Testing the Mertens Theorems and Conjecture

Chris De Corte

chrisdecorte@yahoo.com

March 3, 2015

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CONTENTS CONTENTS

Contents

1 Key-Words 3

2 Introduction 3

3 The theorems 3

3.1 Mertens’ 1st theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3.2 Mertens’ 2nd theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3.3 Mertens’ 3rd theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3.4 Mertens’ Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4 Calculation and results 4

4.1 Mertens’ 1st theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4.2 Mertens’ 2nd theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4.3 Mertens’ 3rd theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4.4 Mertens’ Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

5 References 7

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3 THE THEOREMS

1 Key-Words

Prime numbers, number theory, Riemann, Mertens, Meissel, zahlentheorie, Odlyzko, Riele,

Pintz.

2 Introduction

The following document originated out of my interest for primes.

Several people already told me that my findings were related to Mertens’ out comings.

Recently, there was someone on the Linkedin Group of Number Theory asking whether the

theorems of Mertens have been retested lately.

So, I decided to take the challenge.

3 The theorems

In the following, let p ≤ n mean all primes not exceeding n [1].

3.1 Mertens’ 1st theorem

|∑p≤n

ln(p)

p− ln(n)| ≤ 2 (1)

3.2 Mertens’ 2nd theorem

limn→∞

[∑p≤n

1

p− ln(ln(n))−M

]= 0 (2)

Where M is the Meissel-Mertens constant which is approximately equal to 0.261497212847642783755

[2].

3.3 Mertens’ 3rd theorem

limn→∞

[ln(n)

∏p≤n

(1− 1

p)]

= e−γ (3)

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3.4 Mertens’ Conjecture 4 CALCULATION AND RESULTS

Where γ is the Euler-Mascheroni constant.

3.4 Mertens’ Conjecture

Out of [5]:

M(r) =r∑

n=1

µ(n) ≤√r (4)

Where M(n) is the Mertens function and µ(n) is the Mobius function defined as [5]:

µ(n) =

0 : if n has one or more repeated prime factors

1 : if n = 1

(−1)k : if n is a product of k distinct primes

(5)

Mertens and Stieltjes in general claimed that:

m(r) = |M(r)|/√r ≤ 1. (6)

Odlyzko and the Riele on the other hand proved that the above equation does not always

hold and that there must exist counter examples also according to Pintz.

Moreover, Odlyzko and Riele conjectured that:

limn→∞

sup|M(n)|n−1/2 =∞ (7)

4 Calculation and results

4.1 Mertens’ 1st theorem

The left hand side of formula 1 has been calculated by a self written C++ program. The

results are written in 1000 tables each containing the primes and results over a range of

1000000 numbers. The results are summarized into Table 1. The results of this theorem are

written in column 3 and graphically represented in Figure 1.

Based on the first results, we can assume that the theorem is correct. Probably, the limit of

2 can even become a bit lower: 1.35 in stead of 2?.

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4.2 Mertens’ 2nd theorem 4 CALCULATION AND RESULTS

4.2 Mertens’ 2nd theorem

The first 2 terms on the left hand side of formula 2 has been calculated by the same self

written C++ program. The results are written in the same tables as mentioned before. The

results are summarized into Table 1. The results of this theorem are written in column 4 and

graphically represented in Figure 2. Based on the formula’s of the theorem, we expect the

results to go towards the Meissel-Mertens constant.

Based on the first results, we can assume that the theorem is correct.

4.3 Mertens’ 3rd theorem

I will not separately test this 3rd theorem here as it’s correctness has been shown in [3].

4.4 Mertens’ Conjecture

The left side of formula 6 has been calculated using Pari-gp. Assuming that we can rely on

these figures (as they are demonstrated in figure 4), the final results are shown in table 2 and

figure 3.

Following our first impression of the graph, we have no reason to believe that it would sud-

denly start moving outside the boundaries defined in formula 6.

However:

We also know from formula 5 that M(x) can only increase under one scenario: if the next fig-

ure is a multiple of an even number of distinct primes. Also we notice that M(x) can decrease

under 2 scenario’s: if the next figure is a multiple of an odd number of distinct primes or if

there is a new prime coming since this would also mean an odd number of primes (1 is odd).

Since any multiple can be formed with same chances from an odd as from an even number

of distinct primes, it would mean that we expect the direction of M(x) to be dictated by the

occurrence of new primes only and hence M(x) could further decrease for larger x. However,

contrary to our first impression, it will seem that a simple prime falls in the category of an

odd number of primes and that both scenario’s yield the same result but in opposite direction

(canceling themselves out).

