Riemann Hypothesis and Primorial Number Choe Ryong Gil October 8, 2014 Abstract; In this paper we consider the Riemann hypothesis by the primorial numbers. Keywords; Riemann hypothesis, Primorial number. 1. Introduction and main result of paper Let N be the set of the natural numbers. The function ϕ ( n ) = n ⋅ ∏ p|n (1 − p −1 ) is called Euler’s function of n ∈ N ([3]). Here p | n note p is the prime divisor of n . Robin showed in his paper [5] (also see [4]) [Robin theorem] If the Riemann hypothesis (RH) is false, then there exist constants 0 < β < 1/ 2 and c > 0 such that σ ( n ) ≥ eγ ⋅ n ⋅ log log n + c ⋅ n ⋅ log log n / ( log n ) β holds for infinitely many n ∈ N , where σ ( n ) = ∑ d |n d is the divisor function of n ∈ N ([5]) and γ = 0.577" is Euler’s constant ([3]). From this we have [Theorem 1] If there exists a constant c0 ≥ 1 such that ( n / ϕ ( n ) ≤ eγ ⋅ log log c0 ⋅ n ⋅ exp holds for any n ≥ 2 , then the RH is true. For n ∈ N ( n ≠ 1) ( ( ( log n ⋅ ( log log n ) ( )) we define Φ 0 ( n ) = exp exp e −γ ⋅ n / ϕ ( n ) / n ⋅ exp ( 2 )) (*) log n ⋅ ( log log n ) 2 )) . Then we give [Theorem 2] For any n ≥ 2 we have Φ 0 ( n ) ≤ 24 . 2. Proof of Theorem 1 It is clear that σ ( n ) ⋅ ϕ ( n ) ≤ n 2 for any n ≥ 2 . If (*) holds, but the RH is false, then c ⋅ log log n eγ ⋅ log log n + ( log n ) β ≤ σ (n) n ≤ ( n ≤ eγ ⋅ log log c0 ⋅ n ⋅ exp ϕ (n) ( log n ⋅ ( log log n ) holds for infinitely many n ∈ N . On the other hand, since log (1 + t ) ≤ t ( t > 0 ) , we have ( log log c0 ⋅ n ⋅ exp ( log n ⋅ ( log log n ) 2 )) = log ( log n + log c + 0 log n ⋅ ( log log n ) 2 2 )= ⎛ ⎛ log c ( log log n )2 ⎞ ⎞ ⎛ log c ( log log n )2 ⎞ 0 0 ⎜ ⎟ ⎟ = log log n + log ⎜1 + ⎟≤ = log log n ⋅ ⎜1 + + + ⎜ ⎟⎟ ⎜ ⎜ log n log n log n log n ⎟⎠ ⎝ ⎠ ⎝ ⎝ ⎠ log c0 ( log log n ) + . log n log n Therefore, for infinitely many n ∈ N we have 2 ≤ log log n + −γ e ⋅ c ⋅ log log n ( log n ) β log c0 ( log log n ) . ≤ + log n log n 2 From this we have 0 < e −γ ⋅ c ≤ log c0 1 log log n ⋅ + → 0 (n → ∞) , 1− β 1/ 2 − β log log n ( log n ) ( log n ) but it is a contradiction. 