On the composition of the Euler function and the sum of divisors

On the composition of the Euler function and the sum of divisors function Jean-Marie De Koninck and Florian Luca ´ Edition du 29 d´ecembre 2006 Abstract σ(φ(n)) Let H(n) = , where φ(n) is Euler’s function and σ(n) φ(σ(n)) stands for the sum of the positive divisors of n. We obtain the maximal and minimal orders of H(n) as well as its average order, and we also prove two density theorems. In particular, we answer a question raised by Golomb.

AMS Classification: 11A25, 11N37, 11N56 Key words: sum-of-divisors function, Euler’s function

§1. Introduction Let φ be Euler’s function and let σ be the sum of the divisors function. The composition of the functions σ and φ has been the object of several studies; see for instance M¸akowski and Schinzel [9], Pomerance [11], S´andor [12], Ford [2], Luca and Pomerance [8]. In 1993, Golomb [3] investigated the difference σ(φ(n)) − φ(σ(n)) showing that it is both positive and negative infinitely often, and asked what is the proportion of each. In this paper, we answer this question of Golomb and more, by studying the behavior of the quotient H(n) :=

σ(φ(n)) . φ(σ(n))

In particular, we obtain the maximal and minimal orders of H(n), its average order, and we also prove two density theorems. Given any positive real number x we write log x for the maximum between the natural logarithm of x and 1. If k is a positive integer, we write logk x for the k-th iteration of the function log x. Throughout this paper, p, 1

q and r stand for prime numbers, while γ stands for Euler’s constant. We also use π(x) for the number of primes up to x and ω(n) for the number of distinct prime factors of n. Acknowledgments. The first author was supported in part by a grant from NSERC. The second author was supported in part by Grants SEPCONACyT 46755, PAPIIT IN104505 and a Guggenheim Fellowship. The authors wish to thank the referee for helpful remarks and suggestions which improved the quality of this paper. §2. Main results Theorem 1. The maximal order of H(n) is e2γ log22 n, that is lim sup n→∞

H(n) = e2γ . log22 n

Theorem 2. There exists a positive constant δ such that the minimal order of H(n) is δ/ log2 n, that is lim inf H(n) log2 n = δ. n→∞ Moreover δ ∈ [(1/40)e−γ , 2e−γ ]. Theorem 3. As x → ∞, ´ ³ 1X 3/2 H(n) = c0 e2γ log23 x + O log3 x , x n≤x

where c0 = =

1 X φ(n) x→∞ x n≤x σ(n) lim

Y p

Ã

∞ 3 1 (p − 1)3 X 1 1− + 2 + 2 i p(p + 1) p (p + 1) p i=3 p − 1

2

!

≈ 0.4578.

Theorem 4. For each number u, 0 ≤ u ≤ 1, the asymptotic density of the set of numbers n with H(n) > ue2γ log23 n exists, and this density function is strictly decreasing, varies continuously with u, and is 0 when u = 1. In particular, Theorem 4 shows that σ(φ(n)) − φ(σ(n)) is positive for most n, thus providing an answer to Golomb’s question. Theorem 5. The set {H(1), H(2), H(3), . . .} is dense in [0, +∞). §3. Preliminary results Theorem A (Heath-Brown). Let k and a be coprime positive integers. Then there exists a prime number p ≡ a (mod k) which satisfies p = O(k 11/2 ). Proof. See Heath-Brown [6]. Remark. It has been shown by Alford, Granville and Pomerance [1] that for most values of k, one can replace the constant 11/2 by 12/5 + ε for any fixed ε > 0. It can also be shown that if GRH holds, then the constant 11/2 can be replaced by 2 + ε for any fixed ε > 0. Theorem B. (Pomerance) There exists a constant κ > 0 such that, for all positive integers n, σ(φ(n)) > κ. n Proof. See Pomerance [11]. Remark. This statement relates to a long standing conjecture of M¸akowski σ(φ(n)) 1 and Schinzel [9], which asserts that ≥ . Recently, Ford [2] has n 2 1 . Note also that the conjectured minimum 21 is attained shown that κ ≥ 39.4 when n is a product of the first Fermat primes, such as n = 2, 6, 30, 510, 131070 and 8589934590. 3

Lemma 1. (Mertens’ Theorem) The estimate Y p≤x

Ã

1 1− p

!

