This post has to be read after the previous post on the same topic. https://vbala99.blogspot.com/2019/04/estimating-next-prime-number-conts.html
Having understood Gap and R1, we proceed further. Take the prime 523 which has a high Gap. Note that there are other primes with even higher Gaps. And the only prime N1 whose Gap is higher (though marginally) than square root of N1 is 113 as we saw in the previous post.
523 R1+n*factor (only evens upto 38)
R1
3: 2 8 14 20 26
5: 2 12 22 32
7: 2 16 30
11: 5 16 38
13: 10 36
17: 4 38
19: 9 28
23: 6
Given below is a ready reckoner for prime factors 3,5,7,11,13 and sample data for prime factors 17,19,23.
It lists the even values of R1+n*factor for each combination of factor and R1. Obviously R1 can never exceed the factor.
Let's assume there is a prime number N1 for which we have found out the R1 values for each prime factor.
Take the first row below. When R1 for factor 3 is 1, the even numbers in R1+n*factor are 4, 10, 16, 22, 28, 34, 40 etc upto the square root of N1
Take the fourth row. When R1 for factor 5 is 2, the even numbers are 12, 22, 32 etc.
3: 1 4 10 16 22 28 34 40...
3: 2 2 8 14 20 26 32 38...
5: 1 6 16 26 36...
5: 2 2 12 22 32...
5: 3 8 18 28...
5: 4 14 24 34...
7: 1 8 22 36...
7: 2 16 30...
7: 3 10 24 38...
7: 4 18 32...
7: 5 12 26...
7: 6 20 34...
11: 1 12 34...
11: 2 24...
11: 3 14 36...
11: 4 26...
11: 5 16 38...
11: 6 28...
11: 7 18 40...
11: 8 30...
11: 9 20...
11: 10 32...
13: 1 14...
13: 2 28...
13: 3 16...
13: 4 30...
13: 5 18...
13: 6 32...
13: 7 20...
13: 8 34...
13: 9 22...
13: 10 36...
13: 11 24...
13: 12 38...
17: 1 18...
17: 2 36...
17: 3 20...
17: 4 38...
19: 1 20...
19: 2 40...
19: 3 22...
19: 4
19: 5 24...
19: 6
19: 7 26...
19: 8
19: 9 28...
23: 1 24...
23: 2
23: 3 26...
23: 4
23: 5 28...
23: 6
...
This lookup is provided in the LOOKUP sheet in https://docs.google.com/spreadsheets/d/1nppxKrfkNOpI22NfzWP_ZNerC9wzvIBSJm8WOb9BqMQ/edit?usp=drivesdk. The Prime Factor sheet calculates the next prime when given a prime.
When you have a prime that has an R1 of 1 for factor 3 and R1 2 for factor 5 then take the union of the sets in the first and fourth rows and find the least even number. Of, course you have to consider ALL primes higher than 3,5 also upto the square root of the current prime N1 that you are considering. If N1 is 523, you have to consider prime factors upto 23. For each prime factor, you will take one row depending upon the value of R1. If Oppermann's Conjecture is true, then you will find an even number, less than the square root of N1, that doesn't belong to the union of the sets as explained before. The reverse is also true. If for every N1, you are always able to find such an even, then the Conjecture is true. Incidentally this even number is nothing but the Gap.
The only trivial exception as I mentioned earlier is N1= 113 for which the Gap (=14) is slightly higher than the square root of N1.
Additional Reading
- https://en.wikipedia.org/wiki/Prime_gap
- http://math.info/Arithmetic/Prime_Numbers_List/ List of prime numbers
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