Therefore, we have set out a test in Pari/Gp calculating the number of positive µ(x) and the

number of negative µ(x). We have used the following code in Pari/Gp:

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4.4 Mertens’ Conjecture 4 CALCULATION AND RESULTS

m1=0;p1=0;n=1000;for(x=1,n,if(moebius(x)<0,m1=m1-1,p1=p1+moebius(x)));

print(x,” ”, m1, ” ”, p1)

We have summarized our results in figure 4 and we will discuss them now:

In the first row, I display ”n” which is the number that I used in my calculation (how far I

am calculating). I only displayed a limited amount of n for space reasons. In the second row,

I list the result of the number of negative counts. This means the number of odd multiples

the calculation encountered and this includes the single primes. In the third row, I list the

number of positive counts which represents the even multiples. In row ”5”, I subtracted the

absolute values from the calculations above and divided the result by n. This should give us

a good idea whether there is any tendency to produce more odd or even multiples. As we

guessed, we can see that this discrepancy is in general getting smaller, though we may have

the impression that it temporarily shoots up a bit after we accidentally hit a very good match

between even and odd hits for a certain n. In row ”6” and ”7”, I calculate respectively n

divided by the odd counts and n divided by the even counts. To my surprise, I see that these

values are moving towards 2.ζ(2) = π2/3. This can not be a coincidence! I added an extra

line with this ”target” as row ”8” and displayed all 3 rows in figure 5. In row ”9”, I totaled

the even and the odd occurrences and divided them by n. I call it the ”single” as this count

excludes multiple occurrences of the same prime as a divisor. Of coarse this result now goes

to 1/ζ(2) = 6/π2. In row ”11”, I also calculate 1 − singles = squares which represents the

chance that n has repeated primes as divisors. This chance is now of coarse 1−6/π2. Finally,

In row ”15”, I calculate m(x) which for the cases calculated clearly is < 1.

Conclusion: figure 3 at first gives us no reason to believe that Mertens’ Conjectures (formula

6) would be false.

Following our testing results, we would like to conjecture the following:

limn→∞

n∑nr|k=odd µ(r)

= −2.ζ(2) =−π2

3(8)

limn→∞

n∑nr|k=even µ(r)

= 2.ζ(2) =π2

3(9)

limn→∞

∑nr|k=even µ(r)−

∑nr|k=odd µ(r)

n=

6

π2=

1

ζ(2)(10)

With r|k = odd in the above we mean that we will only sum up the values ik k is odd based

on formula 5.

We see no reason to have confidence in formula 7 of Odlyzko and Riele especially since figure

5 is clearly suggesting the opposite.

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5 REFERENCES

5 References

1. en.wikipedia.org ; Mertens theorems.

2. en.wikipedia.org ; Meissel-Mertens constant.

3. www.researchgate.net profile Chris De Corte ; probabilistic approach to prime counting

; page 7 ; formula (9).

4. mathworld.wolfram.com ; Moebius Function.

5. mathworld.wolfram.com ; Mertens Conjecture.

6. www.linkedin.com groups ; Number Theory.

7. Springer ; Prime Numbers A computational perspective ; Richard Crandall and Carl

Pomerance ; p10

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5 REFERENCES

Table 1: In this table we show prime counts in column 1, the respective prime in column 2, the value

of∑

p≤nln(p)p − ln(n) in column 3, the value of

∑p≤n

1p − ln(ln(n)) in column 4.

Count Prime Mertens1 Mertens2

1 2 -0.3465735902799726 0.8665129205816644

10 29 -1.065279554926441 0.319328773958854

100 541 -1.288709102825433 0.2668375929202569

1000 7919 -1.319426316487266 0.2627432741614228

10000 104729 -1.328852576657907 0.2617825319440525

100000 1299709 -1.331638702778868 0.2615569337934529

1000000 15485863 -1.332437299245298 0.2615044140293952

10000000 179424673 -1.332469668566326 0.2615027849863427

20000000 373587883 -1.332554746979635 0.2614983789475636

30000000 573259391 -1.33254688346041 0.2614987907804349

40000000 776531401 -1.332562588288901 0.2614980277872814

50000000 982451653 -1.332556847781483 0.2614983121936398

50847534 999999937 -1.332548831967259 0.2614986992562249

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5 REFERENCES

Table 2: In this table we show the value of x in column 1, the Mertens’ value M(x) =∑x

n=1 µ(x) for

x in column 2, the value of∑x

n=1 µ(x)/√x in column 3.

x M(x) M(x)/√x

1 1 1

10 -1 -0.316227766

100 1 0.1

1000 2 0.063245553

10000 -23 -0.23

100000 -48 -0.151789328

1000000 212 0.212

1.107 1037 0.327928193

2.107 -953 -0.213097278

3.107 -1157 -0.211238333

4.107 -333 -0.052651923

5.107 -908 -0.128410591

1.109 -222 -0.007020256

2.109 6556 0.146596617

3.109 7452 0.136054283

4.109 -11741 -0.18564151

5.109 12306 0.174033121

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5 REFERENCES

Figure 1: In this figure we show the value for∑

p≤nln(p)p − ln(n) for values from 107 to 5.107.

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5 REFERENCES

Figure 2: In this figure we show the value for∑

p≤n1p − ln(ln(n)) for values from 107 to 5.107.

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5 REFERENCES

Figure 3: In this figure we show the value for∑x

n=1 µ(n)/√x for values from 107 to 5.107.

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5 REFERENCES

Figure 4: In this figure we summarize our testing results and demonstrate our conjecture about

formula’s 8, 9 and 10.

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5 REFERENCES

Figure 5: In this figure we show that the part of the odd and the even number of divisors (scenario

3 in formula 5) are alternating but moving towards the conjectured value.

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