Ã

Ã

e−γ 1 = 1+O log x log x

!!

holds for large values of x. Proof. This result is due to Mertens [10]. φ(n) log2 n = e−γ . n→∞ n Proof. This result, which follows essentially from Mertens’ Theorem, was first obtained by Landau [7]. Lemma 2. lim inf

Lemma 3. lim sup n→∞

σ(n) = eγ . n log2 n

Proof. This result also follows from Mertens’ Theorem and was first obtained by Gronwall [4]. Lemma 4. There exists a positive constant c1 such that for large real numbers x, both φ(n) and σ(n) are divisible by all prime powers pa < c1 log2 x/ log3 x for all positive integers n < x with O(x/ log23 x) exceptions. Proof. The above result for the case of the function φ(n) is Lemma 2 in [8]. To prove the result for the function σ(n), let m be an arbitrary positive integer and write X 1 . S(x, m) = log2 x≤q≤x q m|(q+1)

From the Siegel-Walfisz Theorem (see Theorem 5, Chapter II.8 in Tenenbaum [13]) and partial summation, it follows that there exist positive numbers c1 and x0 such that the inequality S(x, m) ≥

c1 log2 x φ(m)

holds for x > x0 and all m ≤ log x. Let g(x) = c1 log2 x/ log3 x. Using Brun’s sieve, it follows that the set Nm of numbers n ≤ x which have no

4

prime factor q > log2 x congruent to −1 modulo m satisfies Ã

Y

#Nm < c2 x

log2 x x0 , we get that if m = pa < g(x), then c2 x c2 x c2 x < = . a exp(S(x, p )) exp(log3 x) log2 x

#Npa
log2 x

5

1 , p

then (2)

X

hφ (n) ¿

n≤x

x log3 x

and

X

hσ (n) ¿

n≤x

x . log3 x

Proof. Clearly we have X

X x x x x log2 x x log2 x + ¿ 2+ ¿ . 2 p p p p q≤x q

1≤

n≤x p|φ(n)

p|q−1

It now follows that X

hφ (n) =

n≤x

X1 X p≤x

p

1 ¿ x log2 x

X p>log2

n≤x p|φ(n)

1 x ¿ , 2 log3 x x p

where we used (1) with z := log2 x, thus establishing the first assertion in (2). We use a similar argument to establish the second assertion in (2). First of all, note that since ω(n) < log x holds for all n ≤ x provided x is large enough, it follows that hσ (n) ≤

X

1 ¿ log3 x, i≤log x pi

where we used pi to denote the i-th prime number. Let N1 be the set of all positive integers n ≤ x such that there exists a prime q > log23 x whose square divides n. Then, using (1), #N1 ≤

X

x x ¿ . 2 2 q log3 x log4 x q>log2 x 3

Hence, (3)

X

hσ (n) ¿ #N1 log3 x ¿

n∈N1

x . log3 x log4 x

Now let N2 be the set of those n ≤ x which are not in N1 and which are divisible by a prime power q a , with a = bc3 log4 xc + 2, where c3 := 2/ log 2. For a fixed prime number q, the number of such numbers n is ≤ x/q a , and therefore Z ∞ X x x x x dt , ≤ a +x ¿ a ¿ #N2 ≤ a a 2 t 2 2 log23 x q≥2 q 6

which implies that X

(4)

hσ (n) ¿ #N2 log3 x ¿

n∈N2

x . log3 x

Finally, let N3 be the set of positive integers n ≤ x which do not belong to either N1 or N2 . If n ∈ N3 and q αq ||n with αq > 1, then q < log23 x and αq ¿ log4 x, so that q αq ≤ exp(O(log24 x)). Hence σ(q αq ) < exp(O(log24 x)). In particular, for large x, we have that σ(q αq ) < log2 x. Hence, if n ∈ N3 and p > log2 x is a prime dividing σ(n), it follows that there exists a prime factor q||n of n such that p|(q + 1). Now the same argument used for the function hφ tells us that if p > log2 x is a fixed prime, then X

X x

1¿

q≤x p|q+1

p|σ(n) n∈N3

q

¿

x log2 x . p

Therefore X

(5)

X

hσ (n) ≤

n∈N3

p>log2

X 1 X 1¿ p≥log x p p|σ(n)

2

n∈N3

x log2 x x ¿ . 2 p log3 x x

The second estimate (2) then follows from estimates (3), (4) and (5), and the proof of Lemma 5 is complete. Lemma 6. As x → ∞, X φ(n) n≤x

σ(n)

= c0 x + O(x3/4 ),

where c0 is the constant appearing in the statement of Theorem 3. Proof. Given any number s with 1 and letting ζ(s) stand for the Riemann Zeta Function, we have ∞ X φ(n)/σ(n) n=1

ns



=

Y

1 +

p

= ζ(s)

Y p

Ã

p−1 p+1 ps

+

p(p−1) p2 +p+1 p2s



!

