Aj(x)) : F (n) > w3},whereF (n) = v2(n) +∑p |nv2(p− 1) (15)for every positive integer n, andw3 = c3 log2 x log3 x, c3 =c1 + 1log 2.If x is sufficiently large, then#A5(x) ≤ x(log x)2. (16)Proof. For each integer n ∈ A5(x), we have:• P (n) > y1 and P (n)2 ∤ n;• ω(n) ≤ w1;• 2w2 ∤ n, and 2w2 ∤ (p− 1) for every prime p dividing n;• n is not a power of 2;• F (n) > w3.Let n = 2α∏kℓ=1 pβℓℓ be the prime factorization of n, where the primes pℓ areodd and distinct; reordering the odd primes, if necessary, we can assume thatv2(pℓ−1) is a nondecreasing function of ℓ. Then there exists an integer t ≥ 1,integers 1 ≤ α1 < · · · < αt, and integers κ1, . . . , κt ≥ 1 with the followingthree properties:(i) 2αj | (pℓ − 1) if 1 ≤ j ≤ t and κ1 + · · ·+ κj−1 < ℓ ≤ κ1 + · · ·+ κj;(ii) The inequalityt∑j=1κj = k = ω(n)− δ ≤ c1 log2 xholds, where δ = 0 if α = 0, and δ = 1 otherwise;15(iii) The equalityα +t∑j=1κjαj = ⌊c3 log2 x log3 x⌋holds.Here, the integers αi are all the possible values of v2(p − 1) as p runs overthe odd prime factors of n, arranged increasingly, and κi is the multiplicitywith which αi occurs (i.e., the number of prime factors p of n such thatαi = v2(p− 1)).Let D be the set of all (2t+2)-tuples (t, α, α1, . . . , αt, κ1, . . . , κt) for whichthese properties hold for some n ∈ A5(x). Clearly,#D ≤∑0≤α y1/2αj > y1/21hold, and thereforelog(x/(2αjm))≫ log y1 ≫ log x log3 xlog2 x.16Consequently, for a fixed value of m, the number of such primes P is≪ x log2 xlog x· 12αj· 1m=x log2 xlog x· 12α+αj∏1≤ℓ≤kℓ 6=κ1+···+κj−1+11pβℓℓ.We now sum up the above inequality over all possible integers m of type (17)(with j fixed) and deduce that the corresponding contribution to #A5(x) is≪ x log2 xlog x· 12α+αj∏1≤i≤ti6=j1κi! ∑p≤xp≡1 (mod 2αi )∑β≥11pβκi≤ x log2 xlog x· 12α+Ptj=1 αjκj· (c4 log2 x)Ptj=1 κj . (18)Here, c4 is an absolute constant for which the inequality∑1≤p≤xp≡1 (mod 2α)∑β≥11pβ≤ c4 log2 x2αholds for all sufficiently large x and uniformly in α ≤ w2 (see again (8)).Since(c4 log2 x)Ptj=1 κj ≤ exp (c1 log2 x log(c4 log2 x))= exp ((c1 + o(1)) log2 x log3 x) ,and2α+Ptj=1 αjκj = exp (⌊c3 log2 x log3 x⌋ log 2)= exp ((c1 + 1 + o(1)) log2 x log3 x) ,we find that the expression (18) is bounded above byxexp ((1 + o(1)) log2 x log3 x).Summing this over all possible choices of j ∈ {1, . . . , t}, then over all possible(2t+ 2)-tuples in D, we derive the estimate#A5(x)≪ log2 x ·#D ·xexp ((1 + o(1)) log2 x log3 x)≤ x(log x)2for all sufficiently large x.17Now fix n ∈ A(x)\ (∪5j=1Aj(x)), and let n = 2α∏kℓ=1 pβℓℓ be its primefactorization. Let P = {p1, . . . , pk} be the set of odd primes that divide n,P1 = {p1, . . . , pk1} be the subset of P consisting of the odd primes p thatdivide gcd(n, ϕ(n)), and Q be the set of odd prime factors of ϕ(n) that donot divide n. Since n 6∈ A2(x), every prime p ∈ P1 satisfies the boundp ≤ y2 = (log x)2; thus, if q > y2 and q | ϕ(n), then q ∈ Q. Let q1, . . . , qs bethe primes in Q. Factoringpℓ − 1 = 