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(i) If the Riemann Hypothesis is true, then, for each n 5041, k ≥ nk γ > e log log nk ϕ(nk) for all k 1. ≥ (ii) If the Riemann Hypothesis is false, then,

nk γ nk γ < e log log nk and > e log log nk ϕ(nk) ϕ(nk) occur for infinitely many k 1. ≥ The lower and upper bounds for almost every integer are significantly smaller by an iterated factor of log as demonstrated below.

Theorem 1.4. For almost all integers n 1, the ratio n/ϕ(n) has the followings bounds. ≥ (i) n log log log n . ≪ ϕ(n)

(ii) n log log log n. ϕ(n) ≪

Theorem 1.5. Let x 1 be a large number. Then, the Euler totient function has the followings bounds. ≥

(i) There is a constant c0 > 0 for which

ϕ(n) c0 n ≥ log log log n for almost all large integers n 1. ≥ (ii) There is a constant c1 > 0 for which

ϕ(n) c1 n ≤ log log log n for almost all large integers n 1. ≥ Some related and earlier works on this topic include the works of Ramanujan, Erdos, and other on abundant numbers, see [25], [1], and recent related works appeared in [29], [2], [32], [35], and [36]. The first few sections cover some background materials focusing on some finite sums over the prime numbers and some associated and products. The proofs of Theorems 1.3, 1.4 and 1.5 are given in the last few sections respectively. euler totient function inequality 3

2 Prime Numbers Theorems

The omega notation f(x) = g(x)+Ω (h(x)) means that both f(x) > g(x)+ c0h(x) and ± f(x) < g(x) c h(x) occur infinitely often as x , where c > 0 and c > 0 are − 1 → ∞ 0 1 constants, see [20, p. 5], and similar references.

The weighted primes counting functions, psi ψ(x) and theta θ(x), are defined by

θ(x)= log p and ψ(x)= log pk (3) p x pk x X≤ X≤ respectively. The standard prime counting function is denoted by

π(x)=# p x = 1. (4) { ≤ } p x X≤ x 1 This function is usually expressed in term of the logarithm integral li(x)= 2 (log t)− dt. Theorem 2.1. Uniformly for x 2 the psi and theta functions have the followingsR asymp- totic formulae. ≥

(i) Unconditionally, c0√log x θ(x)= x + O xe− .   (ii) Unconditional oscillation,

θ(x)= x +Ω x1/2 log log log x . ±   (iii) Conditional on the RH,

θ(x)= x + O x1/2 log2 x .   Proof. (ii) The oscillations form of the theta function is proved in [20, p. 479], 

The same asymptotics hold for the function ψ(x). Explicit estimates for both of these functions are given in [4], [27], [5, Theorem 5.2], and related literature.

Conjecture 2.1. Assuming the RH and the LI conjecture, the suprema are

ψ(x) x 1 ψ(x) x 1 and lim inf − 2 = − lim sup − 2 = . (5) x √x(log log x) π x √x(log log x) π →∞ →∞ More details on the Linear Independence conjecture appear in [14], [7, Theorem 6.4], and recent literature. The LI conjecture asserts that the imaginary parts of the nontrivial zeros ρ = 1/2+ iγ of the zeta function ζ(s) are linearly independent over the set 1, 0, 1 . n n {− } In short, the equations rnγn = 0, (6) 1 n M ≤X≤ where r 1, 0, 1 , have no nontrivial solutions. n ∈ {− } Theorem 2.2. Let x 1 be a large number. Then ≥ euler totient function inequality 4

(i) Unconditionally, c0√log x π(x) = li(x)+ O xe− .   (ii) Unconditional oscillation,

x1/2 log log log x π(x) = li(x)+Ω . ± log x !

(iii) Conditional on the RH,

π(x) = li(x)+ O x1/2 log x .   Proof. (i) The unconditional part of the prime counting formula arises from the delaVallee c0√log x Poussin form π(x) = li(x)+ O xe− of the prime number theorem, see [20, p. 179]. Recent information on the constantc0 > 0 and the sharper estimate π(x) = 3 5 −2 5 c0 log x / (log log x) / li(x)+ O xe− appears in [10].   (ii) The unconditional oscillations part arises from the Littlewood form π(x) = li(x)+ Ω x1/2 log log log x/ log x of the prime number theorem, consult [15, p. 51], [20, p. 479], ± et cetera.
(iii) The conditional part arises from the Riemann form π(x) = li(x)+ O x1/2 log2 x of the prime number theorem. 
New explict estimates for the number of primes in arithmetic progressions are computed in [3].

