- On Thu, 29 November 2001, d.broadhurst@... wrote:
> Titanic irregular primes: proposal for a Caldwell top-20

[SNIP]

> Kummer was able to prove Fermat's last theorem for all prime

My guess would be that Irregular primes are thus as archivable as Sophie Germains.

> exponents that are "regular".

Phil

Don't be fooled, CRC Press are _not_ the good guys.

They've taken Wolfram's money - _don't_ give them yours.

http://mathworld.wolfram.com/erics_commentary.html

Find the best deals on the web at AltaVista Shopping!

http://www.shopping.altavista.com - Ironies seem to abound in mathematics. It is so easy to show

that almost all numbers are transcendental, but to determine if a

given number is transcendental can be very difficult.

I will indeed grant the request for the "top 20" category irregular

primes, which are both so common (in theory) and so rare (in

lists of proven titanics).

Thank you David for

1. Knowing the criteria, and giving the explicit references.

2. Writing a fine note which I hope I can use (with credit of course)

on the top-20 page and in the glossary

3. Showing there is effort involved in finding them.

The last criteria is never stated explicitly, but consider the category

"odd primes" or

"primes congruent to 7 mod 8." There would be no trouble finding

references for these

(use the dozens of sum of squares articles... for the latter), but surely

they do not

deserve a comment. They are too easy to recognize.

To make your request perfect, all you need to add is one thing: a quick way

for the

parser to estimate the size... they had a nice Fourier expansion don't

they? I need

to get my books out--but right now back to class!

Chris. - Chris Caldwell asked for

> a quick way for the parser to estimate the size...

Oh that one is easy:

|B(2*n)|=2*(2*n)!*zeta(2*n)/(2*pi)^(2*n)

zeta(2*n) = 1 + 1/2^(2*n) + 1/3^(2*n) + ....

so you will (almost) always get the number digits of

a titanic irregular prime from setting the zeta to 1.

and taking logs. Your parser can handle log factorial,

so you only need to tell it log(2*pi).

Best

David - On Fri, 30 November 2001, Chris Caldwell wrote:
> To make your request perfect, all you need to add is one thing: a quick way

The top 25 hits on google between them yield nothing better than:

> for the

> parser to estimate the size... they had a nice Fourier expansion don't

> they? I need

> to get my books out--but right now back to class!

>

numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html

<<<

log|N_2k| = 2klog(k)+O(k).>>>

However, there's no indication of the size of the error term, so this approximation might be useless.

On the same page there are also approximations for the Bernoulli numbers themselves, so as long as submitters factor the denominator back in, you should be OK.

(Don't bother trying to look at that page unless you use a graphical browser and a MS operating system - it's a MS-Frottage-generated non-conformant mess.)

Phil

Don't be fooled, CRC Press are _not_ the good guys.

They've taken Wolfram's money - _don't_ give them yours.

http://mathworld.wolfram.com/erics_commentary.html

Find the best deals on the web at AltaVista Shopping!

http://www.shopping.altavista.com