Goldbach conjecture : Evaluation formula

Creation date November 17, 2004

Copyright Didier van der Straten 2004-2005


As a reminder, let us state what this outstanding conjecture says : every even number beyond 2 is the sum of two prime.

While a multitude of researchers and amateurs are striving to discover a demonstration, allowing to win a substantial reward as well as to transform the conjecture in a theorem, in the same fashion as what happened in the 1990's with the "Last Fermat Theorem" thanks to Andrew Wiles, there seems to be no known attempts to present observations as to the number of solutions, called Goldbach partitions, with successive values of said even number.

Let us restate the conjecture in that perspective :

Every even number number beyond 4 is the sum of two odd primes, in a certain number of solutions, always at least one. The number of solutions as the even number increases evolves with fluctuations, the mean trend being a slower and slower increase.

That trend is fairly well documented for instance here.

To the knowledge of the author, no attempt exists so far to present the reasons of those fluctuations and of that trend to increase.

It is hoped that presenting such an explanation could provide a clue enabling one day to demonstrate the Goldbach conjecture, or at least to find out why it could not be demonstrated, or what else should be demonstrated which would, as a consequence, assert the Goldbach conjecture.

Initial observations :

If the even number (say 2n) is divisible by 3, the number of solutions has such an increase slope that the fluctuations mentioned hereabove cannot outnumber the average trend. This is an observation and does not pretend to be a demonstration.

If 2n is divisible by 5 (and not by 3), the above slope is smaller, yet enabling to pretend the conjecture is true for all multiples of 5.

If 2n is divisible by 3 and 5, the slope is higher than with 3 only...

In that initial version of this document, the author likes to present in a tabulated form some results illustrating those observations, with an attempt to provide a formula predicting the number of solutions for sequential values of 2n, and corroborating this evaluation with the exact number Goldbach partitions.

Goldbach partition count and evaluation table example.

Besides the evaluation formula of Hardy and Littlewood as studied in the reference hereabove and as presented also here another flavor is developped here. There is a potential relationship beween the two. (to be further researched).

Is the same fashion as in the attempt to predict the prime count function pi(x) by eliminating all the composite numbers from x, one can imagine the process to eliminate from the number of partitions of 2n the count of partitions made of two (not necessarily distinct) composite numbers.

Is so doing the elimination of all partitions made of two even numbers is straightforward. The count of remaining partitions (made of two odd numbers) is easy to evaluate :

C(o-o) =... n/2

As a next step, we explore the partitions made of multiples of odd primes and start obviously with the smallest ones say 3 and 5. To be precise we first observe whether n is divisible by 3 and/or by 5, for if n is e.g. divisible by 3, we shall already discount all the partitions made of two multiples of 3.

This is also straightforward to evaluate :

C(om3-om3) =... n/3
Note this seems to include the partition (3, 2n-3), and should therefore subtract one, but what we have counted is in fact the number of intervals between qualifying partitions, both adjustments to be made are vanishing.

If n was divisible by 5 (and not by 3), an analog formula could be obtained, at the risk of counting partitions that would have been already counted when attempting to do so for composite numbers being respectively multiples of 3 and 5.

What indeed happens when attempting to count the partitions of 2n made of one odd number multiple of 3 (and not by 5) and one odd number multiple of 5 (and not by 3) is, in certain cases, to count twice some partitions made of two odd numbers multiple of 3.

In certain cases : i.e. when n is a multiple of 3. If n is not such a multiple, there is no partition made of two such multiple, and the count for "3,5" is following the evaluation :

C(om3-om5) =... n/15 - 1
The term "- 1" takes care of the partitions (3, 2n-3) and (5,2n-5).

What all this reasoning dictates is that we first establish the sequence of couple of odd prime numbers for which we attempt the evaluation :

(3;3), (3;5), (3;7), ... (5;5), (5;7), ... (7;7),...

This shows already the domain in which the above odd prime numbers are selected. The last one to consider is before we reach the square root of 2n.

If n is divisible by 3, we may skip the suite (3,5), (3,7).... as all those partitions are already counted when counting for (3,3). Corresponding remark for (5,5), (7,7) etc.

It is less obvious to grasp how duplicate counting occurs between any other two couples of odd primes. One may recall the evaluation attempt of counting composite numbers between 0 and x.

x/2 - 1,
x/3 - x/6 - 1,
x/5 - x/6 - x/10 - x/15 + x/30 - 1,
....

This leads to another "close" evaluation :

x * (1-1/2) * (1-1/3) * (1-1/5) *... - pi(sqrt(x))

The last prime to be taken in the above product being the highest prime before sqrt(x).

