The first 50 million prime numbers -bonn.mpg.de

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The First 50 Million Prime Numbers*

Don Zagier

To my parents

sin ~F(s)

I -

s ;

sin -s

for the algebraist it is "the characteristic of a finite field"

or "a point in Spec ~"

or "a non-archimedean valuation";

the combinatorist defines the prime numb e r s i n d u c t i v e l y b y the r e c u r s i o n (I)

1 Pn+1 = [I - log2( ~ +

n

(-1)r

)]

r~1 ISil 1200)

it is also valid for 2n+I, completing the induction.

For the bound in the other direction, we need a simple lemma which can be proved easily using the well-known formula for the power of p which divides n! (17) :

Lemma: Let p be a prime. If pVp is the lar-

(~), gest power of p dividing then

pVP < n.

Corollary: Evergbinemial coefficient"(~)

satisfies

(~) = ~ pVp _< n~(n)

p~n

If we add the inequality of the corollary

for all b i n o m i a l c o e f f i c i e n t s (~)

with given n, then we find n

2n = (I+I) n = Z (~) S (n+l).n ~(n) k=O

and taking logarithms gives

~(n) > n log 2 log (n+1)

- 'log n -

log n

>2 n 3 log n

(n > 200).

In closing, I would like to say a few words about Riemann's work. Although Riemann never proved the prime number theo-

Pl = ~I + 1 4 . 1 3 4 7 2 5 i, P2 = ~I + 2 1 . 0 2 2 0 4 0 i, P3 = ~I + 2 5 . 0 1 0 8 5 6 i, P4 = ~I + 3 0 . 4 2 4 8 7 8 i, P5 = ~I + 3 2 . 9 3 5 0 5 7 i,

Pl ~ 89- 14.134725 i, I

P2 = ~ - 21.022040 i, P3 = ~1 - 2 5 . 0 1 0 8 5 6 i,

54 = 89- 3 0 . 4 2 4 8 7 8 i,

55 = 89- 3 2 . 9 3 5 0 5 7 i.

It is easy to show that with each root its complex conjugate also appears. But that the real part of every root is exactly I/2 is still unproved: this is the famous Riemann hypothesis, which would have far-readhing consequences for number theory(20). It has been verified for 7 million roots.

With the help of the Riemann function R(X) introduced above we can write Riemann's formula in the form

~ ( x ) = R ( x ) - Z R ( x p) P

The k th a p p r o x i m a t i o n to ~ (x) w h i c h this formula yields is the function

Rk(X) = R(x) + TI(X) + T2(x) +...+ Tk(X) ,

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