2αℓk1∏i=1pγi,ℓis∏j=1qδj,ℓj , ℓ = 1, . . . , k,we derive thatϕ(n) = ϕ(2α) · 2Pkℓ=1 αℓk1∏i=1pβi−1+Pkℓ=1 γi,ℓis∏j=1qPkℓ=1 δj,ℓj .Defining µj = v2(qj − 1) for j = 1, . . . , s, it follows thatv2 (λ(ϕ(n))) = max{v2(λ(ϕ(2α) · 2Pkℓ=1 αℓ)), max1≤i≤k1{αi − 1}, max1≤j≤s{µj}}= max{α +k∑ℓ=1αℓ, max1≤j≤s{µj}}+O(1),The preceding formula may be rewritten in the formv2 (λ(ϕ(n))) = max{F (n),maxq∈Q{v2(q − 1)}}+O(1). (19)On the other hand, we haveλ(n) = max{λ(2α), 2maxℓ{αℓ}} k1∏i=1pmax{βi−1,maxℓ{γi,ℓ}}is∏j=1qmaxℓ{δj,ℓ}j ,where the maxima in the exponents are taken over ℓ ∈ {1, . . . , k}. From thepreceding relation, we see that v2 (ϕ(λ(n))) is equal tov2(max{ϕ(λ(2α)), 2maxℓ{αℓ−1}})+∑p∈P1v2(p− 1) +∑q∈Qv2(q − 1)= max{v2(n),maxp |n{v2(p− 1)}}+∑p∈P1v2(p− 1) +∑q∈Qv2(q − 1) +O(1).18Combining this result with (19) and the fact that v2(λ(ϕ(n))) = v2(ϕ(λ(n))),we obtain:max{v2(n),maxp |n{v2(p− 1)}}+∑p∈P1v2(p− 1) +∑q∈Qv2(q − 1)= max{F (n),maxq∈Q{v2(q − 1)}}+O(1).(20)We now define A6(x) be the set of those integers n ∈ A(x)\(∪5j=1Aj(x))for which the maximum on the right hand side of (20) is not achieved withthe term F (n).Lemma 5. We have#A6(x)≪ x(log x)2+o(1). (21)Proof. If n ∈ A6(x), there exists a prime q′ ∈ Q such that the maximum onthe right hand side of (20) is achieved with the term v(q′− 1), and it followsthatmax{v2(n),maxp |n{v2(p− 1)}}+∑p∈P1v2(p− 1) +∑q∈Qq 6=q′v2(q − 1)≪ 1,which implies, in particular, that #P1 + #Q ≪ 1. Hence, there exists anabsolute constant c5 > 0 such that for every n ∈ A6(x), the inequalityω(p− 1) ≤ c5 holds for all prime factors p of n.Let R1 = {p : ω(p− 1) ≤ c5}. Our first step is to establish the estimate:#R1(x)≪ x(log2 x)c5+1(log x)2(22)For this, let p ∈ R1(x). We may assume that P (p− 1) > y1, for the numberof primes p ≤ x with P (p− 1) ≤ y1 is no more thanΨ(x, y1) ≤ x exp (−(1 + o(1))u logu)≪ x(log x)2. (23)We may also assume that there does not exist a prime q > y3 = log x suchthat q2 | p− 1. Indeed, the number of primes p ≤ x for which q2 | p− 1 forsome q > y3 is no more than∑y3y31q2≪ xlog x· 1y3≪ x(log x)2, (24)and for the sum S2 we need only the trivial fact that π(x, q2, 1) ≤ x/q2 toderive the bound:S2 ≤ x∑q>x1/31q2≪ x2/3log x≪ x(log x)2. (25)Thus, we may assume that p − 1 = Pm, where P > max{P (m), y1}. Sinceω(p− 1) ≤ c5, it follows that ω(m) ≤ c5 − 1. Fixing one such number m, weapply Brun’s method (see [27]) to deduce that the number of possibilities forthe prime p is≪ xϕ(m)(log(x/m))2.Since x/m > y1, we havelog(x/m) > log y1 ≫ log xlog2 x. (26)Hence, the number of possibilities for p ≤ x when m is fixed, is≪ x(log2 x)2(log x)2· 1ϕ(m).Summing this inequality over all m with ω(m) ≤ c5 − 1, and using theinequalities (23), (24), and (25), we obtain that#R1(x) ≤ x(log2 x)2(log x)2∑m max{P (m), y1}, and m ∈ M1(x). Let m be fixed. According to (22),the prime P ∈ R1(x/m) can be chosen in at most≪ x (log2(x/m))c5+1m(log(x/m))2different ways. By inequality (26), the number of possibilities for P is≪ x(log2 