1 )) 3. Reduction to the primorial number Let p1 = 2, p2 = 3, p3 = 5," be first consecutive primes. Then pm is m − th prime number. The number ( p1 " pm ) is called the primorial number ([1]). Assume n = q1λ " qmλ 1 m is the prime factorization of n ∈ N . Here q1 ," , qm are distinct primes and λ1 ," , λm are nonnegative integers ≥ 1 . Put ℑm = p1 " pm , then it is clear that n ≥ ℑm , ( m n = ∏ i =1 1 − qi−1 ϕ (n) ) −1 ( ≤ ∏ i =1 1 − pi−1 m ) −1 = ℑm and so Φ 0 ( n ) ≤ Φ 0 ( ℑm ) . This ϕ ( ℑm ) shows that the boundedness of the function Φ 0 ( n ) for n ∈ N is reduced to one for the primorial numbers. 4. Some symbols It is known ∑ p ≤t p −1 = log log t + b + E ( t ) by [6], where b = γ + ∑ p ⎡⎣ log (1 − 1/ p ) + 1/ p ⎤⎦ = 0.26" and ( ( E ( t ) = Ο exp −a1 ⋅ log t )) ( a > 0) and t is a real number ≥ 2 . Put F 1 m = ℑm / ϕ ( ℑm ) , then we have log ( Fm ) = −∑ i =1 log (1 − 1/ pi ) = −∑ i =1 ⎡⎣log (1 − 1/ pi ) + 1/ pi ⎤⎦ + ∑ i =11/ pi = m m m = −∑ i =1 ⎡⎣log (1 − 1/ pi ) + 1/ pi ⎤⎦ + log log pm + b + E ( pm ) = m = −∑ i =1 ⎡⎣log (1 − 1/ pi ) + 1/ pi ⎤⎦ + log log pm + γ + ∑ p ⎡⎣ log (1 − 1/ p ) + 1/ p ⎤⎦ + E ( pm ) = m = log log pm + γ + E ( pm ) + ε 0 ( pm ) , where ε 0 ( pm ) = ∑ p > pm ⎡⎣log (1 − 1/ p ) + 1/ p ⎤⎦ . From this we have ( (e −γ ) ( ) ) exp e−γ ⋅ Fm = pm ⋅ e0′ , ⋅ Fm = log pm ⋅ e0 , ( ) where e0 = exp E ( pm ) + ε 0 ( pm ) and e0′ = exp log pm ⋅ ( e0 − 1) . Similarly, we have (e ( −γ ) ( ) ) exp e −γ ⋅ Fm−1 = pm −1 ⋅ e1′ , ⋅ Fm −1 = log pm −1 ⋅ e1 , ( ) where e1 = exp E ( pm −1 ) + ε 0 ( pm −1 ) and e1′ = exp log pm −1 ⋅ ( e1 − 1) . We recall the Chebyshev’s function ϑ ( t ) = ∑ p ≤t log p ([3]). It is known that ϑ ( pm ) = pm ⋅ (1 + θ ( pm ) ) ( ( by the prime number theorem ([3]), where θ ( pm ) = Ο exp − a2 ⋅ log pm )) ( a 2 > 0 ) . Then we see log ℑm = pm ⋅ α 0 and log ℑm −1 = pm −1 ⋅ α1 , where α 0 = 1 + θ ( pm ) and α1 = 1 + θ ( pm −1 ) . Now we put N i = log ℑm −i ⋅ ( log log ℑm −i ) ( i = 0, 1) and Cm = Φ 0 ( ℑm )( m ≥ 1) . 