Some bad bounds on

ω

and related sums

26. Dezember 2006

1 Denition We use

pn

to denote the nth prime number. We dene

ω(pn ) as the smallest integer with

the properties

If

ω(pn )

ω(pn ) > pn+1 − pn

(1)

ω(pn )|pn − (pn+1 − pn )

(2)

does not exist, we dene it as 0. Furthermore we dene

ζ(pn ) = We will call

ω(pn )

pn − (pn+1 − pn ) ω(pn ) pn

the weight of the prime

can be shown that for all

(3)

while we use the term level for

2 < ω(pn ) < pn − 6

2 Basic theorems Theorem We have

Pn lim

n→∞

Proof

We will use the statement

n X i=5

ζ(pn ).

It

n≥5

ω(pi )ζ(pi ) =

n X

i=5 ω(pi )ζ(pi ) n2 ln(n)

=

(4)

1 2

2pn − pn+1 = ω(pn )ζ(pn ) pi − (pi+1 − pi ) =

i=5

n X i=5

pi −

(5)

to start.

n X

(pi+1 − pi )

(6)

i=5

Due to a theorem of Bach and Shallit we have

Pn lim

n→∞

i=1 pn 2 n ln(n)

1

=

1 2

(7)

Therefore we see that

Pn

i=5 ω(pi )ζ(pi ) n2 ln(n)

Pn

i=5 pn

=

n2 ln(n)

Pn −

i=5 (pi+1 − pi ) n2 ln(n)

(8)

Now it is easy to understand that

Pn

i=5 ω(pi )ζ(pi ) n2 ln(n)

lim

n→∞

Pn = lim

n→∞

i=5 pn

n2 ln(n)

Pn

i=5 (pi+1 − pi ) n2 ln(n)

−

1 = − lim 2 n→∞

Pn

i=5 (pi+1 − pi ) n2 ln(n) (9)

pn+1 −11 which The last expression is nothing more than a telescopic sum having the value n2 ln(n) tends against 0 rather quickly.

3 ω(pn) and ζ(pn) in Zone 2/3 primes 3.1

ζ(pn )

is small

The main statement is that, if

ω(pn ) > ζ(pn )

for some

pn ,

the level

ζ(pn )

has to be

relatively small.

Theorem If ω(pn) > ζ(pn), then ζ(pn) ≤ pn+1 − pn. Proof Assume the statement to be false. Then we have a solution of the equation 2pn − pn+1 = ω(pn )ζ(pn ) with

(10)

ω(pn ) > ζ(pn ) > pn+1 − pn which would immediately imply that ω(pn ) cannot be pn as ζ(pn ) does also fulll the requirements and is smaller. Contradiction.

the weight of

We can furthermore show that, if

ω(pn )

is bigger than

ζ(pn ),

then it will be very big. In

fact we will proof the following

Theorem If ω(pn) > ζ(pn), then ω(pn ) ≥

Proof

pn −1 pn+1 − pn

(11)

We have

2pn − pn+1 = ω(pn )ζ(pn )

and we do know that

ζ(pn ) ≤ pn+1 − pn .

(12)

The theorem follows.

3.2 The primality of the weight in Zone2/3 primes There are reasons to state the following

Conjecture If ω(pn) > ζ(pn) then ω(pn) is prime except for p6, p11, p30, p32 and p154. We can show the following weaker

Theorem

Assume that

ω(pn ) > ζ(pn ).

Then, if

ω(pn )

is not a prime number, this

implies that

(pn+1 − pn )3 ≥ pn − (pn+1 − pn )

2

(13)

Proof

ω(pn ) is not a prime number factors of ω(pn ) which will lead us

We will rst assume that

statements about the prime

and then derive some to the theorem stated

above. First of all we dene

ω(pn ) =

m Y

αi

(14)

i=1 Please notice that we are not using the canonic representation of primes but rather list them all (such as in

ω(pn )

20 = 2 ∗ 2 ∗ 5).

We will also implicitly assume that

m≥2

(otherwise

would be a prime anyway). Now we immediately nd that all those prime factors

have to be small. Indeed, we have

αi ≤ pn+1 − pn .

If this was false for some special

αj

we see that 1.

αj

2.

αj > pn+1 − pn

3.

αj < ω(pn )

is a divisor of

pn − (pn+1 − pn )

(because so is

αj does fulll the necessary requirements for ω(pn ). By denition the weight is the smallest

ω(pn ))

Hence

the weight and is actually smaller

than

number fullling the requirements

and thus we have a contradiction. Now, on the other hand, the prime factors cannot be too small. Indeed, we nd

∀1 ≤ i ≤ m : αi ≥

ω(pn ) pn+1 − pn

To see this, we will again assume the equation to be false for some special

(15)

αj .

If the

equation was false, we could conclude that

pn+1 − pn
pn+1 − pn

3.

ω(pn ) αj

< ω(pn )

pn − (pn+1 − pn )

Thus we found again a smaller number fullling the requirements while the weight is dened as the smallest such number and we have again reached a contradiction. Now, we found that for all

αi

we have

ω(pn ) ≤ αi ≤ pn+1 − pn pn+1 − pn

3

(17)

Thus we immediately obtain that

ω(pn ) ≤ (pn+1 − pn )2 Now, we can use the fact that

ζ(pn )

(18)

is small and see that

pn − (pn+1 − pn ) = ω(pn )ζ(pn ) ≤ (pn+1 − pn )2 (pn+1 − pn ) = (pn+1 − pn )3

(19)

3.3 How many Zone2/3 primes are there? In this section we will assume the inequality

Ω(n) > 0.6n,

which is yet nothing more

than a mere conjecture, and show that there are not too few Zone2/3 primes.

