Okay, thanks for providing that.
]]>A simple spreadsheet comes up with the factors
5 * 1375
11 * 625
and
25 * 275
55 * 125
I still need to rebuild my database. Thanks for taking the time to look at the blog.
ray
]]>You have my humble apologies.
ray
]]>Okay, thanks.
]]>ray
]]>These add up to 1665. In addition there is one unprime with only four combinations.
What is that one number?
]]>If you have a few minutes, could you take a quick look at some other ideas that I have put together in a blog on how the primes can be thought of as two sequences. I see that I cannot include links so you would have to do a Google search with the following:
unprime6 A different way of looking at prime numbers
I have found the concept of logically dividing the primes into two sequences intriguing and useful. The 6n-1 and 6n+1 composites are perhaps even more interesting.
In any case, thanks for your help.
ray
]]>I think you are stating it incorrectly. The product of of the primes although it is of the form 6n+1, may not even be a prime. Euclid's proof has nothing to say about that number. Except that it is larger then Pn and does not have a factor in the sequence P1P2P3...Pn.
2*3*5*7*11*13 + 1 is not a prime.
Euclid's proof is for all primes up to Pn.
Let's look at Euclid's famous proof about the number of primes being infinite. It becomes obvious that Euclid's proof applies only to the 6n + 1 series.
It is easy to see that the product of P1*P2*P3...Pn + 1 will always be of the form 6n+1
This is known: All twin primes are of the form 6n + 1 and 6n -1 except for 2 and 3. For that matter, except for 2 and 3 all primes are of that form.
]]>2*3*5*7*11 = 2310. Adding 1 gives 2311. Dividing by 6 gives 385 and +1 remainder, hence 6n+1.
Think of it as 2*3*( any sequence of primes ) + 1
ray
]]>It is well known that all the primes are of the form 6n+1 and 6n-1 if I remember because we used to use that fact in programs.
In more detail, the Euclid proof of multiplying primes together, including the 2 and 3, and then adding 1 always creates a result belonging to the 6n+1 series, prime or otherwise.
This is not true. My primes could be
2*3*5*7*11+1 the 11 is of the form 6n-1. Euclid's proof works for all primes of any type.
]]>ray
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All prime numbers except for 2 and 3 belong to one of two sequences, 6n + 1 and 6n - 1.
Let's look at Euclid's famous proof about the number of primes being infinite. It becomes obvious that Euclid's proof applies only to the 6n + 1 series.
In more detail, the Euclid proof of multiplying primes together, including the 2 and 3, and then adding 1 always creates a result belonging to the 6n+1 series, prime or otherwise.
Therefore Euclids proof applies only to the primes belonging to the 6n+1 series. Are there an infinite number of 6n-1 primes? Could we change Euclids method to subtract 1 instead of adding 1 to include those 6n-1 primes? As far as I can determine, the -1 proof works just as well as the +1 proof.
Essentially for every possible twin prime, there are four possibilities with only option 1 producing a twin prime.
1. 6n-1 prime 6n+1 prime
2. 6n-1 prime 6n+1 composite
3. 6n-1 composite 6n+1 prime
4. 6n-1 composite 6n+1 composite
Now we can look at the problem in a different light, essentially like looking at a pair of scissors with two infinitely long blades instead of a sword with one infinitely long blade. The traditional sword approach presumes that something special happens for a twin prime to exist at higher values. This alternate scissors approach requires nothing special for twin primes to exist at higher values and would actually require some pattern to perpetually disallow twin primes at very high values.
If we can say that Euclids proof applies to both the 6n-1 as well as the 6n+1 series, then unless there is a recurring pattern to the distribution of primes, there is nothing to prevent a twin prime from existing at any stage.
Paradoxically, it is the lack of the so far unobserved recurring pattern for prime numbers that allows the twin primes to exist.
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