NUMBERS WHICH ARE ORDERS ONLY OF CYCLIC GROUPS
PAUL POLLACK
Abstract. We call n a cyclic number if every group of order n is cyclic. It is implicit in work of Dickson, and explicit in work of Szele, that n is cyclic precisely when gcd(n, φ(n)) = 1. With C(x) denoting the count of cyclic n ≤ x, Erd˝osproved that C(x) ∼ e−γ x/ log log log x, as x → ∞. We show that C(x) has an asymptotic series expansion, in the sense of Poincar´e,in descending powers of log log log x, namely ! e−γ x γ γ2 + 1 π2 γ3 + 1 γπ2 + 2 ζ(3) 1 − + 12 − 4 3 + ... . log log log x log log log x (log log log x)2 (log log log x)3
1. Introduction Call the positive integer n cyclic if the cyclic group of order n is the unique group of order n. For instance, all primes are cyclic numbers. It is implicit in work of Dickson [Dic05], and explicit in work of Szele [Sze47], that n is cyclic precisely when gcd(n, φ(n)) = 1, where φ(n) is Euler’s totient. (In fact, this criterion had been stated as “evident” already by Miller in 1899 [Mil99, p. 235].) If C(x) denotes the count of cyclic numbers n ≤ x, Erd˝os proved in [Erd48] that
(1) C(x) ∼ e−γx/ log log log x, as x → ∞, where γ is the Euler–Mascheroni constant. Thus, the relative frequency of cyclic numbers decays to 0 but “with great dignity” (Shanks).
Several authors have investigated analogues of (1) for related counting functions from enumerative group theory. See, for example, [May79, MM84, War85, Sri87, EMM87, EM88, NS88, Sri91, NP18]. Our purpose in this note is somewhat different; we aim to refine the formula (1). Begunts [Beg01], optimizing the method of [Erd48], showed that C(x) is given by e−γx/ log log log x up to a multiplicative error of size 1 + O(log log log log x/ log log log x) (the same result appears as Exercise 2 on p. 390 of [MV07]). We improve this as follows.
Theorem 1.1. The function C(x) admits an asymptotic series expansion, in the sense of Poincar´e (see [dB81, §1.5]), in descending powers of log log log x. Precisely: There is a sequence of real numbers c1, c2, c3,... such that, for each fixed positive integer N and all
2010 Mathematics Subject Classification. Primary 11N37; Secondary 20D60. The author is supported by the National Science Foundation (NSF) under award DMS-2001581. 1 2 PAUL POLLACK large x, e−γx c c c C(x) = 1 + 1 + 2 + ··· + N log log log x log log log x (log log log x)2 (log log log x)N x + O . N (log log log x)N+2
Our proof of Theorem 1.1 yields the following explicit determination of the constants ck. Write the Taylor series for the Γ-function, centered at 1, in the form Γ(1 + z) = 2 1 + C1z + C2z + ... . Then the coefficients c1, c2,... are determined by the formal relation 2 3 2 3 1 + c1z + c2z + c3z + ··· = exp(0!C1z + 1!C2z + 2!C3z + ... ).
For computations of the Ck and ck, it is useful to recall that ∞ ! X (−1)k (2) Γ(1 + z) = exp −γz + ζ(k)zk . k k=2 (This is one version of a well-known expansion for the digamma function; see, e.g., entries 5.7.3 and 5.7.4 in [OLBC10].) The first few ck are given by 1 π2 1 2 c = −γ, c = γ2 + ζ(2) = γ2 + , c = − γ3 + γπ2 + ζ(3) . 1 2 2 12 3 4 3
Owing to (2), each ck belongs to the ring Q[γ, ζ(2), ζ(3), . . . ζ(k)]. From the fact that the coefficients of log Γ(1 + z) are alternating in sign, one deduces that both the Ck and the ck are alternating as well. Moreover,
|ck| ≥ (k − 1)!|Ck| ≥ (k − 1)!ζ(k)/k ≥ (k − 1)!/k 2 for each k ≥ 2. It follows that the series 1 + c1/ log log log x + c2/(log log log x) + ... is purely an asymptotic series, in that it diverges for all values of x. The proof of Theorem 1.1 has many ingredients in common with the related work cited above (see also [PP, Pol]). But we must be more careful about error terms than in earlier papers, and somewhat delicate bookkeeping is required to wind up with a clean result.