2 5. Some numerical estimates 5.1. An estimate of e1 and e1′ We put p = pm −1 , p0 = pm below. For the theoretical calculation we assume p ≥ e14 . The discussion ( ) for p ≤ e14 is supported by MATLAB. Since e −γ ⋅ Fm −1 = log p ⋅ e1 < log p + 1/ log p ( p ≥ 2 ) by (3.30) of ( ) ( [6], we respectively have e1 < 1 + 1/ log p < 1.0052 p ≥ e14 , e1′ < exp(1/ log p ) < 1.075 p ≥ e14 2 ( ) ) and e1 ⋅ e1′ < 1.08 p ≥ e14 . 5.2. An estimate of ( e1 ⋅ e1′ ) Since if e1 < 1 then e1′ < 1 , we have e1 ⋅ e1′ < 1 . On the other hand, It is known that by (3.17), (3.20) of [6] ( −1/ log t ) ≤ E ( t ) = ∑ 2 p ≤t p −1 − b − log log t ≤ (1/ log 2 t ) ( t > 1) . Hence if e1 > 1 , then 0 < a := E ( p ) + ε 0 ( p ) < 1/ log 2 p ≤ 0.0052 , since ε 0 ( p ) < 0 , and so e1 = 1 + a + ∑ n=2 a n / n! ≤ 1 + a + a 2 / ( 2 ⋅ (1 − a ) ) ≤ 1 + a + 0.51⋅ a 2 . ∞ 2 ( ) ( ) We have e1 ⋅ e1′ = exp a + log p ⋅ ( e1 − 1) ≤ 1 + b + b2 / 2 ⋅ (1 − b ) , where b = (1 + log p ) ⋅ a + 0.51 ⋅ log p ⋅ a 2 ( ) and b ≤ 0.09 p ≥ e14 . Therefore we have e1 ⋅ e1′ ≤ 1 + (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) + 0.59 ⋅ (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) 2 2 ( e > 1, p ≥ e ) . 14 1 5.3. An estimate of K 0 := p0 ⋅ ( e0′ − α 0 ) − p ⋅ ( e1′ − α1 ) It is clear that E ( p0 ) − E ( p ) = (∑ m ) (∑ 1/ pi − log log pm − b − i =1 = 1/ pm − log log pm + log log pm −1 = ) m −1 i =1 1/ pi − log log pm −1 − b = ⎛ log p0 ⎞ 1 − log ⎜ ⎟ p0 ⎝ log p ⎠ and ε 0 ( p0 ) − ε 0 ( p ) = − log (1 − 1/ p0 ) − 1/ p0 . From this we have ⎛ log p ⋅ e1 ⎞ e0 ⎛ log p ⎞ ⎛ e0′ 1 ⎞ p =⎜ = ⋅ exp ⎜ ⎟ ⋅ ⎜1 + ⎟ and ⎟. e1 ⎝ log p0 ⎠ ⎝ p0 − 1 ⎠ e1′ p0 ⎝ p0 − 1 ⎠ Thus we have ⎛ ⎛ log p ⋅ e1 ⎞ ⎞ ⎛ p ⋅ e′ ⎞ K 0 = p ⋅ e1′ ⋅ ⎜ 0 0 − 1⎟ − log p0 = p ⋅ e1′ ⋅ ⎜⎜ exp ⎜ ⎟ − 1⎟⎟ − log p0 = log p0 ⋅ ( μ ⋅ e1′ − 1) , − p 1 ⎝ p ⋅ e1′ ⎠ 0 ⎝ ⎠ ⎝ ⎠ where μ = ⎛ log p ⋅ e1 ⎞ ⎞ p ⎛ ⋅ ⎜⎜ exp ⎜ ⎟ − 1⎟⎟ . Hence we get log p0 ⎝ ⎝ p0 − 1 ⎠ ⎠ 2 ⎛ log p ⋅ e1 ⎞ ⎞ p ⎛ p ⎛ log p ⋅ e1 1 ⎛ log p ⋅ e1 ⎞ ⎛ log p ⋅ e1 ⎞ ⎞ ⋅ ⎜ exp ⎜ ⋅⎜ + ⋅⎜ μ≤ ⎟ − 1⎟ ≤ ⎟ / ⎜1 − ⎟⎟ ≤ p p p ⎠ ⎝ p ⎠ ⎟⎠ log p ⎝ 2 ⎝ ⎝ ⎠ ⎠ log p ⎜⎝ 1 log p ⋅ e1 ⎛ log p ⋅ e1 ⎞ log p ≤ e1 + ⋅ e1 > 1, p ≥ e14 ) / ⎜1 − ( ⎟ ≤ e1 + 0.5053 ⋅ p p p 2 ⎝ ⎠ log p e1 > 1, p ≥ e14 ) . and μ ⋅ e1′ − 1 ≤ ( e1 ⋅ e1′ − 1) + 0.55 ⋅ ( p 5.4. An estimate of G0 := ( log p0 ⋅ R ( ℑm−1 ) − ( N 0 − N1 ) ) / N 0 Here R ( ℑm −1 ) := ( log log ℑm−1 ) 2 ⋅ log ℑm −1 2 + 2 ⋅ log log ℑm −1 . log ℑm −1 Since log (1 + t ) ≥ t ⋅ (1 − t / 2 ) ( t > 0 ) , first we have N 0 − N1 = ( ) ( log ℑm − log ℑm −1 ⋅ ( log log ℑm ) + log ℑm −1 ⋅ ( log log ℑm ) − ( log log ℑm −1 ) 2 2 ≥ log p0 2 ⋅ ( log log ℑm −1 ) + log ℑm−1 ⋅ 2 ⋅ log log ℑm −1 ⋅ ( log log ℑm − log log ℑm −1 ) = 2 ⋅ log ℑm = ⎛ log p0 log p0 ⎞ 2 ⋅ ( log log ℑm −1 ) + log ℑm −1 ⋅ 2 ⋅ log log ℑm −1 ⋅ log ⎜ 1 + ⎟≥ 2 ⋅ log ℑm ⎝ log ℑm−1 ⎠ ≥ log p0 log p0 2 ⋅ ( log log ℑm −1 ) + 2 ⋅ log log ℑm −1 ⋅ 2 ⋅ log ℑm log ℑm −1 and 3 ⎛ log p0 ⎞ ⋅ ⎜1 − ⎟ ⎝ 2 ⋅ log ℑm −1 ⎠ 2 )≥ log p0 ⋅ R ( ℑm −1 ) − ( N 0 − N1 ) ≤ log p0 ⋅ + log p0 ⋅ ≤ ≤ log p0 2 ( log log ℑm−1 ) 2 ⋅ log ℑm −1 2 − log p0 2 ⋅ ( log log ℑm −1 ) + 2 ⋅ log ℑm 2 ⋅ log log ℑm−1 2 ⋅ log log ℑm −1 ⎛ log p0 ⎞ − log p0 ⋅ ⋅ ⎜1 − ⎟≤ log ℑm −1 log ℑm −1 ⎝ 2 ⋅ log ℑm −1 ⎠ ⎛ 1 1 ⋅⎜ − ⎜ log ℑ log ℑm m −1 ⎝ log 2 p0 ( log ℑm−1 ) 3/ 2 ⎞ log 2 p0 2 ⋅ log log ℑm −1 ≤ ⎟⎟ ⋅ ( log log ℑm −1 ) + 3/ 2 ( log ℑm−1 ) ⎠ ⎞ 1 2 ⎛1 ⋅ ( log log ℑm −1 ) ⋅ ⎜ + ⎟. ⎝ 4 log log ℑm −1 ⎠ On the other hand, it is known that pk2+1 ≤ 2 ⋅ pk2 for pk ≥ 7 by 247p. of[2] and t − t / log t < ϑ ( t )( t ≥ 41) by (3.16) of [6]. So if p ≥ e14 then we have α1 ≥ (1 − 1/14 ) and G0 ≤ ≤ log 2 p0 ( log ℑm−1 ) ⎞ 1 1 2 ⎛1 ⋅ ( log log ℑm −1 ) ⋅ ⎜ + ≤ ⎟⋅ ⎝ 4 log log ℑm −1 ⎠ N 0 ⎛1 ⎞ 1 ⋅⎜ + ⎟≤ ⎝ 4 log log ℑm −1 ⎠ log 2 p0 ( log ℑm−1 ) 3/ 2 2 2 ⎞ log 3 p ⎛ log 2 ⎞ ⎛ 1 1 1 0.01 ≤ ⋅ ⎜1 + ≤ p ≥ e14 ) . ( ⎟⎟ ⋅ ⎜ + ⎟⋅ 2 ⎜ log p ⎠ ⎝ 4 log p + log α1 ⎠ p ⋅ log p p ⋅ log p p ⋅ α1 ⎝ 5.5. An estimate of S ( p′ ) := ∑ 1/ ( p ⋅ log p ) p ′≤ p ≤+∞ Put s (t ) = ∑ p ≤t 1/ p = log log t + b + E (t ) . Then we have S ( p′ ) = ∫ =∫ +∞ p′ ≤− +∞ p′ +∞ 1 ⎛ dt ⎞ 1 ⋅ ds ( t ) = ∫ ⋅⎜ + dE ( t ) ⎟ = ′ p log t log t ⎝ t ⋅ log t ⎠ E ( t ) +∞ +∞ E ( t ) dt | p′ + ∫ + ⋅ dt ≤ 2 p ′ t ⋅ log 2 t t ⋅ log t log t 1 +∞ E ( t ) +∞ +∞ 1 | p′ + | p′ + ∫ ⋅ dt ≤ p ′ t ⋅ log 4 t log t log t . +∞ E ( p′ ) 1 1 − +∫ ⋅ dt ≤ ′ p log p′ log p′ t ⋅ log 4 t 1 4 1 1 1 + |+∞ ≤ + − p′ = 3 3 log p′ 3 ⋅ log 3 p′ log p′ log p′ 3 ⋅ log t ≤ and 1 +∞ E ( t ) +∞ +∞ 1 1 4 ⋅ dt ≥ − | p′ + | p′ − ∫ . 4 p ′ t ⋅ log t log t log t log p′ 3 ⋅ log 3 p′ If p′ is a first prime ≥ e14 , then p′ = 1202609 and it is 93118-th prime. And we have 0.06 ≤ S ( p′ ) ≤ 0.08 . S ( p′ ) ≥ − Now we are ready for the proof of the following lemma. Lemma. For any m ≥ 4 we have Cm < 1 . proof. Let Dm = pm ⋅ ( e0′ − α 0 ) / ( ) pm ⋅ α 0 ⋅ log 2 ( pm ⋅ α 0 ) ( m ≥ 4 ) . Then Cm < 1 is equivalent to Dm < 1 . 4 And we here have Dm < 1 for 7 ≤ pm ≤ e14 and Dm ≤ am := 1 − 11 ⋅ S ( pm ) for any pm ≥ e14 . In fact, it is easy to see that for 7 ≤ pm ≤ e14 by MATLAB (see the table 1 and the table 2 ) ( ( ) ℜm := log e −γ ⋅ Fm − log log log ℑm + log ℑm ⋅ ( log log ℑm ) 2 ) < 0. Next, p′ = 1202609 then we have D93118 = 0.01038" ≤ 0.1 ≤ 1 − 11 ⋅ S ( p′ ) ≤ 0.4 < 1 . Now assume p ≥ e14 and Dm −1 ≤ am −1 . Let us see Dm ≤ am . We have Dm = p0 ⋅ ( e0′ − α 0 ) 1 N K = ⋅ ( p ⋅ ( e1′ − α1 ) + K 0 ) = Dm −1 ⋅ 1 + 0 ≤ N0 N0 N0 N0 ≤ am −1 ⋅ ( N1 1 + ⋅ log p0 ⋅ ( μ ⋅ e1′ − 1) ≤ am −1 + bm −1 , N0 N0 ) where bm −1 = log p0 ⋅ ( μ ⋅ e1′ − 1) − am −1 ⋅ ( N 0 − N1 ) / N 0 . We have to obtain bm −1 ≤ 11/ ( p ⋅ log p ) . By the assumption Dm −1 ≤ am −1 , we have ⎛ p ⋅ α1 ⋅ log 2 ( p ⋅ α1 ) log 2 ( p ⋅ α1 ) ⎞ = α1 ⋅ ⎜1 + am −1 ⋅ ⎟ ⎜ p p ⋅ α1 ⎟⎠ ⎝ e1′ < α1 + am −1 ⋅ and by taking logarithm of both sides ⎛ log 2 ( p ⋅ α1 ) ⎞ log 2 ( p ⋅ α1 ) . log e1′ = log p ⋅ ( e1 − 1) < log α1 + log ⎜ 1 + am −1 ⋅ ⎟ ≤ θ ( p ) + am −1 ⋅ ⎜ p ⋅ α1 ⎟⎠ p ⋅ α1 ⎝ From this we also have log 2 ( p ⋅ α1 ) ⎞ 1 ⎛ e1 < 1 + ⋅ ⎜ θ ( p ) + am −1 ⋅ ⎟, log p ⎜⎝ p ⋅ α1 ⎟⎠ E ( p) + ε0 ( p) < Thus we see log 2 ( p ⋅ α1 ) ⎞ 1 ⎛ ⋅ ⎜ θ ( p ) + am −1 ⋅ ⎟. log p ⎜⎝ p ⋅ α1 ⎟⎠ log p ⋅ E ( p ) − θ ( p ) < δ 