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If one transforms sample data of prime gaps by $g_n := (-1)^n (p_{n+1} - p_n)$ then tests the means of positive and negative gaps one finds the mean difference between groups to be 0. Taking the mean value of the gaps $\mu_n$ to give $p_n + \mu_n$ one can derive a +-$1.96\sigma_n$ confidence interval, assuing normal distribution, such that $p_{n+1}\in I$ for arbitrarily large n. Here are my results using 3000 gaps:

Welch Two Sample t-test

data: positivegaps and negativegaps t = 74.083, df = 2999, p-value < 2.2e-16 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 17.82724 18.79656 sample estimates: mean of x mean of y 9.148568 -9.163333

My question is - why do the prime gaps seemingly sum to zero when transformed this way?

B. Jenkins
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  • Have a look at : https://math.stackexchange.com/questions/2314467/on-the-regularity-of-the-alterning-sum-of-prime-numbers – Lourrran Oct 07 '23 at 07:39
  • Thank you so much for the link, i got similar results. I find it so strange that if one takes the mean of the prime gaps up to some arbitrary number of gaps, n, that this mean(n) + p(n) is extremely close to p(n+1), even for arbitrarily large n? – B. Jenkins Oct 08 '23 at 00:49
  • Density of primes is very high. For number like $10^{20}$ the gap between 2 consecutives primes is in average $40$ or $50$ ; if we get an example where the gap is $400$, it is very very big compared to the average $40$, but in your process, it is very very small compared to $10^{20}$. – Lourrran Oct 08 '23 at 07:41
  • Agreed. The Central limit theorem will imply that most of the gaps are close to the sample mean when you take an infinity-sized sample. 'Big gaps' will be relative to whatever sample size is taken. I would've thought the best estimator would remain the mean regardless of sample size? – B. Jenkins Oct 10 '23 at 04:27

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