Notation. The letters p and q are reserved for primes. We use K0,K1,K2, etc. for absolute positive constants. To save space, we write logk for the kth iterate of the natural logarithm.
2. Lemmata
We will use Mertens’ theorem in the following√ form, which is a consequence of the prime number theorem with the classical x exp(−K0 log x) error estimate of de la Vall´eePoussin.
Lemma 2.1. There is an absolute constant c such that, for all X ≥ 3, X 1 = log X + c + O(exp(−K plog X)). p 2 1 p≤X Moreover, for all X ≥ 3, Y 1 e−γ 1 − = 1 + O(exp(−K plog X)) . p log X 2 p≤X NUMBERS WHICH ARE ORDERS ONLY OF CYCLIC GROUPS 3
The following sieve result is a special case of [HR74, Theorem 7.2].
Lemma 2.2. Suppose that X ≥ Z ≥ 3. Let P be a set of primes not exceeding Z. The number of n ≤ X coprime to all elements of P is Y 1 1 log X X 1 − 1 + O exp − . p 2 log Z p∈P
The final estimate of this section was proved independently by Pomerance (see Remark 1 of [Pom77]) and Norton (see the Lemma on p. 699 of [Nor76]).
Lemma 2.3. For every positive integer m and every X ≥ 3, X 1 log X log (2m) = 2 + O . p φ(m) φ(m) p≤X p ≡ 1 (mod m)
3. Proof of Theorem 1.1 3.1. Outline. We summarize the strategy of the proof, deferring the more intricate calcu- lations to later sections. Put log2 x p y = and z = (log2 x) · exp( log3 x). 2 log3 x
Let us call the prime p a standard divisor of gcd(n, φ(n)) if there is a prime q ≤ x1/ log2 x dividing n with q ≡ 1 (mod p). Clearly, each standard divisor of (n, φ(n)) is a divisor of gcd(n, φ(n)).
Let S0 be the set of n ≤ x with no prime factor in [2, y]. For each positive integer k, let Sk be the set of n ∈ S0 having exactly k distinct prime factors from the interval (y, z], all of which divide n to the first power only, and at least one of which is a standard divisor of gcd(n, φ(n)). We will estimate C(x) by [ X (3) # S0 \ Sk = #S0 − #Sk.
1≤k≤log3 x 1≤k≤log3 x
Suppose n is counted by C(x) but not by (3). Then n has a prime factor p ≤ y. Since n is counted by C(x), it must be that p - φ(n), so that n is not divisible by any q ≡ 1 (mod p). Q By Lemma 2.2, for a given p the number of those n ≤ x is x q≤x, q≡1 (mod p)(1 − 1/q) ≤ P x exp(− q≤x, q≡1 (mod p) 1/q). And by Lemma 2.3, X 1 1 = log x + O(1) ≥ 2 log x + O(1). q p − 1 2 3 q≤x q≡1 (mod p) 2 Thus, the number of n corresponding to a given p is x exp(−2 log3 x) = x/(log2 x) . Summing on p ≤ y, we deduce that the total number of n counted by C(x) but not (3) is O(x/ log2 x). Working from the opposite side, suppose that n is counted by (3) but not by C(x). Then at least one of the following holds: 4 PAUL POLLACK
(i) there is a prime p > y for which p2 | n, (ii) there is a prime p > z that divides n and φ(n), (iii) there is a prime p in (y, z] dividing n and a prime q ≡ 1 (mod p) dividing n with q > x1/ log2 x, (iv) n has more than log3 x different prime factors in (y, z]. P 2 The number of n ≤ x for which (i) holds is x p>y 1/p x/y log y x/ log2 x. In order for (ii) to hold but (i) to fail, there must be a prime q ≡ 1 (mod p) dividing n. Clearly, there are most x/pq such n corresponding to a given p, q. Thus, the number of n that arise this way is X 1 X 1 X log x + log p x log x x x x 2 2 = . p q p2 z p p>z q≤x p>z exp( log3 x) q≡1 (mod p) For similar reasons, the number of n ≤ x for which (iii) holds is