0 , δ 0 = δ1 + δ 2 , δ1 = am −1 ⋅ log 2 ( p ⋅ α1 ) / p ⋅ α1 , δ 2 = − log p ⋅ ε 0 ( p ) . On the other hand, by the Abel’s identity ([3]) and ϑ ( p ) , we see ⎛ ⎞ dt + dE ( t ) ⎟ = ⎝ t ⋅ log t ⎠ ϑ ( p ) = p + p ⋅ θ ( p ) = ∑ i =1 log pi = ∑ i =1 pi log pi / pi = ∫ t ⋅ log t ⋅ ⎜ m −1 m −1 p 2 = p − 2 + p ⋅ log p ⋅ E ( p ) − 2 ⋅ log 2 ⋅ E ( 2 ) − ∫ (1 + log t ) ⋅ E ( t ) ⋅ dt p 2 and p ⋅ log p ⋅ E ( p ) − p ⋅ θ ( p ) = ∫ (1 + log t ) ⋅ E ( t ) ⋅ dt + η1 , p 2 where η1 = 2 + 2 ⋅ log 2 ⋅ E ( 2 ) . From this we get ∫ (1 + log t ) ⋅E ( t ) dt + η p 1 2 And we have p ⋅ δ1 = ∫ p 2 < p ⋅δ0 . Rα ( t ) ⋅ dt + ηα , where Rα ( t ) = am −1 ⋅ ⎞ log 2 (α1 ⋅ t ) ⎛ 4 2 ⋅ ⎜⎜1 + ⎟⎟ , ηα = am −1 ⋅ 2 ⋅ log ( 2 ⋅ α1 ) / α1 . 2 ⋅ α1 ⋅ t ⎝ log (α1 ⋅ t ) ⎠ 5 Therefore we obtain ∫ p 2 f 0 ( t ) ⋅ dt < δ 3 , where δ 3 = p ⋅ δ 2 − η1 + ηα and f 0 ( t ) = (1 + log t ) ⋅ E ( t ) − Rα ( t ) . From the integration by parts we have ∫ p 2 Since ∑ (1/ p ) − b is p ≤t f0 ( t ) ⋅ dt = p ⋅ f0 ( p ) − 2 ⋅ f 0 ( 2 ) − ∑i =1 m−2 ( ∫ pi+1 pi t ⋅ df0 ( t ) . ) a constant and log log t + Rα ( t ) / (1 + log t ) is increasing function in each interval ( pi , pi +1 ) , the function E ( t ) − Rα ( t ) / (1 + log t ) is decreasing. And f 0 ( t ) ≤ (1 + log t ) E ( t ) − Rα ( t ) ⎛ 1 1 ⎞ ≤⎜ + + R (t ) . (1 + log t ) ⎝ log t log 2 t ⎟⎠ α Thus the function f 0 ( t ) is decreasing and so it is a bounded variation function. From this we have ∫ pi+1 pi t ⋅ df 0 ( t ) ≤ 0 and (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) ≤ Rα ( p ) + δ 4 / p , where δ 4 = ( 2 ⋅ f 0 ( 2 ) + ηα − η1 ) . On the other hand, it is known ϑ ( t ) ≤ t + t ( t > 1) by (3.15) of [6]. From this we obtain 2 ⋅ log t α1 ≤ 1 + 1/ ( 2 ⋅ log p ) ≤ 1.036 ( p ≥ e14 ) and δ 4 = 2 ⋅ (1 + log 2 ) ⋅ E ( 2 ) − 2 ⋅ Rα ( 2 ) + ηα − 2 − 2 ⋅ log 2 ⋅ E ( 2 ) ≤ ⎛1 ⎞ ≤ 2 ⋅ E ( 2 ) + ηα − 2 ≤ −2 + 2 ⋅ ⎜ − b − log log 2 ⎟ + am −1 ⋅ 2 ⋅ log 2 ( 2 ⋅ α1 ) / α1 ≤ ⎝2 ⎠ ≤ −0.7899 + 2 ⋅ log 2 ( 2 ⋅ α1 ) / α1 ≤ −0.7899 + 2 ⋅ log 2 ( 2 ⋅ (1 + 1/ 2 /14 ) ) / 1 − 1/14 ≤ ≤ −0.7899 + 0.7784 ≤ −0.0115 < 0. ( ) Consequently, we have (1 + log p ) ⋅ E ( p ) + ε 0 ( p ) ≤ Rα ( p ) . If e1 > 1 , then we also have (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) 2 2 ≤ Rα2 ( p ) ≤ 2 ⎞ log 4 ( p ⋅ α1 ) ⎛ 1 log 4 ( p ⋅ α1 ) 2 ≤ ⋅ ⎜⎜ + . ⎟⎟ ≤ 0.4143 ⋅ p ⋅ α1 p ⋅ α1 ⎝ 2 log ( p ⋅ α1 ) ⎠ and log p0 ⋅ ( μ ⋅ e1′ − 1) − am −1 ⋅ ( N 0 − N1 ) ≤ ≤ log p0 ⋅ ( e1 ⋅ e1′ − 1) + 0.55 ⋅ log p0 ⋅ log p − am−1 ⋅ ( N 0 − N1 ) ≤ p ≤ log p0 ⋅ (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) − am −1 ⋅ ( N 0 − N1 ) + 0.55 ⋅ log 2 p0 + p +0.59 ⋅ log p0 ⋅ (1 + log p ) ⋅ ( E ( p ) + ε 0 ( p ) ) ≤ 2 2 ≤ G0 ⋅ N 0 + 0.25 ⋅ log p0 ⋅ log 4 ( p ⋅ α1 ) log 2 p0 . + 0.55 ⋅ p ⋅ α1 p Finally, we have bm −1 ≤ G0 + 0.25 ⋅ log p0 ⋅ log 4 ( p ⋅ α1 ) p ⋅ α1 ⋅ N1 log 2 p0 + 0.55 ⋅ ≤ p ⋅ N1 ( log p + log 2 ) ⋅ log p ⋅ ( log p + log α ) ≤ G + 0.25 ⋅ 1 0 p ⋅ log p p ⋅ α13/ 2 6 2 + 2 ⎛ log 2 − log α1 ⎞ 1 ⋅ ⎜⎜1 + ≤ ⎟⎟ ⋅ log p + log α1 ⎠ p ⋅ log p ⎝ 0.01 10.01 0.008 11 p ≥ e14 ) . ≤ + + ≤ ( p ⋅ log p p ⋅ log p p ⋅ log p p ⋅ log p log p +0.55 ⋅ p ⋅ α1 Next, if e1 ≤ 1 then it is very easy to give the proof of the lemma. Indeed, in this case we have bm −1 ≤ 0.55 ⋅ log 2 p0 0.008 p ≥ e14 ) . ≤ ( p ⋅ N1 p ⋅ log p 6. Proof of Theorem 2 Let n = q1λ1 " qmλm be the prime factorization of any natural number n ≥ 2 . Then it is clear pm ≤ qm . And if 7 ≤ pm ≤ e14 , then we have Cm < 1 , since ℜm < 0 (see the table 1 and the table 2), and if pm ≥ e14 then we have Cm < 1 by the Lemma. Therefore we have Φ 0 ( n ) ≤ Φ 0 ( ℑm ) = Cm ≤ max {Cm } ≤ 24 . m ≥1 (Note) The table 1 shows the values Cm = Φ 0 ( ℑm ) and ℜ m to ω ( n ) = m of n ∈ N . There are only values of Cm and ℜm for 1 ≤ m ≤ 10 here. But it is not difficult to verify them for 31 ≤ pm ≤ e14 . Note, if more informations, then it should be taken ℜm < 0 , not Cm < 1 , for 263 ≤ pm ≤ e14 , by reason of the limited values of MATLAB 6.5. The table 2 shows the values ℜ m for 93109 ≤ m ≤ 93118 . Of course, all the values in the table 1 and the table 2 are approximate. = Appendix = The algorithm for ℜm to ω ( n ) = m by matlab is as follows: Function Pi-Index, clc, gamma=0.57721566490153286060; format long P= [2, 3, 5, 7,…,1202609]; M=length(P); for m=1:M; p=P(1:m); q=1-1./p; F=-gamma+log(prod(1./q)); N1=sum(log(p.