Theorem Assuming Ω(n) > 0.6n; if B is dened as the set of zone 2/3 primes among the set of prime numbers

{p5 , p6 , p7 , ..., pn }, |B| ≥

then we have

√ √ 0.6n2 − n pn + 4 pn √ pn − pn

(20)

Proof

We will rst dene two sets A and B with the properties A ∪ B = {5, 6, ..., n} and A ∩ B = {}. the set A contains all those numbers i with the property ω(pi ) ≤ ζ(pi ) while in B we nd those i for which ω(pi ) > ζ(pi ). Then we apparently have |A| + |B| = n − 4. We will use this to split the sum into

n X

ω(pi ) =

i=5

X

ω(pi ) +

i∈A

X

ω(pi )

(21)

i∈B

The following inequality is apparent:

X

ω(pi ) ≤

i∈A

Xp X√ X√ √ pi − (pi+1 − pi ) ≤ pi ≤ pn = |A| pn i∈A

i∈A

(22)

i∈A

Furthermore we also have

X

ω(pi ) ≤

i∈B Now, using the assumption

0.6n2 ≤

i∈A

ω(pi ) +

pi ≤ |B| pn

(23)

i∈B

Ω(n) > 0.6n X

X

we can derive, that

X

√ ω(pi ) ≤ |A| pn + |B| pn

(24)

i∈B

|A| = n − 4 − |B| which leads us to √ √ √ √ √ 0.6n2 ≤ n pn − 4 pn − |B| pn + |B| pn = |B| (pn − pn ) + n pn − 4 pn

Now consider that

√

Now we can isolate

|B|

(25)

and eectively obtain that

√ √ 0.6n2 − n pn + 4 pn ≤ |B| √ pn − pn

4

(26)

4 The average functions 4.1 Average of the weight:

Ω(n)

As it follows from the denition the value

ω(pn )

is rather irregular. Therefore it does

seem appropriate to study the average value, which we dene as

n

1X Ω(n) := ω(pi ) n

(27)

i=5

Theorem It is quite easy to show that Ω(n) is unbounded. Proof n

n

i=5

i=5

Now, given the fact that

pn > n ln(n)

1X 1X pn+1 − 11 n − 4 + ω(pi ) ≥ (pi+1 − pi + 1) = n n n n

Ω(n) =

n ≥ 15 Ω(n) >

(28)

due to the theorem of Rosser, we nd that for

pn+1 − 11 n − 4 pn + ≥ > ln(n) n n n

On the other hand some numerical experiments (for

(29)

1 ≤ n ≤ 5000)

indicate that for

almost all n we nd

Ω(n) < 0.7n 4.2 Average of the level:

(30)

ξ(n)

We dene

n

1X ξ(n) = ζ(pi ) n

(31)

i=5

Pn

4.3 A bound on

i=5

ω(pi ) + ζ(pi )

In this section we will attempt to show an inequality that gives us a lower bound for the growth of the sum of the average functions, by far better than

ln(n).

In fact we will

attempt to prove the following

Theorem

For all

n≥5

we nd

n X

Proof

i=5

2 3 16 ω(pi ) + ζ(pi ) ≥ n 2 − 3 3

(32)

We will start with applying the easiest case of the AGM inequality. For positive

√ a and b it can be shown easily that a + b ≥ 2 ab. Therefore we nd that (using pn+1 ≤ 3p2n ) n n p n r X X X pi ω(pi ) + ζ(pi ) ≥ 2 pi − (pi+1 − pi ) ≥ 2 (33) 2

numbers that

i=5

i=5

i=5

5

Now we do know that

pn > n ln(n) 2

n X

r

i=5

and therefore it is easy to see that

n

n

√ Xp pi √ X √ = 2 pi ≥ 2 n ln(n) 2 i=5

(34)

i=5

Now it should prove rather dicult to evaluate this sum, which is why we are using integrals now.

p x ln(x)

is a monotone increasing function, which immediately leads us

to

p

n

Z n ln(n) ≥

p x ln(x) dx

(35)

n−1 Applying this inequality several times and adding the integrals gives us

n p √ X √ Z 2 n ln(n) ≥ 2

np

x ln(x) dx

(36)

4

i=5

This integral can be evaluated exactly (Mathematica, for example, can do this) but in the result we nd some nasty functions. Hence we will try to nd a lower bound for that integral. Omitting the logarithm we get an easy integral, which we can evaluate without any problems.

Z

np

n√

Z x ln(x) dx ≥

4

4

2 3 16 x dx = n 2 − 3 3

(37)

Hence we can nally say with certainty that

n X

2 3 16 ω(pi ) + ζ(pi ) ≥ n 2 − 3 3

(38)

2 3 16 ω(pi ) + ζ(pi ) = n(Ω(n) + ξ(n)) ≥ n 2 − 3 3

(39)

i=5 This immediately leads us to

n X i=5

6