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• A humbling addendum: I ve worked out what the three at the end of every cycle really means. Lets just say it is far from intriguing now! Simply put, if p# =
Message 1 of 2 , May 4, 2005
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I've worked out what the "three" at the end of every cycle really
means. Lets just say it is far from intriguing now! Simply put, if
p# = 2*3*5*7*...*p, and for n>0, then between n*p# - p and n*p# + p
there are exactly two numbers having factors all greater than p. Of
course, these numbers are n*p# - 1 and n*p# + 1 . (!) There goes any
sense of mystery.... :)

Along the path of working that out however, I arrived at a nice tidy
discovery (it seems vaguely familiar though, maybe rediscovery? It
would be well known to number theorists I presume). It is this:

If p# means 2*3*5*7*...*p, and (p-1)# means (2-1)*(3-1)*(5-1)*(7-
1)*...*(p-1) then given any interval of exactly p# numbers and the
interval starting at p or greater, there will be exactly (p-1)#
numbers with factors all greater than p.

For example, if p = 3 then p# = 2*3 = 6 and (p-1)# = (2-1)*(3-1) =
2. Thus any interval of 6 numbers (the interval starting at 3 or
more) will have exactly two numbers with factors all greater than
3. For example examine the intervals (3,4,5,6,7,8)
(14,15,16,17,18,19) (90,91,92,93,94,95) etc.

Similarly, if p = 5 then any interval of 5*3*2 = 30 numbers (interval
starting at 5 or greater) will have exactly 4*2*1 = 8 numbers having
factors all greater than 5. For instance in the interval from 10 to
39 one will find exactly 8 numbers having factors all greater than
5. (Of course in this case all eight of these numbers must be
prime!) In the interval from 10001 to 10030 one will find exactly 8
numbers having factors all greater than 5.

It may be that all the intervals can start at 2 rather than p, I'll
have to look into that.

hmmm, all this may come in handy for some proof in the future :)

Mark
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