^1)); N2=(N1)^(1/2); N3=(log(N1))^2; N4=N2*N3; N5=N1+N4; m, Pm, Rm=F-log(log(N5)) Table 1 m pm Cm ℜm 1 2 3 4 5 6 7 8 9 10 2 3 5 7 11 13 17 19 23 29 9.66806133818849 23.15168798263150 7.73864609733096 0.83171792006862 0.01114282713904 1.102119966548700e-004 3.834259945131073e-007 1.397561045763582e-009 2.821898264763264e-012 2.081541289212468e-015 0.73259862957209 0.14633620860732 -0.00636141995881 -0.09308687002330 -0.12730939385590 -0.15077316854133 -0.15960912308179 -0.16612788105591 -0.17415284347098 Table 2 m pm ℜm 93109 1202477 93110 93111 93112 93113 93114 93115 93116 93117 93118 1202483 1202497 1202501 1202507 1202549 1202561 1202569 1202603 1202609 -0.01154791933871 -0.01154786567870 -0.01154781201949 -0.01154775835370 -0.01154770468282 -0.01154765103339 -0.01154759738330 -0.01154754372957 -0.01154749009141 -0.01154743644815 7 References [1] P. Sole, M. Planat, Robin inequality for 7-free integers, arXiv: 1012.0671v1 [math.NT] 3 Dec 2010. [2] J. Sandor, D. S. Mitrinovic, B. Crstici, Handbook of Number theory 1, Springer, 2006. [3] H. L. Montgomery, R. C. Vaugnan, Multiplicative Number Theory, Cambridge, 2006. [4] J. C. Lagarias, An elementary problem quivalent to the Riemann hypothesis, Amer. Math. Monthly 109 (2002), 534-543 [5] G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Rimann, Journal of Math. Pures et appl. 63 (1984), 187-213 [6] J. B. Rosser, L. Schoenfeld, Approximate formulars for some functions of prime numbers, IIlinois J. Math. 6 (1962), 64-94. 2010 Mathematics Subject Classification; 11M26, 11N05. Post Address; Choe Ryong Gil, Department of Mathematics, University of Sciences, Gwahak-1 dong, Unjong District, Pyongyang, D.P.R.Korea. Email; [email protected] 8

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