3 Sums Over The Primes

The most basic finite sum over the prime numbers is the prime harmonic sum p x 1/p. The refined estimate of this finite sum, stated below, is a synthesis of various results≤ due P to various authors.

Lemma 3.1. Let x 2 be a large number, then ≥ (i) Unconditionally, 1 c0√log x = log log x + B + O e− . p 1 p x X≤   (ii) Unconditional oscillation,

1 log log log x = log log x + B1 +Ω . p ± x1/2 log x p x   X≤ (iii) Conditional on the RH,

1 log x = log log x + B1 + O . p x1/2 p x   X≤ euler totient function inequality 5

where B1 = 0.2614972128 ..., is Mertens constant, and c0 > 0 is an absolute constant. x 1 Proof. Replace the logarithm integral li(x) = 2 (log t)− dt, and the appropriate prime counting measure π(x) in Theorem 4.1 into the Stieltjes integral representation R 1 x 1 = dπ(t) (7) p t p x 2 X≤ Z and evaluate it.

(i) The unconditional part of the prime counting formula arises from the delaVallee Poussin c0√log x form π(x) = li(x)+ O xe− of the prime number theorem, see [20, p. 179].   (ii) The unconditional oscillations part arises from the Littlewood form π(x) = li(x)+ Ω x1/2 log log log x/ log x of the prime number theorem, consult [15, p. 51], [20, p. 479], ± et cetera.
(iii) The conditional part arises from the Riemann form π(x) = li(x)+ O x1/2 log2 x of the prime number theorem. 
The asymptotic order 1/p log log x is due to Euler, confer [8, Chapter 15?]. The p x ∼ earliest version including≤ error term 1/p = log log x + B + O(1/ log x) is due to P p x 1 Mertens, see [28]. The qualitative form of≤ the oscillations of the differences P 1 log p (log log x + γ) and (log x + γ) (8) pk − pk − pk x pk x X≤ X≤ seems to be due to Phragmen, confer [24, p. 182].

The Euler constant and Mertens constant occur very frequently in analysis. The former is defined by 1 γ = lim log x = 0.577215665 ..., (9) x  n −  →∞ n x X≤ and the later is defined by  

1 B1 = lim log log x = 0.2614972128 .... (10) x  p 1 −  →∞ p x X≤ −   Other definitions of these constants are available in the literature, confer [33].

Lemma 3.2. The constants γ and B1 satisfy the linear relation 1 B = γ . (11) 1 − npn p 2 n 2 X≥ X≥ Proof. This relation stems from the power series expansion 1 1 B γ = log 1 + (12) 1 − − p p p 2     X≥ via the power series for log(1 + z) with z < 1. This leads to this identity, see [13, p. 466], | | [20, p. 182].  euler totient function inequality 6

Lemma 3.3. Let x 2 be a large number, then ≥ 1 1 1 = γ B1 + O . (13) npn − − x log x x log2 x p x n 2   X≤ X≥ Proof. Rearrange the power series expansion as

1 1 1 1 = γ B1 = γ B1 + O . (14) npn − − npn − − x log x x log2 x p x n 2 p>x n 2   X≤ X≥ X X≥ The estimate for the last two terms on the right follows from Lemma 3.4 computed below. 

Lemma 3.4. Let x 1 be a large number, then ≥ 1 1 1 = + O . (15) npn x log x x log2 x p>x n 2   X X≥ Proof. Split the infinite sum into two subsums: 1 1 1 = + npn 2p2 npn p>x n 2 p>x p>x n 3 X X≥ X X X≥ 1 1 = + O . 2p2 x2 log x p x   X≥ Employ the prime counting measure π(t)=# p t to evaluate the first subsum using { ≤ } the integral

1 1 = ∞ dπ(t) p2 t2 p x x X≥ Z π(x) π(t) = + 2 ∞ dt − x2 t3 Zx 1 1 = + O . x log x x log2 x   

3.1 Problems 1. Use the identity B = γ (npn) 1 to prove or disprove that B and γ are 1 − p 2 n 2 − 1 linearly independent over the rational≥ numbers≥ Q. P P 1+α 2. Evaluate the finite sum n x(n log n)− , where α is a real number. ≤ P 4 Products Over The Primes

The asymptotics for a variety of interesting products are simple applications of the results for prime harmonic sums in the previous section.

