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#56 |
Dec 2008
you know...around...
11000010112 Posts |
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Just an intermediate result that made me go "hmmmm...". Suppose we assume \(CSG=1+O(1)\) for the gaps between non-consecutive primes, then, if I did the math right, this would imply that we also assume \(\pi(x)=Li(x)+O(\sqrt{x})\), i.e. the error term is smaller by a factor log x compared to the RH prediction. Correct [y/n]?
Code:
Outline from my train of thought: p_1 = 2 (or set p_0 = 0, say) p_k = x k = pi(x)-1 ~ pi(x) gap = x-2 ~ x m = Gram(x)-Gram(2)-k+1 ~ Gram(x)-pi(x) CSG = m*|m|/gap - but for simplicity suppose that m is positive (means we assume a scarcity instead of an abundance of primes; the error term works both ways anyway): CSG = m^2/gap ~ (Gram(x)-pi(x))^2/x CSG ~ 1 --> (Gram(x)-pi(x))^2 ~ x --> Gram(x)-pi(x) ~ sqrt(x) OTOH, if Gram(x)-pi(x) = O(sqrt(x)*log(x)), then CSG = m^2/gap ~ O((x*log²x)/x) ~ O(log²x) |
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#57 |
May 2018
26610 Posts |
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That is correct. By the way, you should submit the sequences in this thread to OEIS.
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