## zetaprime puzzle

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• zeta(-k)/zeta(-1) is prime for k=25,33,37,73,117,673, what, if anything, comes next?
Message 1 of 9 , Sep 1, 2002
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zeta(-k)/zeta(-1) is prime for
k=25,33,37,73,117,673,
what, if anything, comes next?
• ... I am puzzled,... probabilities?
Message 2 of 9 , Sep 2, 2002
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> zeta(-k)/zeta(-1) is prime for
> 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
Message 3 of 9 , Sep 2, 2002
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I wrote:

> zeta(-k)/zeta(-1) is prime for
> k=25,33,37,73,117,673,
> what, if anything, comes next?

"ttpi314159" wrote:

> 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

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

zm4305.txt

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

Henri: Can you handle that in your database?

David
• Purely for amusement - don t burn too many brain cells thinking about it: When is zeta(a+ib)/zeta(c+id) prime? Jon Perry perry@globalnet.co.uk
Message 4 of 9 , Sep 2, 2002
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Purely for amusement - don't burn too many brain cells thinking about it:

When is zeta(a+ib)/zeta(c+id) prime?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• In a message dated 02/09/02 11:50:29 GMT Daylight Time, ... Congrats on a really spendid puzzle, David, and on a big new prime and an even bigger PrP! I had
Message 5 of 9 , Sep 3, 2002
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In a message dated 02/09/02 11:50:29 GMT Daylight Time,

> I wrote:
>
> > zeta(-k)/zeta(-1) is prime for
> > k=25,33,37,73,117,673,
> > what, if anything, comes next?
>
>
> zeta(-4305)/zeta(-1)=6*Bern(2*2153)/2153
> is probably prime.
>
>
Congrats on a really spendid puzzle, David, and on a big new prime and an
even bigger PrP!

I had tumbled to the fact that B[k+1]*12/(k+1) must be prime, which also
entails B[k+1] = 1/6 mod 1, and was wondering how best to tackle the
computation of huge B[n]'s, when your solution thundered into my in-tray.

The problem with the most straightforward method, which uses the recurrence
relation
(n+2)B[n+1] = - sigma{k=0 to n}(n+2,k)B[k]
where (n+2,k) = (n+2)!/(k!*(n+2-k)!)
is that you have to store all the B[n]. This limits the max. index to about
n=15000 on a m/c with 500Mb memory, doesn't it?

The following URL
http://numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html
gives a mass of info on Bernoulli numbers.

And
http://www.cecm.sfu.ca/CAG/Projects/High_Precision_Numerics/H1.html
gives a nice way of computing a single B[-k], by evalutating zeta(k+1) to
"sufficient" precision, once B[-k] mod 1 is known exactly (which is an easier
computation).

So memory would be no longer a limitation, only time. And one could restrict
attention to the subset for which the fractional part = 1/6.

Why not go for the world's biggest prime?!

Mike Oakes

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• I received message from Thomas Nicely about a new record prime gap: From: Thomas R. Nicely To: Selected parties Re: A new maximal prime
Message 6 of 9 , Sep 3, 2002
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I received message from Thomas Nicely about a new record prime gap:

From: Thomas R. Nicely <trnicely@...>
To: Selected parties
Re: A new maximal prime gap of 1184
Date: 2002.09.03.0856 GMT

(1) Dr. Bertil Nyman <bertil.nyman@...> has discovered a new first
occurrence and maximal prime gap of measure 1184 following the prime
43841547845541059.
(2) Dr. Nyman first detected the gap 31 August 2002.
It was confirmed as a first occurrence and maximal prime gap 2 September 2002,
when his exhaustive scan of prime gaps was completed to 4.4445*10^16.
(3) This is the first maximal gap found since the maximal gap of 1132 following
the prime 1693182318746371, also discovered by Dr. Nyman, 24 January 1999, and
confirmed as maximal 18 February 1999 by the work of Dr. Nyman and Thomas R.
Nicely. The gap of 1132 constitutes the first occurrence of any gap of 1000 or
greater.
(4) The gap of 1184 has a merit, M=G/ln(P_1), of 30.90, second among known gaps
only to the maximal gap of 1132, whose merit is 32.28. These are the only two
gaps currently known (to me) with merits >= 30.
(5) These gaps will be among the subjects of a paper (in preparation) on first
occurrence and maximal prime gaps, authored by Dr. Nyman and Thomas R. Nicely.
(6) Further information is available at Nicely�s website
<http://www.trnicely.net>; see in particular the listing of all presently known
first occurrence and maximal prime gaps at
<http://www.trnicely.net/gaps/gaplist.html>.
(7) Feel free to distribute this information as you wish.
(8) I do not wish to spam your e-mail with unwanted information.
I have attempted to restrict this mailing list to parties believed to have an
interest in the message. If you do not wish to receive such messages
distribution list.
Regards,

Thomas R. Nicely

__________________________________________________
Do You Yahoo!?
Yahoo! Finance - Get real-time stock quotes
http://finance.yahoo.com
• ... Here is a numerical workaround; for large p, it computes twice as many terms in zeta(2*p) as needed, so the _relative_ error in ... =6*|Bern(2*p)|/p
Message 7 of 9 , Sep 3, 2002
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Mike Oakes wrote:

> The problem with the most straightforward method,
> which uses the recurrence relation
> (n+2)B[n+1] = - sigma{k=0 to n}(n+2,k)B[k]
> where (n+2,k) = (n+2)!/(k!*(n+2-k)!)
> is that you have to store all the B[n].

Here is a numerical workaround;
for large p, it computes twice as many terms in
zeta(2*p) as needed, so the _relative_ error in

|zeta(1-2*p)/zeta(-1)|
=6*|Bern(2*p)|/p
=24*(2*p-1)!/(2*Pi)^(2*p)*zeta(2*p)

is O(1/2^(2*p))

But you have to be careful in choosing the precision,
which should be comfortably more than

log(abs(zeta(1-2*p)))/log(10) decimal digits

where

|zeta(1-2*p)|=O((p/Pi/exp(1))^(2*p))

Example, with p=337:

-----

\p1200

{fn(p)=local(a,m,r);m=2*ceil(p/Pi/exp(1));
a=24*(2*p-1)!/(2*Pi)^(2*p)*sum(k=1,m,1./k^(2*p));
r=round(a);if(abs(r-a)<1/10^10&&ispseudoprime(r),print(r))}

fn(337)

------

OK, it's crude and slow, but it sure solves the storage problem.

PS: Of course one should not use Pari's "ispseudoprime".
Better to output the integer to Pfgw.

David
• Whoops! I said ... which was just the opposite of what I meant. The _absolute_ error is O(1/2^(2*p)) if you take ... terms in zeta(2*p), with p Pi*exp(1)
Message 8 of 9 , Sep 3, 2002
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Whoops! I said

> so the _relative_ error

which was just the opposite of what I meant.

The _absolute_ error is O(1/2^(2*p)) if you take

> m=2*ceil(p/Pi/exp(1))

terms in zeta(2*p), with p >> Pi*exp(1)

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:-
Message 9 of 9 , Sep 8, 2002
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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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