+

p3s

p−1 1 Y 1− s 1 + p+1 + p ps p

7



p2 (p−1) p3 +p2 +p+1

+ . . .

p(p−1) p2 +p+1 p2s

+

p2 (p−1) p3 +p2 +p+1 p3s



+ . . .



= ζ(s)

Y

p−1 p+1

1 +

−1

+

ps

p

p(p−1) − p−1 p2 +p+1 p+1 p2s

p2 (p−1)

+

p3 +p2 +p+1

−

p(p−1) p2 +p+1

p3s

= ζ(s)R(s), say. Expanding the product R(s) into a Dirichlet series, say R(s) =

∞ X an n=1

ns

, 3 . 4

then it converges absolutely in the half-plane x

Since

X d≤x

and

X |ad | d>x

it follows that

d

=

|ad | =



+O



X

|ad | .

d≤x

X |ad | d3/4 = O(x3/4 ) 3/4

d≤x

X |ad | 1 d>x



X ad X x =x + O  |ad | d d≤x d d≤x

n≤x

= R(1)x + O x



· ¸

d3/4 d1/4

X φ(n) n≤x σ(n)

d

≤ x−1/4

X |ad | d>x

d3/4

³

´

= O x−1/4 ,

= R(1)x + O(x3/4 ),

which completes the proof of Lemma 6, since R(1) = c0 . Lemma 7. There exists a constant c4 such that the set of positive integers n ≤ x such that ω(φ(n)) > c4 log32 x contains at most O(x/ log22 x) elements. The same holds when the φ function is replaced by the σ function. Proof. First let D1 be the set of all n ≤ x such that k = ω(n) > 3e log2 x. A well-known result of Hardy and Ramanujan (see [5]) asserts that #{n ≤ x : ω(n) = k} ¿

1 x · · (log2 x + O(1))k−1 , log x (k − 1)! 8



+ . . .

an inequality which together with Stirling’s formula implies that Ã

x e log2 x + O(1) #{n ≤ x : ω(n) = k} ¿ · log x k−1

!k−1


3e log log x − 1 and x is assumed to be large. Thus, x X 1 x x ¿ ¿ . k log x k 2 log x log22 x

#D1 = #{n ≤ x : ω(n) > 3e log2 x} ¿

Assume now that D2 is the set of all n ≤ x which are divisible by the square of a prime p > log22 x. Then X x x #D2 ≤ ¿ . 2 p log22 x p>log2 x 2

Let D3 be the set of those n ≤ x which are divisible by a prime number p such that ω(p − 1) ≥ b := be2 log2 xc. Then  X

#D3 ≤

p≤x ω(p−1)≥b

¿ x

X k≥b

Ã

k

X 1 X 1 x   ≤x a p k≥b k! q a ≤x q

e log2 x + O(1) k

!k

¿x

X 1 k≥b

2k

¿

x x x ¿ ¿ , 2b log x log22 x

2

where we used the facts that e > 2 and 2e > e. Let D4 be the set of those n ≤ x which are divisible by a prime number p such that ω(p + 1) ≥ b. The same argument as above shows that x #D4 ¿ . log22 x Let D5 be the set of those n ≤ x which do not belong to D1 ∪ D2 ∪ D3 ∪ D4 and such that there exists a prime power pa |n, where a = bc3 log3 xc, where c3 = log2 2 . By an argument similar to the one used in the proof of Lemma 5, we get that Z ∞ x x x dt ¿ a +x ¿ a ¿ . #D5 ≤ x a a 2 t 2 2 log22 x p≥2 p X 1

Thus, D = ∪5i=1 Di contains O(x/ log22 x) elements, as claimed.

9

§4. The maximal order of H(n) We will show that for n sufficiently large, H(n) ≤ (1 + o(1)) e2γ log22 n.

(6)

Then clearly the proof of Theorem 1 will follow if we can also show the following result. Claim. There exists an infinite sequence of integers n for which H(n) is bounded below by (1 + o(1)) e2γ log22 n. To prove (6), first observe that it follows from Lemma 2 that (7)

σ(φ(n)) ≤ (1 + o(1))eγ φ(n) log2 φ(n) ≤ (1 + o(1))eγ n log2 n.

