Saturday, 5 November 2011

The n-factor grid: twin primes, Polignac primes and the Goldbach conjecture.

So, having introduced the grid of n-factor composite numbers in the previous post, how does this help us visualise prime problems?

First, let's look at twin primes. If we are looking for twin primes, we need to look at the pairs of numbers 6k-1 and 6k+1. In this image, the highlighted boxes contain the products of possible twin prime candidates (5*7, 11*13 etc). This diagonal, continued to infinity would contain all such products.

The reason for looking at the products of these numbers is this: where the numbers along this diagonal are semi-prime (yellow), the two factors on the horizontal axes must be twin primes. This is kind of obvious, since only the prime numbers on the central axes will produce a semi-prime when multiplied together. But it does provide some visual explanation of why you are likely to get a proportion of primes along a diagonal of this grid. Put in basic terms, it is obviously going to be pretty hard (especially at low numbers) to draw a diagonal that doesn't hit one of the vertical or horizontal stripes of semi-primes. It also doesn't suffer from the same problem as the original grid I was using as it does continue to distinguish primes from semi-primes and other composites.

Next, we can clearly see on this grid why, as we go up through the Sieve of Eratosthenes, the prime N will remove a maximum of 2/N of remaining twin prime candidates: Since the diagonal will pass through 1/N multiples of N on the horizonal axis and 1/N on the vertical axis, it will encounter up to 2/N multiples of N altogether*.

The twin prime conjecture is true if there is an infinite number of semi-primes on this diagonal, continued to infinity. (I'll talk about this a bit more on the next post).

Incidentally, the other diagonals on the grid all relate to other prime problems. In this diagram, the diagonals highlighted show the products of prime pairs with larger gaps, cousin primes (N, N+4), sexy primes (N, N+6) and some larger examples. If Polignac's conjecture that "for any positive even number n , there are infinitely many prime gaps of size n" is true, then any diagonal on this grid (continued to infinity) contains an infinite number of semi-primes. (Only top right is coloured).

 On the grid below, the highlighted diagonals relate to the Goldbach conjecture. In this case the chains are finite - each chain contains all the pairs of 6n+/-1 numbers that could be added together to get a particular sum, for instance for 78 the potential pairs are 73+5, 67+11, 61+17....65+13, 71+7. If  each diagonal that stretches from one axis to the other (or from an axis to the diagonal boundary of square numbers to the bottom left) contains at least one semi-prime then the Goldbach conjecture is true.

Edit: I later realise that while this is true, it would demonstrate a slightly stronger conjecture than Goldbach, since Goldbach would allow pairs of primes where one of the pair is 3 and the grid above ignores products of 3.

* 2/N is a maximum because, as previously noted, the first instance of N is not a composite so may be part of a twin prime pair. Also, for higher gaps between the primes, the maths changes slightly - for instance for primes separated by a gap of 30, the number 5 will only remove 1/5 of candidate pairs, since the diagonal will hit the 5-multiples on the horizontal and vertical axis on the same square.

Incidentally this post is a bit wrong - I should have been clearer that the prime N will remove a maximum of 2/N of remaining twin prime candidates in any infinitely repeating chain of numbers separated by the same interval - so for instance after we have sieved to 11, we know that the pair 101/103 is a twin prime. Within any higher primorial length pattern, there will be repetitions of this "twin prime candidate" gap, for instance in 13#, the gap repeats at 2310+101/3, 4620+101/3, 6930+101/3 and on to infinity. Some of these candidates will be eliminated by higher primes, but the prime P will only eliminate 2/P of all instances. All a bit more complicated then...


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  4. Basit matematik ile goldbach sanısının çözümü

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