Greatest Primorial Divisor of N-1 or minimal gap function of P(n)
- I think the name says a lot. I will use H(N) to symbolize it. But what use is it? I will use N=91 as an example. You want to know the gap after N. Then N-1 = 90 = 3 * 2*3*5 so H(91) = 5 is the answer. It is the largest prime in the primorial. For even numbers, H(1) = 1. I used 91 for a reason. 91/7 = 13. So now I need to find the first prime before as to start the addition to find the prime gap. We find 89 is the greatest P < 91. so how big is the prime gap? g(p) = 2, H(p+H(p)) = H(p+2) = 5 H(p) + H(p+H(p)) = 7. The gap is 8. So we need to change the function to include an addition of mod 2. The reason is the sum of two odds is even which has a H(N) =1. So,
g(p) =H(p) + [H(p+H(p)) + H(p+H(p)) mod 2] + ... until the next prime.
The reason for H(N) works is the gap after an M! + 1 is M. Because there is a largest P(n) which P(n)# | M! then the gap after P(n)# + 1 is also M. I know I have not stated all of the steps. But they have been found. Take M = 6 for example, then M! = 720. 720/30 = 24. And 30 = 2*3*5, so the gap after 30 = 6.
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