Thread: Covering sets View Single Post
2016-01-09, 11:06   #4
robert44444uk

Jun 2003
Oxford, UK

2·7·137 Posts

Quote:
 Originally Posted by axn 1,1,3,4,5,9 ,1,1 : 1,1,3,4,5,9,16,3 (you had it as 0,2,3,4,5,9,16,3. is that correct?) .
You are right, I had transposed my results line incorrectly.

Quote:
 Originally Posted by axn All the lines differ from the next one by 1. i.e. They just represent shifting the starting point by 1. And within a pair, the first five modular classes are same - i.e these are primes that hit more than one point in the 30. 2,3, and 5 together eliminate 22 out of every 30 consecutive numbers (always). Of the remaining 8, the best we can do is to hit 2 each with 7,11, and 13. That leaves 2 points. Any prime > 15 can only hit at most 1 point. Therefore the last two requires two primes (any two primes > 15), which yields a family of solutions
Nice observations! A great way to explain the outcome, and maybe this is food for thought in designing a slightly more sophisticated approach to finding the minimum for 2310. I have to think why 7 can only hit 2 gaps in the 30 range, i.e. smaller primes can always hit 2 out of the 4 in the range where 7 is a factor.

Also this disproves the hypothesis I put forward, as written.

I wonder though, for a range where the total number of range members is > p#, then if 2,3,5...p have to be members of all minimum covering sets.

Last fiddled with by robert44444uk on 2016-01-09 at 11:07