X 1 X 1 X log3 x log3 x x x 2 x . p q p log2 x y k>log3 x y Collecting estimates, we conclude that [ p C(x) = # S0 \ Sk + O(x/ exp( log3 x)). 1≤k≤log3 x N+2 Since the error term is ON (x/(log3 x) ) for any fixed N, for the sake of proving Theorem S 1.1 we may replace C(x) by #(S0 \ Sk). 1≤k≤log3 x In §3.2 we prove suitable estimates for the numbers #Sk and in §3.3 we tie everything together and complete the proof of Theorem 1.1. 3.2. Estimating #Sk. The case k = 0 is easy to dispense with. By Lemmas 2.1 and 2.2, x (4) #S = e−γ + O(x/ exp(K plog x)). 0 log y 4 3 Now suppose that 1 ≤ k ≤ log3 x. In order for the integer n ≤ x to be counted by Sk, it is necessary and sufficient than n = p1 ··· pkm, where (a) p1, . . . , pk are distinct primes belonging to (y, z], (b) the integer m is free of prime factors in [2, z], and (c) m has a prime 1/ log x factor q ≤ x 2 with q ≡ 1 (mod pi) for some i = 1, 2, . . . , k. Fix distinct primes p1, . . . , pk ∈ (y, z]. We will count the number of n ∈ Sk for which p1, . . . , pk are the prime divisors of n in (y, z]. To get at this, we count all n = p1 . . . pkm ≤ x NUMBERS WHICH ARE ORDERS ONLY OF CYCLIC GROUPS 5 where condition (b) holds and then subtract the contribution from n for which (b) holds but (c) fails. By Lemma 2.2, this is approximately ! x Y 1 Y 1 (5) 1 − 1 − 1 − . p1 ··· pk p q p≤z z Now collect estimates. We find that the number of n ∈ Sk where p1, . . . , pk are the prime divisors of n from (y, z] is −γ k ! ! e 1 Y exp(−(log x)/pi) x (6) x − 2 + O . log z p ··· p p p 1 k i=1 i p1 ··· pk exp(K5 log3 x) Finally, we sum (6) over all sets of distinct primes p1, . . . , pk ∈ (y, z]. The O-terms contribute p O(x/ exp(K5 log3 x)). Next we look at the contribution from the 1/p1 ··· pk terms. On the one hand, the multinomial theorem immediately implies that X 1 1 k X 1 ≤ σ0 , where σ0 := . p1 ··· pk k! p y We can estimate the sum over p1, . . . , pk−1 in a similar way. Iterating, we find that k−1 2 k X 1 Y i 2(log3 x) ≥ σ0 − ≥ σ0 − , p1 ··· pk y log2 x p1,...,pk∈(y,z] i=0 distinct so that 2 k X 1 1 2(log3 x) ≥ σ0 − . p1 ··· pk k! log2 x y Qk The contribution from the terms of the form i=1 exp(−(log2 x)/pi)/pi can be handled similarly. Put X exp(−(log x)/p) σ := 2 . 1 p y k 4 X Y exp(−(log2 x)/pi) 1 k 1 (log3 x) = σ1 + O . pi k! k! log2 x y Piecing everything together, we conclude that k k 4 ! −γ x σ0 σ1 x x (log3 x) (7) #Sk = e − + O p + . log z k! k! exp(K5 log3 x) k! log2 x 3.3. Denouement. Summing (7) over positive integers k ≤ log3 x, keeping in mind that p σ0, σ1 1/ log3 x, we find that ! X −γ x x #Sk = e (exp(σ0) − exp(σ1)) + O p . log z exp(K6 log3 x) 1≤k≤log3 x log z p By Mertens’ theorem, exp(σ0) = log y 1 + O(1/ exp(K7 log3 x)) . So recalling (4), X x #S − #S = e−γ exp(σ ) + O(x/ exp(K plog x)). 