The following reasoning does not disprove this conjecture, but it does cast serious doubt about it.

Consider the family of natural numbers which begins as 2, 2×3, 2x3x5, 2x3x5x7, 2x3x5x7x11, … and continues indefinitely.

The nth member of this family will be the product of the first n prime numbers and may be denoted Pn. They may be called ‘dense prime evens’. The nth prime number may be denoted un.

[There is a page (211) of the masterpiece The Mathematical Experience (Philip Davis and Reuben Hersh 1980) which prints the first 2,500 prime numbers. The value of u2500 is 22307. But the dense prime even P2500 would be a natural number —the product of all these primes— with about 11,250 digits, so we may never know the largest prime number which falls within P2500, i.e. the prime number ut immediately below P2500. P2500 itself would be relatively easily to compute using today’s computer power, though printing the answer would require quite a few pages packed with digits.]

The identification of two primes which add-up to an even number 2y may be called it’s “breakdown”. The “breakdown point” is at the smaller of these two primes.

It is clear that the breakdown of Pn cannot involve any prime number which is a factor of Pn, because if it were, uj , the other breakdown “putative prime”, would have to be Pn – uj.. Unfortunately this has the factor uj, so it is not a prime number.

This means that the partition cannot occur in the range of primes between 0 and un.

Nor can Pn/2 + uj be involved for the same reason. There are three other ranges in which the partition can’t fall: between Pn/2 and Pn/2+un, between Pn/2 and Pn/2 – un and between Pn and Pn-un. All the numbers in these ranges have a factor in common with Pn.

The situation may be shown by a diagram:

0….—————————————,,,,Pn/2….—————————————,,,,Pn.

Here the numberline depicts the dense prime even number Pn . Its partition into two prime numbers cannot occur in the bits marked ,,,, or ….which represent the ranges mention above. [Because Pn is a product of all the first n prime numbers, it is very large, and this diagram does not realistically show its size relative to that of the nth prime number.]

Now as n increases indefinitely, Pn will swallow up more and more prime numbers which cannot be part of any Goldberg breakdown of very large dense prime evens, Pns. If we assume that the likely breakdown point for a very large Pn is in some random way located across Pn, it seems likely that eventually it would have to fall into one of the four impossible regions marked ,,,, and …. above. In other words, it wouldn’t happen.

So the family of dense prime evens is a family on numbers which poses the greatest strain for Goldach’s conjecture. It is an infinite family, and we can only assume that patterns in mathematics continue to infinity when some actual logical repeated occurrence points to this. So to establish Goldbach’s Conjecture it would be necessary to find a logical condition which points to this. The initial presumption must switch, surely, to the likelihood that a breakdown will eventually fail to occur.

When n is greater than 2500 the only primes involved in possible Goldberg breakdowns of dense prime evens are very large indeed, and much thinner on the ground than the examples given in Davis and Hersh’s book (pages 178-179). All the smaller prime numbers have long since been disqualified: but these were the prime numbers most likely needed to fill small gaps.

So a challenge for the future will be to find the smallest value of Pn which cannot be equated to the sum of two prime numbers.

**CHRISTOPHER ORMELL 1st August 2024 If you would like a free online copy of the P E R Narrative Maths Manifesto, send an email requesting this to per4group@gmail.com Also comments on the reasoning in this or earlier blogs in this series can be submitted by email to the same address. This includes any counter-argument submitted as a bid for the prize offered in blog 49.**