Lemma 4.1. Let x 2 be a large number, then ≥ euler totient function inequality 7

(i) Unconditionally,

1 1 − γ c0√log x 1 = e log x + O e− . − p p x   Y≤   (ii) Unconditional oscillation,

1 1 − log log log x 1 = eγ log x +Ω . − p x1/2 p x     Y≤ (iii) Conditional on the RH,

1 1 − log x 1 = eγ log x + O . − p x1/2 p x     Y≤ where B1 = 0.2614972128 ..., is Mertens constant, and c0 > 0 is an absolute constant. The results for products over arithmetic progression are proved in [18], et alii.

The nonquantitative unconditional oscillations of the error of the product of primes is implied by the work of Phragmen, refer to equation (8), and [24, p. 182]. Since then, various authors have developed quantitative versions, see [26], [6], [17], [16], et alii. The specific quantitative form

1 1 − f(x) 1 = eγ log x +Ω , (16) − p ± x1/2 p x     Y≤ where f(x) is a slowly increasing function, was proved in [6].

Theorem 4.1. (Martens, 1874) The following asymptotic formulas hold: (i) 1 1 1 − lim 1 = eγ . x log x − p →∞ p x   Y≤ (ii) 1 γ 1 1 − 6e lim 1+ = . x log x p π2 →∞ p x   Y≤ 5 Logarithm Sums of Primes

The logarithm of a product can be derived from the estimate of the product itself. However, an independent proof based on the power series of the logarithm will be used here to obtain these estimates. Lemma 5.1. Let x 1 be a large number, then ≥ (i) Unconditionally,

1 c0√log x 1 − e log 1 = log log x + γ + O − . − p log x p x   ! Y≤ euler totient function inequality 8

(ii) Unconditional oscillation,

1 1 − log log log x log 1 = log log x + γ +Ω . − p x1/2 log x p x     Y≤ (iii) Conditional on the RH,

1 1 − log log log log x log 1 = log log x + γ + O . − p x1/2 log x p x     Y≤ where p x is the largest prime divisor of n. ≤ Proof. (i) The product form of the ratio n/ϕ(n) has the equivalent form

1 1 1 1 − 1 − 1 − 1 1 = 1 1+ 1+ − p − p p p p x   p x       Y≤ Y≤ 1 1 − 1 = 1 1+ . (17) − p2 p p x     Y≤ Taking logarithms and replacing the power series expansions return

1 n+1 1 − 1 1 ( 1) log 1 + log 1+ = + − (18) − p2 p np2n npn p x    ! p x n 1 p x n 1 X≤ X≤ X≥ X≤ X≥ 1 ( 1)n+1 1 = + − + np2n npn p p x n 1 p x n 2 p x X≤ X≥ X≤ X≥ X≤ 1 ( 1)n+1 1 = + − + .  np2n npn  p p x n 1 n 2 p x X≤ X≥ X≥ X≤   By Lemma 3.1, the last finite sum

1 log log log x = log log x + B +Ω (19) p ± x1/2 log x p x   X≤ 1 log log log x = log log x + γ +Ω . − npn ± x1/2 log x p 2 n 2   X≥ X≥ The last line follows from Lemma 3.3. Replacing back and combining the infinite series, euler totient function inequality 9 yield

1 ( 1)n+1 1 = + − + (20)  np2n npn  p p x n 1 n 2 p x X≤ X≥ X≥ X≤   1 ( 1)n+1 1 log log log x = + − + log log x + γ +Ω  np2n npn  − npn ± x1/2 log x p x n 1 n 2 p 2 n 2   X≤ X≥ X≥ X≥ X≥   1 ( 1)n+1 1 = + −  np2n npn − npn  p x n 1 n 2 n 2 X≤ X≥ X≥ X≥   1 log log log x + log log x + γ +Ω − npn ± x1/2 log x p>x n 2   X X≥ 1 log log log x = log log x + γ +Ω . − npn ± x1/2 log x p>x n 2   X X≥ The triple series on the third line vanish, and the series on the fourth line is small enought to be absorved into the omega term. 