0 k log z 1 8 3 1≤k≤log3 x By another application of the prime number theorem with the de la Vall´eePoussin error term, Z z Z z exp(−(log2 x)/t) exp(−(log2 x)/t) p σ1 = dθ(t) = dt + O(1/ exp(K9 log3 x)), y t log t y t log t and thus (8) Z z X −γ x exp(−(log2 x)/t) p #S0 − #Sk = e exp dt + O(x/ exp(K10 log3 x)). log z y t log t 1≤k≤log3 x We proceed to analyze the integral appearing in this last estimate. Making the change of variables u = (log2 x)/t, −1 Z z exp(−(log x)/t) 1 Z log3 x exp(−u) log u 2 dt = 1 − du. t log t log x u log x y 3 (log2 x)/z 3 p p Here (log2 x)/z = exp(− log3 x). Inside the domain of integration, log u log3 x, and so for each fixed positive integer M, −1 2 M log u log u log u log u −(M+1)/2 1 − = 1 + + + ··· + + OM ((log3 x) ). log3 x log3 x log3 x log3 x Thus, −1 1 Z log3 x exp(−u) log u 1 − du log x u log x 3 (log2 x)/z 3 M X 1 Z log3 x exp(−u) = logk u du (log x)k+1 u k=0 3 (log2 x)/z ! 1 Z log3 x exp(−u) + O du . (log x)(M+3)/2 u 3 (log2 x)/z 8 PAUL POLLACK 1 1 − (M+3) R log3 x −1− M The O-term here is (log3 x) 2 du/u (log3 x) 2 . To handle the (log2 x)/z main term, we integrate by parts to find that u=log3 x Z log3 x exp(−u) logk+1 u logk u du = exp(−u) u k + 1 (log2 x)/z u=(log2 x)/z 1 Z log3 x + exp(−u) logk+1 u du. k + 1 (log2 x)/z For each 0 ≤ k ≤ M, and all large x, u=log3 x logk+1 u −1 log xk+1 2 p exp(−u) = log + OM (1/ exp(K11 log3 x)), k + 1 k + 1 z u=(log2 x)/z while 1 Z log3 x exp(−u) logk+1 u du k + 1 (log2 x)/z Z ∞ 1 k+1 p = exp(−u) log u du + OM (1/ exp(K12 log3 x)) k + 1 0 1 = Γ(k+1)(1) + O (1/ exp(K plog x)) k + 1 M 12 3 p = k!Ck+1 + OM (1/ exp(K12 log3 x)). Assembling our results, Z z exp(−(log x)/t) 2 dt y t log t M k+1 M X 1 log((log2 x)/z) X k!Ck+1 −1− 1 M = − + + O ((log x) 2 ) k + 1 log x (log x)k+1 M 3 k=0 3 k=0 3 M log((log2 x)/z) X k!Ck+1 −1− 1 M = log 1 − + + O ((log x) 2 ) log x (log x)k+1 M 3 3 k=0 3 M log z X k!Ck+1 −1− 1 M = log + + O ((log x) 2 ). log x (log x)k+1 M 3 3 k=0 3 We now choose M = 2N, where N is as in Theorem 1.1. In the last displayed sum on k, the terms of the sum with k ≥ N may be absorbed into the error. Doing so and exponentiating, Z z exp(−(log x)/t) exp 2 dt y t log t log z X (k − 1)!Ck = exp 1 + O ((log x)−1−N ) , log x (log x)k N 3 3 1≤k≤N 3 NUMBERS WHICH ARE ORDERS ONLY OF CYCLIC GROUPS 9 so that x Z z exp(−(log x)/t) e−γ exp 2 dt log z y t log t x X (k − 1)!Ck = e−γ exp 1 + O ((log x)−1−N ) log x (log x)k N 3 3 1≤k≤N 3 x X (k − 1)!Ck = e−γ exp + O (x(log x)−2−N ). log x (log x)k N 3 3 1≤k≤N 3 S This expression describes #(S0 \ Sk), by (8), and so also describes C(x), by 1≤k≤log3 x the discussion in §3.1. Theorem 1.1 follows, along with the description of the constants ck appearing in the introduction. References [Beg01] A. V. Begunts, On an asymptotic formula for F1(x), Vestnik Moskov. Univ. Ser. I Mat. Mekh. (2001), no. 5, 57–59, 71–72 (Russian). [dB81] N. G. de Bruijn, Asymptotic methods in analysis, third ed., Dover Publications, Inc., New York, 1981. [Dic05] L. E. 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