x)c5+3(log x)2· 1m.Summing this inequality over all m ∈M1(x), we derive that#A6(x) ≪ x(log2 x)c5+3(log x)2∑m∈M1(x)1m≪ x(log2 x)c5+3(log x)2=x(log x)2+o(1),which finishes the proof.Now let n be an integer in A(x)\ (∪6j=1Aj(x)). Then, by (20), we havemax{v2(n),maxp |n{v2(p− 1)}} +∑p∈P1v2(p− 1) +∑q∈Qv2(q − 1)= F (n) +O(1).21In particular, it follows that∑q∈Qv2(q − 1) ≤∑p∈Pv2(p− 1) +O(1). (28)Since, for every prime factor q > y2 of ϕ(n), there exists a unique p | n suchthat q | p− 1 (because n 6∈ A2(x)), from (28) we deduce that∑p |n∑q>y2q | p−1v2(q − 1) ≤∑p |nv2(p− 1) +O(1). (29)For any real number z > 0, letf(z) = exp(exp(log2 z(log3 z)2)).For any real number z and positive integer n, we write ω≤z(n) and ω>z(n) forthe number of distinct prime factors of n that are ≤ z and > z, respectively.Consider the following set of prime numbers:R2 = {p : ω>f(p)(p− 1) ≤ log2 p/ log3 p}.We claim that the estimate#R2(x) ≤ x(log x)2+o(1)(30)holds. Indeed, let p ∈ R2(x) be fixed. Since π(x/ log x)≪ x/(log x)2, we canassume that p > x/ log x. As in the above estimate for the counting function#R1(x) of R1, we can assume that P = P (p− 1) > y1 and that q2 does notdivide p−1 for any prime q > y3 = log x, since the size of the exceptional setis ≪ x/(log x)2. Finally, we can assume that ω(p − 1) ≤ w1, for otherwisep − 1 ∈ A3(x), and we have already seen that #A3(x) ≪ x/(log x)2. Now,write p−1 = Pm, where P = P (p−1) > max{P (m), y1}, m satisfies furtherω(m) ≤ w1 = c1 log2 x, and letM2(x) be the set of all integers m obtained inthis way. For fixed m ∈ M2(x), we apply Brun’s method once again, whichshows that the number of possibilities for p is≪ xϕ(m)(log(x/m))2≪ x(log2 x)3(log x)2· 1m22(for the second estimate, we use (26) and the bound ϕ(m) ≫ m/ log2m).Clearly, for m ∈ M2(x), we have ω>f(x)(m) < log2 x/ log3 x. Thus, if k2 andk3 denote the number of prime factors of m which are ≤ f(x) and > f(x),respectively, then k2 ≤ w1, and k3 ≤ w4 = ⌊log2 x/ log3 x⌋. Therefore,#R2(x) ≤ x(log2 x)3(log x)2∑m∈M2(x)1m+O(x(log x)2), (31)and∑m∈M2(x)1m≤∑k2≤w1k3≤w41k2! ∑p≤f(x)∑β≥11pβk2 · 1k3!(∑p≤x∑β≥11pβ)k3.Note that for k2 ≤ w1, using Stirling’s formula, we have1k2! ∑p≤f(x)∑β≥11pβk2 ≪ (e log2 f(x) +O(1)k2)k2=(e log2 x/(log3 x)2 +O(1)k2)k2≤ exp(e log2 x(log3 x)2+O(1))= (log x)o(1),uniformly in k2 ≤ w1. Similarly, for k3 ≤ w4, we have1k3!(∑p≤x∑β≥11pβ)k3≤(e log2 x+O(1)k3)k3≪(e log2 x+O(1)w4)w4≪ exp(log2 x log4 xlog3 x)= (log x)o(1).Since the pair (k2, k3) can be chosen in at most O((log2 x)2) different ways,we get that ∑m∈M2(x)1m≪ (log x)o(1), (32)which together with (31) implies our claim (30).23In particular, it follows from (30) that∑p∈R21p= c7.Now, let n ∈ A(x)\ (∪6j=1Aj(x)). Using (29) together with the fact thatn 6∈ A5(x), we see that∑p |nω>y2(p− 1) ≤∑p |n∑q>y2q | p−1v2(q − 1) ≤∑p |nv2(p− 1) +O(1)≤ w3 +O(1)≪ log2 x log3 x. (33)Puty4 = exp(exp((log3 x)3)).Clearly, every prime factor p of n satisfies precisely one of the followingproperties:(i) p ∈ R2 and p ≥ y4;(ii) p < y4;(iii) p lies in the interval [y4, x] but not in R2.Suppose that n has k4 primes of type (i), k5 primes of type (ii), and k6primes of type (iii). Note thatf(y4) = exp(exp(log2 y4(log3 y4)2))= exp(exp((log3 x)39(log4 x)2))> y2if x is sufficiently large. Thus, primes p of type (iii) have the property thatω>y2(p− 1) ≥ ω>f(p)(p− 1) >log2 plog3 p≥ log2 y4log3 y4=(log3 x)39(log4 x)2,From (33), it follows thatk6 · (log3 x)3(log4 x)2≪ log2 x log3 x,and thereforek6 ≪ log2 x (log4 x)2(log3 x)2.24Let c8 be the constant implied in the preceding inequality. Putw5 =log2 xlog3 x, w6 = c8log2 x (log4 x)2(log3 x)2, w7 = w5 − w6,and consider the setA7(x) ={n ∈ A(x)\ (∪6j=1Aj(x)) : ω(n) > w5} .Lemma 6. The following estimate holds:#A7(x)≪ x(log x)2+o(1). (34)Proof. Let R3(x) denote the set of primes p ≤ y4 together with the set ofprimes p ∈ R2(x). Clearly,∑p∈R3(x)∑β≥11pβ≤∑p≤y41p+O(1) ≤ log2 y4 +O(1) = (log3 x)3 +O(1). (35)As before, write n = Pm, where P > max{P (m), y1}, and denote byM3(x)the set of all integers m obtained in this way. Since n ∈ A7(x), we know thatm has k6 ≤ w6 primes p ≤ x that are not in R3(x), andk7 := k4 + k5 ≥ w5 − k6 ≥ w5 − w6 ≥ w7primes in R3(x). Further, k7 ≤ w1 since n 6∈ A3(x). For fixed m ∈ M3(x),the prime number P can be chosen in at mostπ(x/m)≪ xϕ(m)(log(x/m))2≪ x(log2 x)3(log x)2· 1mdifferent ways, where we have again used the inequality (26) together with thefact that ϕ(m) ≫ m/ log2m. Summing this estimate over all m ∈ M3(x),we derive that#A7(x)≪ x(log2 x)3(log x)2∑m∈M3(x)1m, (36)where∑m∈M3(x)1m≤∑k6≤w6w7≤k7≤w11k6!(∑p 1 is fixed, then the function t 7→ (A/t)tis increasing for t ≤ A/e and decreasing for t > A/e. For example, above weused this argument with A = e log2 x+O(1) and t = k6 ≤ w6 < A/e, once xis sufficiently large.On the other hand, for fixed k7 in the interval [w7, w1], using Stirling’sformula again and the estimate (35), we have1k7! ∑p∈R3(x)∑β≥11pβk7 ≤ (e(log3 x)3 +O(1)k7)k7≤(e(log3 x)3 +O(1)w7)w7= exp (−(1 + o(1)) log2 x) = (log x)−1+o(1),where now we have used the fact that (B/t)t is decreasing for the fixedB = e(log3 x)3 +O(1) and t ≥ w7 > B/e, once x is sufficiently large.Since the pair (k6, k7) can be chosen in at most O((log2 x)2) distinctways, we obtain that ∑m∈M3(x)1m≤ 1(log x)1+o(1),which together with (36) leads to the proof of (34).From now on, we consider only integers n ∈ A(x)\ (∪7j=1Aj(x)). If weagain write every such n in the form n = Pm, and let M4(x) be the set ofintegers m that arise in this way, then it follows that∑m∈M4(x)1m≤ (log x)o(1).26Indeed,∑m∈M4(x)1m≪∑k≤w51k!