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Topic: Request for help with short argument
Replies: 8   Last Post: Nov 15, 2012 2:52 PM

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manontheclaphamomnibus

Posts: 31
Registered: 11/12/12
Re: Request for help with short argument
Posted: Nov 15, 2012 2:52 PM
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Thanks for taking the trouble to explain this, trj.

I thought I'd found a way round congruency conditions by giving each p_n the freedom to divide into any of the p_n - 2 numbers in the neighbourhood of n not equal to n-1 or n+1, but I'm obviously not thinking clearly on this point.

You are saying, if I understand, that I am making an (unwarranted) assumption that multiples of various primes are scattered around each of these ranges in a 'smooth' enough distribution to use probability to predict their quantities within the range? Because if so, I thought this was avoided by having the distributions of composites dividing by p_n, p_n-1 etc all relative to each n separately, rather than sieving the whole range for mults of 11, of 13, of 17, etc. And by then considering ratios of those n inside the range, not directly the numbers of multiples of different primes inside the range, since they are in each case only considered relative to one n at a time.

For example, in my argument as I see it, some of those n-x, n+y, n-z values can be outside the range. So p_n^2 - 5 might be one of our n, even if many of the composites being checked for (such as 29 dividing into (p_n^2) + 11) are outside the range I am restricting the search for the actual n to.

Anyway, shall have to think about that for a bit. Don't want to sound stubborn or stupid. I probably need to reread material on congruences.

Thanks for explaining, and please don't feel the need to reply if this is boring for you and I am obviously just not getting this basic point.

Will reread on congruences.



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