Dear Greg

Thank you for your prompt feedbacks.

First I'll try to answer to this one made of 4 questions :

A "Goldbach partition" of an even number 2n is a set of two prime numbers

the sum of which equals 2n. If n is prime, the set (n, n) is also a Goldbach

partition.

I used this vocabulary after reading lots of good internet pages about the

Goldbach conjecture.

The "count" I show in my table is the number of distinct Goldbach partitions

for an even number 2n. One can find an applet on internet that gives the

list

of Goldbach partitions for "any" proposed even number.

The "evaluation" is an attempt to evaluate that said count by a predicting

formula. Observing how "precise" the result is may give a clue as to "solve"

the Goldbach conjecture.

For the moment I do not disclose that formula. I might do so in the future

depending on further feedbacks and feelings.

Feel free to continue fruitful interaction. Offline or online as you wish.

I will be glad to visit again the links you suggested and maybe add

something

here about them.

Rgds.

Didier

-----Message d'origine-----

De :

Greg@... [mailto:

Greg@...]

Envoye : mercredi 17 novembre 2004 20:39

A :

vdstrat@...;

primenumbers@yahoogroups.com
Objet : Re: [PrimeNumbers] Goldbach (again)

In a message dated 11/17/04 9:42:05 AM Eastern Standard Time,

vdstrat@... writes:

> As a starter I propose a method to evaluate the count of Goldbach

> partitions, based on heuristic formulas

> similar to those about twin primes and about pi(x).

>

> I have created a page and a table about this, which is draft version for

the

> time being :

> http://www.geocities.com/dhvanderstraten/gldbevtxt.html

> http://www.geocities.com/dhvanderstraten/gldbevtbl1.html

I didn't understand your table.

2n = 530 and its count is 14.

What are you counting? What is "eval."

Where is the Excel spreadsheet formula?

A set of 2n elements has 2^2n subsets including the empty set and 2n itself.

Where r combinations are expressed as (2^2n/r) then

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

Eg. 2^5 = 32 = 1 + 5 + 10 + 10 + 5 + 1

Alternatively, an unrestricted partition of 5 includes these 5 sums:

4 + 1,

3 + 2,

3 + 1 + 1,

2 + 2 + 1,

2 + 1 + 1 + 1,

1 + 1+ 1 + 1 + 1

Regardless of whatever it is you are counting, I don't see the

"clue" that links such counts to the Goldbach Conjecture.

See if my Stepladder approach offers any clue at:

http://gmazur.freeyellow.com/Goldbach.com
(At least to keep the thread from fading into obscurity)

Greg

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