Thread: Covering sets
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Old 2016-01-09, 11:06   #4
robert44444uk's Avatar
Jun 2003
Oxford, UK

22×13×37 Posts

Originally Posted by axn View Post
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.

Originally Posted by axn View Post

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
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