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Old 2019-03-07, 09:34   #6
robert44444uk
 
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Jun 2003
Oxford, UK

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Quote:
Originally Posted by Dr Sardonicus View Post
One observation: If N is an interprime, N is not divisible by 3, and the smallest prime in the interval is greater than 3, then all the gaps must have length divisible by 3.

For if N =/= 0 (mod 3), p is one of the primes, p =/= N (mod 3), and p =/= 0 (mod 3), then the symmetrically located number 2*N - p will be == 0 (mod 3).

This condition might narrow things down a bit.
Neat! For 3565765770 with prime factors 2,5,356576577 the gaps are 6,6,24,6,18,18,18,6,24,6,6, all 0mod3.

It is easy to demonstrate in general that the majority of gaps are not 0mod3 in length - for example in the first 100000 gaps, the sum of 1mod3 and 2mod3 gaps is about 38.4% more than 0mod3 - and hence the longer (larger n) the n-interprime, the greater the requirement for all prime gaps involved in the interprime to be 0mod3, and subsequently the less likely the interprime is 1 or 2mod3. Using the 41.9% 0mod3 gap density in the first 100000 gaps as a given, the chances of 11 gaps in a row 0mod3 is about 14157:1 - this would provide for less 1mod3 and 2mod3 6-interprimes than are actually turning up 6 out of 896 is 149:1 - this could of course just be chance - but I wonder if I am again missing something here.

Does the ratio of the count of successive prime gaps (1mod3+2mod3)/0mod3 tend towards a constant? Can't see why not. Where can I look for work done on that?

I've tested up to 1.5e10 and, in addition to the 9-interprime mentioned above, there are two others:

74422046685
81661695390

There are 33 8-interprimes in the range 0 to 1.5e10

But no 10-interprime yet

Last fiddled with by robert44444uk on 2019-03-07 at 21:30
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