(∑p≤x∑β≥11pβ)k≪ log2 xlog3 x(e log2 x+O(1)w5)w5≤ exp(O(log2 x log4 xlog3 x))= (log x)o(1). (37)We now putA8(x) = {n ∈ A(x)\(∪7j=1Aj(x)) : F (n) > w2},where F (n) is defined by (15).Lemma 7. The following estimate holds:#A8(x)≪ x(log x)2+o(1). (38)Proof. To prove (38), we follow the same arguments used to bound #A5(x).Suppose that we are given an element (t, α, α1, . . . , αt, κ1, . . . , κt) of D andthat this element encodes the powers of 2 in n and in p−1 for the odd primesp | n, as in the proof of the upper bound (16) for #A5(x). Assume furtherthat P = pi for some i ∈ {1, . . . , t}. When all these data are fixed, thenumber of corresponding integers n is bounded above (see inequality (18))byx log2 xlog x· 12α+Ptj=1 αjκj· (c4 log2 x)Ptj=1 κj .Since α +∑tj=1 αjκj > w2 = c2 log2 x, and∑tj=1 κj = ω(m) ≤ w5, it followsthat12α+Ptj=1 αjκj· (c4 log2 x)Ptj=1 κj ≤ exp (w5 log(c4 log2 x)− w2 log 2)= exp(−(1 + o(1)) log2 x) =1(log x)1+o(1).Therefore,#A8(x)≪ #D · log2 x ·x(log x)2+o(1)=x(log x)2+o(1),which is the desired result.27In order to continue our argument, we shall need the following technicalresult:Lemma 8. Uniformly for 2 ≤ d ≤ x, the following estimate holds:∑p∈R2(x)p≡1 (mod d)1p≪ log2 dϕ(d).Proof. Let d ≥ 2. For x ≥ d, we first determine an upper bound on thecounting function of the set R2(x; d, 1) of primes p ∈ R2(x) in the arithmeticprogression p ≡ 1 (mod d). Without loss of generality, we can assume thatx ≥ exp ((log d)2), for otherwise the inequality asserted by the lemma followsfrom the fact that ∑p≤yp≡1 (mod d)1p≪ log2 yϕ(d),which is valid for all 1 ≤ d ≤ y (see again (8)). We can also assume that dis large enough for our purposes at hand. Now, puty = exp(log x log3 x6 log2 x).Then, if d is sufficiently large,y ≥ exp((log d)2 (log3 d+ log 2)12 log2 d)> d.For each prime p ∈ R(x; d, 1), write p − 1 = dm. We first remark that thenumber of primes p for which P (m) ≤ y (and thus, P (p − 1) ≤ y) cannotexceedΨ(xd, y)≤ xdexp (−(1 + o(1))v log v) ,wherev =log(x/d)log y>12· log xlog y=3 log2 xlog3 x.Hence, v log v ≥ (3 + o(1)) log2 x, therefore the number of such primes p is≪ xd(log x)3+o(1)≤ xd(log x)228if x is sufficiently large.Now, consider primes p ∈ R(x; d, 1) such that P (m) > y. From thissubset, we discard those primes p for which there exists a prime q > y suchthat q2 | p − 1; it is easy to see, using an argument similar the one used toanalyze the counting function #R1(x), that the number of discarded primesis≪ xdy log y≪ xd(log x)2.For the remaining primes, write p− 1 = dm1P , where p is a prime satisfyingP = P (p − 1) > max{P (m1), y}. Let M(x) be the set of integers m1 thatoccur in this way. By Brun’s method, for every fixed value of m1, the numberof possibilities for the prime P is≪ xϕ(dm1)(log(x/dm1))2≪ x log2 xϕ(d)(log(x/dm1))2· 1m1.Since x/(dm1) ≥ P (p− 1) > y, we get thatlog(x/dm1) > log y ≫ log x log3 xlog2 x,therefore the number of possibilities for P is≪ x(log2 x)3ϕ(d)(log x)2· 1m1. (39)Recalling that M(x) is contained in the set of those positive integers thathave at most log2 x/ log3 x primes > f(x), the argument used in the analysisof the counting function #R2(x) (see estimate (32)) shows that∑m1∈M(x)1m1≤ (log x)o(1).Therefore, summing up the inequality (39) over all the possible values form1 ∈M(x), and combining the result with our previous estimates, we