 # 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 (). Here p | n note p is the prime divisor of n . Robin showed in his paper  (also see )
[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 () and γ = 0.577" is Euler’s constant ().
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 (). 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 , 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 (). It is known that ϑ ( pm ) = pm ⋅ (1 + θ ( pm ) )
( (
by the prime number theorem (), 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
(
)
(
, 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 
( −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 and t − t / log t < ϑ ( t )( t ≥ 41)
by (3.16) of . 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 () 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 . 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
 P. Sole, M. Planat, Robin inequality for 7-free integers, arXiv: 1012.0671v1 [math.NT] 3 Dec 2010.
 J. Sandor, D. S. Mitrinovic, B. Crstici, Handbook of Number theory 1, Springer, 2006.
 H. L. Montgomery, R. C. Vaugnan, Multiplicative Number Theory, Cambridge, 2006.
 J. C. Lagarias, An elementary problem quivalent to the Riemann hypothesis, Amer. Math. Monthly 109
(2002), 534-543
 G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Rimann, Journal of Math.
Pures et appl. 63 (1984), 187-213
 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
``` # Guest editor Chiara Bucciarelli-Ducci, M.D., Ph.D., FESC, FRCP # ESC/EACVI Cardiac MRI Course Jeddah Heart Institute, Erfan Hospital, Jeddah KSA # Bounded Fatou components of composite transcendental entire functions with gaps 