## The Two Barrier.

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• Hi all Consider the set of natural numbers. We remove from this the number 1 and all integers with factors of 2 and we are left with
Message 1 of 2 , May 4 1:21 PM
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Hi all

Consider the set of natural numbers. We remove from this the number
1 and all integers with factors of 2 and we are left with

3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51
....

Obviously (so it seems), numbers with factors of p are exactly p
numbers apart. Now we remove all integers with factors of 3 and we
are left with

5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59,61,65 ...

Now we see that numbers with factors of p are no longer exactly p
numbers apart. We focus on the prime one up from 3, that is p = 5.
Numbers with factors of 5 occur at positions 1,8,11,18,21, and so on.
We quickly see a pattern, that numbers with factors of 5 are
distanced from each other by 7,3,7,3,7,3 numbers. This has a
repeating cycle of 2.

Now we remove all integers with factors of 5 and we are left with

7,11,13,17,19,23,29,31,37,41, 43,47,49,53....

As it turns out, numbers with factors of 7 are distanced from each
other by

12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3, ect.
numbers. This has a repeating cycle of 8.

Similarly, when numbers with factors of 7 are removed, numbers with
factors of 11 have a repeating cycle of 48. Here is one cycle:

26,5,10,5,11,15,4,16,10,5,9,15,16,5,15,10,5,15,11,14,20,11,5,9,5,11,20
,14,11,15,5,10,15,5,16,15,9,5,10,16,4,15,11,5,10,5,26,3.

(Yes there is a formula for the length of the repeating cycle: (2-1)*
(3-1)*(5-1)*(7-1)*(11-1)*...*(p(n-1)-1) Found that out the hard
way :))

What is interesting about each cycle for each prime p? For one, the
first number of the cycle. This number (minus one) is the number of
numbers between p and p^2 having no factors below p. In other words
this is the number of primes between p and p^2.

Secondly, we might ask if the first number of the cycle is always the
largest in the cycle. The answer is no. For instance when p = 13, the
first cycle number is 34, but there is a larger number in the cycle,
38. When p = 17, the first cycle number is 55 but there are three
larger numbers in the cycle: 57, 58 and 71. etc.

Thirdly, (and very intriguingly to me) each cycle ends in a 3, as far
as I have checked. And each cycle (as far as I have checked) has
only one 3. Also, if this three is removed, the cycle is symmetrical
from each end. I have not grokked why each cycle appears to end in a
3.

Fourthly, as far as I have checked, this 3 is the lowest number in
the cycle. In other words, in each cycle there always appears to be
at least two numbers not having factors below p between numbers
having a factor of p. If such is actually the case, why is this?
The (exact) two barrier seems to occur exactly once every cycle of p,
and thus infinitely often for every p. Will it ever be broken? Maybe
the picture of the big number 2 on the Yahoo Prime Numbers homepage
has more significance than we thought... :)

Mark
• 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 2 of 2 , May 4 8:19 PM
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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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