obtainthe bound:#R2(x; d, 1) ≤ xϕ(d)(log x)2+o(1) w8},then the above argument and the estimate (40) immediately implies that#A9(x) ≤ x(log x)1+log 2+o(1). (44)Finally, we come to the last set A10(x) = A(x)\(∪9j=1Aj(x)). From nowon, we consider only integers n ∈ A10(x).Lemma 10. The following estimate holds:#A10(x) ≤ x(log x)3/2+o(1). (45)Proof. For n ∈ A10(x), write n = Pm, where P = P (n) and m ∈ M4(x).From the inequality (29), we see thatω>y2(P − 1) ≤∑p |nω>y2(p− 1) ≤ v2(n) +∑p |nv2(p− 1) +O(1)= F (n) +O(1).Let c9 be the constant implied by O(1) in this estimate. Let ν = F (n) ≤ w8,and fix the integer ν. Put g(p) = (log p)2, let α ≤ w2 be a positive integer,and letR4,α,ν = {p : 2α ‖ p− 1 and ω>g(p)(p− 1) ≤ ν + c9}.32Then, P = P (n) ∈ R4,α,ν(x/m) for some positive integer α. We now showthat the estimate#R4,α,ν(x/m) ≤ x2αm(log x)h(δ)+o(1)+xm(log x)2+o(1)(46)holds uniformly for all ν ≤ w8, α ≤ w2, and m ∈ M4(x), where we puth(δ) = 2− δ log(e/δ), with δ = ν/ log2 x. To do this, we first observe that itsuffices to prove only the weaker assertion that#R4,α,ν(x) ≤ x2α(log x)h(δ)+o(1)+x(log x)2(47)holds uniformly for all ν ≤ w8 and α ≤ w2. Indeed, since x/m > y1, itfollows that log(x/m) = (log x)1+o(1) and that δm = ν/ log2(x/m) = δ+ o(1),thus (47) implies (46) uniformly for all m ∈ M4(x). To establish (47), let pbe a prime in R4,α,ν(x). As in the estimation of R1(x) and R2(x), we canassume that P (p−1) > y1, that there does not exist a prime q > y3 for whichq2 | p − 1, and that ω(p − 1) < w1. Let us write p − 1 = 2αPm, where mis odd and P > max{P (m), y1}, and let M5(x) be the set of all integers mobtained in this way. Using Brun’s method again, we see that for each fixedvalue of m ∈M5(x), the number of possibilities for the prime p is≪ x2αϕ(m)(log(x/m))2≪ x(log2 x)32α(log x)2· 1m(see inequality (26) again). Every m ∈ M5(x) has k8 ≤ w9 = δ log2 x + c9prime factors larger than y5 = (log x)3, and k9 ≤ w1 remaining primes whichare ≤ y5; therefore,∑m∈M5(x)1m≤∑k8≤w9k9≤w11k8!(∑p≤x∑β≥11pβ)k8· 1k9!(∑p≤y5∑β≥11pβ)k9.For fixed k9 ≤ w1, we have by Stirling’s formula:1k9!(∑p≤y5∑β≥11pβ)k9≤(e log2 y5 +O(1)k9)k9≤(e log3 x+O(1)k9)k9≤ exp (log3 x+O(1))≪ log2 x.33Similarly, for fixed k8 ≤ w9, we have:1k8!(∑p≤x∑β≥11pβ)k8≤(e log2 x+O(1)k8)k8≪(e log2 x+O(1)w9)w9≪ exp (z(δ) log2 x) = (log x)z(δ),where z(δ) = δ log(e/δ). Thus, since the pair (k8, k9) can be chosen in atmost O((log2 x)2) ways, we get that∑m∈M5(x)1m≪ (log x)z(δ)(log2 x)3.Consequently,R4,α,ν(x) ≪ x(log2 x)32α(log x)2∑m∈M5(x)1m+x(log x)2≤ x(log2 x)62α(log x)2−z(δ)+o(1)+x(log x)2=x2α(log x)h(δ)+o(1)+x(log x)2,which establishes (47).Returning to the integers n ∈ A10(x), we see that each one has the formn = Pm, where P > max{P (m), y1} belongs to the set R4,α,ν(x/m), andm ∈ M4(x). For fixed values of m, ν ≤ w8, and α ≤ w2, the number ofpossibilities for P is at most#R4,α,ν(x/m)≪(x2α(log x)h(δ)+o(1)+x(log x)2)1m.Now an argument similar to the one used to prove estimate (40) (and