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Re: [PrimeNumbers] prime chain,curios!

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• ... These formulas have been commonly known for a long time. +---------------------------------------------------------+ ...
Message 1 of 4 , Oct 1, 2001
At 11:03 PM 10/1/2001 +0200, Norman Luhn wrote:
>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.

These formulas have been commonly known for a long time.

+---------------------------------------------------------+
| Jud McCranie |
| |
| Programming Achieved with Structure, Clarity, And Logic |
+---------------------------------------------------------+
• ... 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
Message 2 of 4 , Oct 1, 2001
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.
• ... 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
Message 3 of 4 , Oct 2, 2001
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
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