# Document 323290

```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.
Choe Ryong Gil,
Department of Mathematics,
University of Sciences,
Gwahak-1 dong, Unjong District,
Pyongyang, D.P.R.Korea.
Email; [email protected]
8
```