- I wrote:

> zeta(-k)/zeta(-1) is prime for

"ttpi314159" wrote:

> k=25,33,37,73,117,673,

> what, if anything, comes next?

> I am puzzled,... probabilities?

Here zeta is Riemann's zeta function,

which vanishes at the even negative integers

and is rational at the odd negative integers,

with zeta(-1)=-1/12.

Primo has proven that zeta(-673)/zeta(-1) is prime.

In terms of the Bernoulli number Bern(674) it is

> 5604 2*3*Bern(674)/337 1077 c4 2002 Irregular, ECPP (notes)

with notes at

http://groups.yahoo.com/group/primenumbers/message/4197

I was wondering: what is the next k=2*p-1 for which

zeta(-k)/zeta(-1)=6*Bern(2*p)/p

is probably prime?

Yesterday, I set a Pari-GP job running on a fast unix box

with 500MB of core:

nm=3000;b=bernvec(nm);

{for(k=1,nm,bv=abs(6*b[k+1]/k);

if(denominator(bv)==1&&ispseudoprime(bv),

print([k,2*k-1])))}

[caution: set a smaller value of nm, say

nm=100, to test this code quickly, with

a small core memory requirement]

After running it overnight, I found this output:

[13, 25]

[17, 33]

[19, 37]

[37, 73]

[59, 117]

[337, 673]

[2153, 4305]

which means that

zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153

is probably prime.

As it has 10,342 decimal digits,

we will have to wait a long while for

hardware or software to develop

to the stage where a proof of primality

is feasible.

The number in question is filed in

http://groups.yahoo.com/group/primenumbers/files/Prime%20Tables/

zm4305.txt

> zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153 is PrP

Henri: Can you handle that in your database?

David - David's puzzle continues to haunt me...

von Staudt's theorem [Hardy & Wright, pp. 90 et seq.] says:-

B[2*k] = - sigma{p:(p-1)|2*k} (1/p) MOD 1

or, in words:-

the fractional part of B[2*k] is the negative of the fractional part of the

sum (1/2+1/3+1/5+1/7+...),

the sum being taken over all primes p such that (p-1) divides (2*k).

Any straightforward approach to solving David's puzzle calls for a search

over those B[2*k] with denominator 6.

The sum on the r.h.s. of the theorem must therefore be _only_ over the primes

2 and 3, giving B[2*k] = -(1/2+1/3) = -5/6 = 1/6 MOD 1.

The search can be restricted to a subset of the even numbers (2*k) by using

the following result, which may be new - at any rate, I have not come across

it in a fairly extensive literature/web search on Bernoulli numbers.

Theorem: If k has any Sophie-Germain prime factor f, then the denominator of

B[2*k] is > 6.

Proof: Suppose f is a Sophie-Germain prime factor of k, 2 <= f <= k.

Then g = 2*f+1 is a prime, 5 <= g <= (2*k+1); and (g-1) = 2*f, which divides

(2*k), so the 1/g must be included in the sum, so the denominator of B[2*k]

must be divisible by the prime g >= 5, and so must be > 6.

QED

In particular, taking f=2 shows that we can restrict attention to 2*k = 2 MOD

4, i.e. k odd; and taking f=3 shows that 2*k = 2 or 4 MOD 6.

This test reduces the number of B[2*k] which need to be evaluated by about

50%, at least for the "small" k (<= 500) which I have looked at.

Note that the converse of this theorem is false. The first countexample

occurs (I believe) for k=119=7*17, as neither 7 nor 17 is a Sophie Germain,

yet B[238] has denominator 1434=6*239.

Mike Oakes

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