similarto the one used to find upper bounds on A5(x) and A8(x)), leads easily tothe conclusion that in formula (40) we may replace the first exponent 1+o(1)by h(δ) + o(1) once δ < 1, in particular, when we are counting numbers inA10,ν(x) = {n ∈ A10(x) : F (n) = ν}. Thus, summing up over all possiblevalues of α and m, we obtain that#A10,ν(x) ≤ x(log x)h(δ)+δ log 2+o(1), (48)34since h(δ) < 2 for δ < 1. Finally, summing up over all possible values of ν(at most O(log2 x) of them), and noticing that the minimum of the functionh(δ) + δ log 2 = 2− δ log(e/δ) + δ log 2occurs at δ = 1/2 with a value of 3/2, we obtain the stated result.Theorem 2 now follows at once from the estimates (9), (12), (13), (14),(16), (21), (34), (38), (44), and (45).4 The normal order of ϕ(λ(n))/λ(ϕ(n))Theorem 3. The estimateϕ(λ(n))λ(ϕ(n))= exp((1 + o(1))(log2 n)2 log3 n)holds on a set of positive integers n of asymptotic density one.Proof. Clearly,ϕ(λ(n)) =ϕ(λ(n))λ(n)· λ(n).Since the inequalities1 ≥ ϕ(λ(n))λ(n)≫ 1log2 λ(n)≥ 1log2 nhold for all n, and the estimateλ(n) = n exp(−(1 + o(1)) log2 n log3 n) (49)holds for almost all n (see [19]), it follows thatϕ(λ(n)) = n exp(−(1 + o(1)) log2 n log3 n) (50)holds for almost all positive integers n.We also have:λ(λ(n)) ≤ λ(ϕ(n)) = λ(ϕ(n)λ(n)· λ(n))≤ λ(λ(n)) · ϕ(n)λ(n).35Here, we used the fact that the prime factors of λ(n) and ϕ(n) are the same,together with the (easily proved) fact that if m = ab and every prime factorof b divides a, then λ(ab) | λ(a)b. Now, writingϕ(n)λ(n)=ϕ(n)n· nλ(n)and using the estimate 1 ≥ ϕ(n)/n ≫ 1/ log2 n together with (49), we seethatλ(ϕ(n)) = λ(λ(n)) exp((1 + o(1)) log2 n log3 n)holds for almost all n. Applying the result (1) of Martin and Pomerance [36],it follows that the estimateλ(ϕ(n)) = n exp(−(1 + o(1))(log2 n)2 log3 n) (51)holds for almost all positive integers n. The result now follows from (50)and (51).References[1] L. M. Adleman, C. 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Tenenbaum, Introduction to analytic and probabilistic number the-ory , University Press, Cambridge, UK, 1995.[48] P. Tura´n, ‘On a theorem of Hardy and Ramanujan’, J. London Math.Soc. 9 (1934), 274–276.[49] R. Warlimont, ‘On the iterates of Euler’s function’, Arch. Math.(Basel) 76 (2001), 345–349.40IndexIntroduction, 2–6Notation, 6Proof of Theorem 1, 7–10Proof of Theorem 2, 10–35Lemma 1, 11Lemma 2, 11Lemma 3, 12Lemma 4, 15–18Lemma 5, 19–21Lemma 6, 25–26Lemma 7, 27–28Lemma 8, 28–30Lemma 9, 30–32Lemma 10, 32–35Proof of Theorem 3, 35–36Parameters definitions, Theorem 2k1, 18k2, 23k3, 23k4, 24k5, 24k6, 24k7, 25k8, 33k9, 33w1, 14w2, 14w3, 15w4, 23w5, 25w6, 25w7, 25w8, 32w9, 33y1, 12y2, 13y3, 19y4, 24y5, 33A1(x), 12A2(x), 13A3(x), 14A4(x), 14A5(x), 15A6(x), 19A7(x), 25A8(x), 27A10(x), 32A9(x), 32M1, 21M2(x), 22M3(x), 25M4(x), 26M5(x), 33R1, 19R2, 22R3(x), 25R4,α,ν , 3241

Compositions with the Euler and Carmichael Functions

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