Lemma 5.2. Let n 1 be a large highly composite integer, then ≥ n log log log log n log = log log log n + γ +Ω . (21) ϕ(n) ± (log n)1/2 log log n     Proof. Given a highly composite integer, the magnitude of the largest prime divisor p n | is determined in Lemma 6.1, which gives

log log log log n p x = log n 1+Ω , (22) ≤ ± (log n)1/2    and log log log log n log log x = log log log n 1+Ω ± (log n)1/2     log log log log n = log log log n + log 1+Ω (23) ± (log n)1/2     log log log log n = log log log n +Ω . ± (log n)1/2 log log n   By the prime number theorem, the error term in (23) changes sign infinitely often.

Applying Lemma 5.1, and replacing (23), yield

n log log log x log = log log x + γ +Ω (24) ϕ(n) ± x1/2 log x     log log log log n = log log log n + γ +Ω , ± (log n)1/2 log log n   this proves the claim.  euler totient function inequality 10

6 Highly Composite Numbers

Let p be the kth prime in increasing order, and let v = max m : pm n is the p-adic k p { | } valuation. Extremely abundant integers, and colossally abundant integers are related to the primorial integers n = 2v2 3v3 pvp , but the exponents have certain multiplicative · · · · structure 1 v v v . ≤ p ≤···≤ 3 ≤ 2 Definition 6.1. Let d(n) = 1. An integer n N is called highly composite if and d n ∈ only if d(m) < d(n) for all integers| m 0

These numbers are studied in [25], [1], [32], [29], [35], et alii.

Lemma 6.1. ([1, Theorem 2]) Let n be a large highly composite integer, then ≥ (i) Unconditionally, the largest prime divisor p n has the asymptotic | 1 p = (log n) 1+ O . (log log n)2    (ii) Unconditionally, the largest prime divisor p n has the asymptotic oscillations | log log log n p = (log n) 1+Ω . ± (log n)1/2    (iii) Modulo the Riemann hypothesis, the largest prime divisor p n has the asymptotic | log log n p = (log n) 1+ O . (log n)1/2    Proof. The asymptotic part p log n follows from Theorem 2 in [1], and the three forms ∼ of the error term R(n) follows from Theorem 2.1. 

7 The Main Result

Proof. (Theorem 1.1) Let Nk be the product of the first k primes pk in increasing order. On the contrary, suppose that

N /ϕ(N ) eγ log log N (26) k k ≤ k for infinitely many k 1. Taking logarithm on both sides yields ≥ N log k γ + log log log N . (27) ϕ(N ) ≤ k  k  euler totient function inequality 11

By Lemma 5.2, the left side has the equivalent form

Nk log log log log Nk log = log log log Nk + γ +Ω . (28) ϕ(N ) ± (log N )1/2 log log N  k   k k  Comparing the right side and the left sides yields

log log log log Nk log log log Nk + γ +Ω γ + log log log Nk. ± (log N )1/2 log log N ≤  k k  Rearrange the inequality yields equivalent form log log log log N Ω k 0. (29) ± (log N )1/2 log log N ≤  k k  By the prime number theorem, the error term in (28) changes sign infinitely often. For example, log log log log N c k 0 (30) 1/2 (log Nk) log log Nk ≤ for infinitely many integers, and some constant c > 0. Clearly, this is a contradiction for infinitely many integers N as k .  k →∞ The unconditional result in Theorem 1.1 implies that for large highly composite integers such that ω(n) log n/ log log n, the ratio ≫ n c eγ log log n + 0 , (31) ϕ(n) ≥ (log n)β where c > 0 is a constant, and 1/2 β < 1, confer (30) for some clarifications. The 0 ≤ conditional result in [23] provides an exact formula n c = eγ log log n + 1 (32) ϕ(n) (log n)1/2 for n = 2 3 p , where p is the kth prime in order, and c > 0 is a very specific constant. · · · · k k 1 However, equation (32) appears to contradict the unconditional result in equation (16), and Montgomerry conjecture 2.1.

8 Average Lower And Upper Bounds

The subset of integers that satisfy inequality (2) is a very thin subet of integers. In contrast, almost every integers has a large lower and upper bounds. Proof. (Theorem 1.3) By the Erdos-Kac theorem, (confer [9], [13, Theorem 431], [12], et cetera), almost every integer n 1 has ω(n) log log n prime divisors. In addition, the ≥ ≪ interval [1, (log log n)2] contains π((log log n)2) log log n primes. Thus, ≫ 1 n 1 − = 1 (33) ϕ(n) − p p n   Y| 1 1 − 1 ≪ − p p (log log n)2   ≪ Y log log log n. ≪ The reverse inequality is similar.  The proof of Theorem 1.5 uses the same technique. euler totient function inequality 12

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