Let
(n,n+1,n+2) represent a triplet of consecutive numbers, each having
5 distinct prime factors.
Find tne value of the lowest n.
So to get at least 5 prime factors for each of (n, n+1, n+2), one needs numbers >~10^6. Also notice that for Charlie's program, there are no triplets of 6 prime factors each under 10^7. I wonder if there is some type of trend here? Meaning, <= q factors for some value of n. I'm not saying this is necessarily related to powers of 10 at all, but I wonder what the curve looks like. Also for (n,... n+m), where m<>3.
Edited on July 2, 2015, 6:51 pm

Posted by Kenny M
on 20150702 18:49:15 