- Hi all!

Perhaps,i have found a new methode to generate

primes. Here is the largest example,where i found it.

We get (41)! primes for this equal px=41+(x^2-x)

for x=0 to 40. I think it is fantastic.

A smaller example is 17 primes for

px=17+(x^2-x), x=0 to 16.

best wishes

Norman.

__________________________________________________________________

Do You Yahoo!?

Gesendet von Yahoo! Mail - http://mail.yahoo.de - At 11:03 PM 10/1/2001 +0200, Norman Luhn wrote:
>Hi all!

These formulas have been commonly known for a long time.

>Perhaps,i have found a new methode to generate

>primes. Here is the largest example,where i found it.

>

>We get (41)! primes for this equal px=41+(x^2-x)

>for x=0 to 40. I think it is fantastic.

>

>A smaller example is 17 primes for

>px=17+(x^2-x), x=0 to 16.

+---------------------------------------------------------+

| Jud McCranie |

| |

| Programming Achieved with Structure, Clarity, And Logic |

+---------------------------------------------------------+ - Jud McCranie asked:
> The class number problem, right?

Precisely! There is an informative note by Christian Radoux in

http://www.math.niu.edu/~rusin/known-math/98/163

and Ribenboim p.200 completes the link to Euler's 41. - Thomas Hadley asked:

> Is 744 a fudge factor?

No, it's very much part of the theory:

744=240*3+24

where 240 comes from Rademacher and 24 (as ever) from Ramanujan.

[BTW: guess why bosonic string theory started in 24+2=26 dimensions.]

Here's a bit of Pari:

\p1000

q=-1/exp(Pi*sqrt(4*41-1));

J(m)=(1+240*sum(k=1,m,sigma(k,3)*q^k))^3/prod(j=1,m,1-q^j)^24/q;

forstep(m=1,61,10,print((-J(m))^(1/3)));

You were taking only m=1 term in the Rademacher sum and Ramanujan

product. If you take m=61 terms, then you get to within better than

1 part in 10^1000 of the integer

N=640320=2^6*3*5*23*29,

where(-N)^3 is the modular invariant.

To see how Norman's (Leonard's, actually)

other examples fare, compare the approximations

(exp(Pi*sqrt(4*41-1))-744)^(1/3) \approx 640320

(exp(Pi*sqrt(4*17-1))-744)^(1/3) \approx 5280

(exp(Pi*sqrt(4*11-1))-744)^(1/3) \approx 960

(exp(Pi*sqrt(4* 5-1))-744)^(1/3) \approx 96

(exp(Pi*sqrt(4* 3-1))-744)^(1/3) \approx 32

(exp(Pi*sqrt(4* 2-1))-744)^(1/3) \approx 15

with the last being as miserable as observing that

n^2+n+2 is composite with n>0.

The connection with n^n+n+p is that neither the

prime generation with p=41 nor the modular trickery with

-D=4*41-1 can be improved upon, since |D|=163 gives the

biggest (in magnitude) discriminant of an imaginary

quadratic number field with unique factorization

(class number=1).

David

PS: Here's a place where folk factorize Laurent coefficients of J:

http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha135.htm