Prime-pair count using the coefficient-five reciprocal sieve
ProvedOddGoldbach19.prime_pair_count_base_fivelarge-sievenumber-theoryprime-pairs
Let x,d be natural numbers with x>=5 and d positive and even. Count the integers n for which n and n+d are prime and n+d<=x. Put z=sqrt(x)/sqrt5. Let K(d) be the product over odd prime factors p of d of (p-1)/(p-2). For a squarefree integer q define g(q) as the product over prime factors p of q of1 when p=2 and2/(p-2) otherwise; set g(q)=0 for non-squarefree q. The prime-pair count is at most K(d)*x divided by the sum of g(q)/(1+q/z) over1<=q<=floor(z), plus z+1. This replaces the previous cutoff sqrt(x)/4 with a larger cutoff justified by the new sieve coefficient5.
Preamble
import Mathlib open scoped BigOperators
Formal statement
theorem OddGoldbach19.prime_pair_count_base_five
(x d : ℕ) (hx : 5 ≤ x) (hd0 : 0 < d) (hd : 2 ∣ d) :
(∑ n ∈ Finset.range (x+1),
if Nat.Prime n ∧ Nat.Prime (n+d) ∧ n+d ≤ x then (1 : ℝ) else 0) ≤
((∏ p ∈ d.primeFactors, if 2 < p then ((p : ℝ)-1)/((p : ℝ)-2) else 1) * x) /
(∑ q ∈ Finset.Icc 1 ⌊Real.sqrt (x : ℝ)/Real.sqrt (5 : ℝ)⌋₊,
(if Squarefree q then ∏ p ∈ q.primeFactors,
if p = 2 then (1 : ℝ) else 2 / ((p : ℝ)-2) else 0) /
(1+(q : ℝ)/(Real.sqrt (x : ℝ)/Real.sqrt (5 : ℝ)))) + Real.sqrt (x : ℝ)/Real.sqrt (5 : ℝ) + 1 := by sorrySource
Adapted from accepted Prove2Me submissioned34ad35-bd22-4e5d-bf6f-b8b128b6e6c1 for PrimePairSieve_prime_pair_count_base_reciprocal_bound (public Apache2.0 contribution). Uses new proved coefficient-five weighted sifted inequality and existing divisor-kernel comparison. Cutoffsqrt(x/5) derived here.