Without developping all the details now, that last evaluation formula "leads" to a set of analogous formulae for the count of partitions of 2n made of two composite numbers, the factors of which are only odd primes :

c(om3,om3) = (n/2)/3 * d(3;3)
with d(3;3) = 1 if n multiple of 3, 0 if not.

c(om3,om5) = (n/(3*5)-1) * (1 - d(3;3))

c(om5,om5) = (n/2)/5 * d(5;5) * (1 - t(3;5;5)/3)
with d(5;5) = 1 if n multiple of 5, 0 if not,
with t(3;5;5) = 1, to discount partitions involving the prime 3.

c(om3,om7) = (n/(3*7)-1) * (1 - d(3;3)) * (1 - t(3;3;7)/3) * (1 - t(5;3;7)/5)
with t(3;3;7) = 1
and with t(5;3;7) = 1, to discount partitions involving 3 and/or 5.
c(om5,om7) = (n/(5*7)-1) * (1 - d(5;5) * (1 - t(3;5;7)/3)
with t(3;5;7) = 1 if n multiple of 3, and 2 if not.

These findings show that if the rule determining the terms d(op;op) is easy, the task to determine the terms t(op1;op2;op3) is rather substantial.

An Excel table depicts this analysis for 2n = 838.

Observing now a "ococ partitions" count evaluation table (under construction) we may now better understand its fluctuations : they have actually two independent sources :

One is due to the (rare or rich) presence of odd prime divisors in the number 2n. The smallest counts are for numbers 2n for which n is prime or a power of 2 or the product of a power of 2 and one prime number higher than this power of 2.

The other is due to a relative unpredictability of "partitions collisions" encountered when developping the prediction terms : i.e. when we consider how the values "r" are obtained in table2, it suggests an expected mean number of "partitions collisions" : partitions associated with a couple (p1;p2) which are already counted with a preceeding couple.

This latter source is analog to the one creating the pi(x) fluctuations around a monotonous curve.

Sofar the curious reader might wonder why a prediction of the count of partitions made of two composite numbers the factors of which are only odd primes is so interesting.

These are the two reasons :

1. By the developments approached above, we see that such a prediction is possible a priori, as opposed to an attempt to directly predict the count of Goldbach partitions. The numerous existing studies seem rather good at corroborating heuristic formulae with a posteriori itemization of all the Goldbach partitions.

2. Once the above prediction is made, one can deduct an evaluation of the count of Goldbach partitions by simple subtraction :

C(op;op) = G(2n) = C(oc;oc) - (n/2 - pi(2n) + 1/2)

This results directly from analyzing the total set of partitions of 2n using only odd numbers from 3 onwards :

Let us denote by C(oc;op) the count of partitions where one of the number is an odd prime and the other an odd composite, in any order.

Counting all the odd numbers between 2 and 2n we get (1) :

(1) : 2 * C(oc;oc) + 2 * C(oc;op) + 2 * C(op;op) = n - 1

Counting all the odd prime numbers up to 2n we get (2) :

(2) : C(oc;op) + 2 * C(op;op) = pi(2n) - 1

Counting all the odd composite numbers up to 2n we get (3) :

(3) : 2 * C(oc;oc) + C(oc;op) = n - 1 - (pi(2n) - 1)

Subtracting (3) from (2) :

(4) : 2 * (C(oc;oc) - C(op:op)) = n - 1 - 2 * (pi(2n) - 1)

Hence we obtain the above formula for C(op;op) versus C(oc;oc).

Note for the attentive reader : this reasoning is only exact for even n. If n is odd a correction is to be developped taking into account whether n is odd prime of odd composite. For the moment let us consider this is a good enough approximation.

Our purpose is to seek evidence that C(op;op) is always positive and to do so we are faced with fluctuations in estimating pi(2n) as well as fluctuations in evaluating C(oc;oc) as per the above detailed approach.

A separate work discusses the pi(x) fluctuations around several trial formulae and will be pursued as time allows.

The study of the C(oc;oc) fluctuations will isolate the two different sources as presented hereabove. More precisely we will focus on the values of 2n which produce "minimal" C(oc;oc) counts, i.e. numbers of the forms 2^m and 2^m * p where m is any positive integer and p is a prime number higher than 2^m.

By "minimal" we speculate that the C(oc;oc) of that selected 2n is not higher than the ones for even numbers in the neighbourhood of 2n and not qualifying for the same selection i.e. those having one or several small odd prime as divisor(s).

The above relationship between C(op;op) and C(oc;oc) shows indeed that if C(oc;oc) passes by "minimal" values, it is likely (but not certain) that C(op;op) also passes by a corresponding minimal value.

So if we can show that minimal C(op;op) values are always positive we have a direction so as to "prove" that C(op;op) for other even numbers is also always positive.

However we might discover that the above fluctuations can defeat this speculation, and that C(op;op) for non qualifying even numbers is lower than C(op;op) for qualifying even numbers in a close neighbourhood

(to be continued)

(last update february 17, 2005)

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