On the other hand, it follows from Lemma 1 that φ(n) ≥ (1 + o(1))

e−γ n , log2 n

so that (8)

φ(σ(n)) ≥ (1 + o(1))

e−γ σ(n) n ≥ (1 + o(1))e−γ . log2 σ(n) log2 n

Combining (7) and (8), we obtain (6). Hence, in order to complete the proof of Theorem 1, it remains to prove our Claim. So let x be a large integer, and let P and Q be the smallest primes such that P ≡ 1 (mod M (x))

and

Q ≡ −1 (mod M (x)),

where M (x) = LCM[1, 2, . . . , x], and set n = P Q. From the Prime Number Theorem, it is clear that M (x) = e(1+o(1))x < e2x , 10

say. Hence, from Theorem A, it follows that P ¿ e11x ,

Q ¿ e11x , so that n = P Q ¿ e22x .

Thus, n < e23x holds for large x. For this particular integer n, we have, since φ(n) = (P − 1)(Q − 1), σ(φ(n)) = φ(n)

Ã

Y

1 1 1 1 + + 2 + . . . + αp p p p

pαp k(P −1)(Q−1)

Ã

Y

≥

1 1 1 1 + + 2 + . . . + αp p p p

pαp k(P −1)

Y

≥

Ã

pβp ≤x

1 1 1 1 + + 2 + . . . + βp p p p

!

!

!

,

where each exponent βp is the unique positive integer satisfying pβp ≤ x < pβp +1 . Therefore, Ã

Y σ(φ(n)) 1 ≥ 1+ φ(n) p−1 p≤x

(9)

!

Ã

Y

Ã

1+O

pβp ≤x

1

!!

.

pβp +1

However, Y pβp ≤x

Ã

Ã

1+O

 

!!

1





(

Ã

π(x) 1  O = exp = exp O  β +1   pp x pβp ≤x

pβp +1

X

Ã

!

1 . = 1+O log x Using this in (9), we obtain that, by Lemma 1, (10)

Y p σ(φ(n)) ≥ (1 + o(1)) = (1 + o(1))eγ log x. φ(n) p≤x p − 1

On the other hand, σ(n) = (P + 1)(Q + 1), so that (11)

φ(σ(n)) = σ(n) ≤

Ã

Y p|(P +1)(Q+1)

Y p|M (x)

Ã

1 1− p

1 1− p !

=

!

≤

Y p≤x

11

Ã

Y

Ã

p|(P +1)

1 1− p

!

1 1− p

!

e−γ = (1 + o(1)) , log x

!)

where we used Lemma 1. Gathering (10) and (11), we get that (12)

H(n) ·

σ(n) ≥ (1 + o(1))e2γ log2 x. φ(n)

Since by our choice of n, we have exp{(1+o(1))x} < n < exp{23x}, it follows that (1 + o(1))x < log n < 23x and therefore that log2 n = log x + O(1), which means that (12) can be replaced by (13)

H(n) ·

σ(n) ≥ (1 + o(1))e2γ log22 n. φ(n)

Observing now that, for large x (that is, large P and Q), σ(n) (P + 1)(Q + 1) = = 1 + o(1), φ(n) (P − 1)(Q − 1) we conclude that our Claim follows immediately from (13), since then by varying x one obtains infinitely many such integers n. The proof of Theorem 1 is thus complete. §5. The minimal order of H(n) It follows from Theorem B and Lemma 3 that, for n sufficiently large, σ(φ(n)) >κ n

n (1 + o(1))e−γ ≥ . σ(n) log2 n

and

Combining these with the trivial inequality get that (14)

H(n) log2 n =

σ(n) ≥ 1, we immediately φ(σ(n))

σ(n) n σ(φ(n)) · · · log2 n ≥ e−γ κ. n φ(σ(n)) σ(n)

To complete the proof of Theorem 2, we shall use an argument developed by M¸akowski and Schinzel in [9].

