Positive density of large prime gaps

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Positive density of large prime gaps

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Recursos

You can trace why, for every fixed C, a positive proportion of the first N prime gaps exceed C log p_n. You follow the paper's chain from a weight on two adjacent intervals to an actual count of consecutive gaps, and you see why that answers the Erdős–Prachar question about p_n/n.

The question and the corollary

Record-gap results (Westzynthius, Erdős–Rankin, Ford–Green–Konyagin–Maynard–Tao) make single gaps enormous. They say nothing about how often gaps are large. The theorem gives #{n ≤ N : p_{n+1} − p_n > C log p_n} ≥ c(C)N for every large N. [0:49] Erdős–Prachar: p_n/n < p_{n+1}/(n+1) holds exactly when p_{n+1} − p_n > p_n/n. Since p_n/n ~ log p_n, the case C = 2 is enough once finitely many indices are dropped.

Why counting empty intervals is not enough

[1:25] With h = 6, the single gap from 113 to 127 contains 8 starts of empty windows. A longer gap contains even more, so a sparse set of very long gaps can account for many empty windows. [1:55] The fix is to ask for a prime just before the empty stretch. Take h = ⌊λ log X⌋, a block J = (m, m+h] and the next block I = (m+h, m+2h]. The weight W(m) ≥ 0 has mass tending to 1, a prime count in J tending to λ (detects), a prime count in I of at most ε (suppresses), and a second-moment constant K_λ that does not depend on ε.

[2:32] W is a normalized square Z², where Z is a signed sum of smooth divisor sums over shift sets S ⊆ I. A prime in J lies outside every S, so the detector stays intact. A prime at m + a in I forces that divisor to equal 1. Deleting a then couples level j with level j + 1. With alternating signs, neighboring levels nearly cancel, so w → 1 and v → 0.

From weight to gaps

[3:25] Two Cauchy–Schwarz steps turn this into δX ordinary starts with a prime in J and none in I. For each such start, take p(m), the last prime in J. It begins a gap longer than h. [3:57] A prime p can be chosen only by starts with p − h ≤ m ≤ p − 1, so at most h starts choose it, however long its gap is. That gives at least δX/h ≥ δX/(λ log X) distinct primes. Choose λ > C and X = ⌊p_N/3⌋, and you get c(C) = δ/(4λ).

Rule to reuse: to get a gap count from an interval statement, pick one canonical prime per start and bound how many starts can choose it.

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