12

Q

Let x be large and let N (x) = p e2 log2 x cannot hold for some prime divisor p of n because n is not in D3 . Write n = n0 · n00 , where n0 =

Y

pαp and n00 =

pαp ||n αp >1

Y

p.

p||n

Since n 6∈ D, the same argument as above shows that ω(σ(n00 )) ¿ log22 x. Finally, note that n0 < exp(O((log3 x)π(log22 x)) = exp(O(log22 x)), so that σ(n0 ) < exp(O(log22 x)), which shows that ω(σ(n0 )) = o(log22 x). Thus, if n ∈ E, then ω(φ(n)) and ω(σ(n)) are both O(log22 x). In particular, for large x, we have that X

max{hφ (n), hσ (n)} ≤ log2

1 ¿ 1. p 3 x xx >

eγ δ log x 2

holds for all x sufficiently large, we get that k+1 Yµ i=1

1 1+ ri

Ã

¶

∈

!

eγ δ log x eγ δ log x , +1 . 2 2

Hence, we can take R = {ri : i = 1, . . . , k + 1}. Note that since k+1 Yµ

1+

i=1

1 ri

¶

= exp(log2 rk+1 − log2 r1 + o(1)) > exp(log2 rk+1 − 3 log x), 4

it follows that the inequality rk+1 < ex holds for large x, for if not, then 4 rk ≥ ex /2, in which case k µ Y i=1

1 1+ ri

¶

> exp(log2 rk − log2 r1 + o(1)) > exp(log x) = x >

eγ δ log x +1 2

which contradicts the definition of k. We now let y be a parameter that depends on x and such that z := 4 log2 y > rk+1 . This inequality is fulfilled if we choose log2 y > ex , which in 21

turn holds if log3 y > x4 . Then let P be the set of all primes p ≤ y such that p ≡ 13 (mod 72), p ≡ 1 + ri (mod ri2 ) for all i = 1, . . . , k + 1, and both p − 1 and p + 1 are coprime to all primes r ≤ z which are ≥ 5 and which do not belong to R. Observe that the above conditions certainly put p in an arithmetic progression a (mod b), where b = 72

k+1 Y

ri2 ,

i=1

and a ≡ 13 (mod 72) and a ≡ 1 + ri (mod ri2 ) for i = 1, . . . , k + 1. Now let Y T := r, 5≤r≤z r6∈R

and, for each d|T , let A(d) := {ad

(mod bd) : d|a2d − 1 and ad ≡ a

(mod b)},

so that #A(d) = 2ω(d) . By the principle of inclusion and exclusion, the cardinality of the set of primes P is none other than X

µ(d)

d|T

X

π(y; ad , bd),

ad ∈A(d)

where, as usual, π(y; s, t) stands for the number of primes p ≤ y satisfying Q p ≡ s (mod t). Observing that bT ¿ r≤z r2 ≤ e2(1+o(1))z < e3z < y 1/3 , we get, by the Bombieri-Vinogradov Theorem, that Ã

!

µ ¶ π(y) Y 2π(z) y 2 P= . 1− +O φ(b) 5≤r≤z r−1 log10 y r6∈R

Since 2π(z) = exp(O(log2 y/ log3 y)) = (log y)o(1) , while φ(b) ¿

Y

r2 < exp(3 log2 y) = log3 y,

r≤z

22

and since Yµ r≤z



¶



 X 2  2 1 1 1− À exp − À exp {−2 log log z} = ≥ , 2   r−1 log y log z r≤z r − 1

it follows that

Ã

µ ¶ 2 y π(y) Y 1− +O P= φ(b) 5≤r≤z r−1 log9 y

!

À

π(y) y À . 4 log y log5 y

r6∈R

0

Finally, let P be the subset of those primes p ∈ P such that neither of ω(p − 1) or ω(p + 1) is larger than e2 log2 y. From the estimates due to Hardy and Ramanujan (see [5] and the proof of Theorem 3), we know that #{n ≤ y : ω(n) > e2 log2 y} ¿

y log y

y ¿ log y

X k>e2 log2

1 (log2 y + O(1))k (k − 1)! y

X k>e2 log2 y

Ã

e log2 y + O(1) k

!k

!

Ã

y 1 y ¿ , · e2 log y = o 2 log y 2 log5 y because e2 log 2 + 1 > 5. Thus, #P 0 À

y . log5 y

In particular, P 0 is non empty for large y. Select P in P 0 and let n = Q N (x)P , where N (x) = pz

Ã

= eγ log2 x 1 + O

Ã

1 log x

!!

23







γ

e δ log x  X · · exp  O  2

r|(P −1) r>z

1   . r

Noting that P − 1 has no more that e2 log2 y prime factors, it follows easily that X r|(P −1) r>z

1 r

X

≤

log2 y