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Erdős problem 726 As n → ∞ n\to \infty n → ∞ ranges over integers
∑ p ≤ n 1 n ∈ ( p / 2 , p ) ( m o d p ) 1 p ∼ log log n 2 \sum_{p\leq n}1_{n\in (p/2,p)\pmod{p}}\frac{1}{p}\sim \frac{\log\log n}{2} ∑ p ≤ n 1 n ∈ ( p /2 , p ) ( mod p ) p 1 ∼ 2 l o g l o g n ?
A conjecture of Erdős, Graham, Ruzsa, and Straus [EGRS75].
By
n ∈ ( p / 2 , p ) ( m o d p ) n\in (p/2,p)\pmod{p} n ∈ ( p /2 , p ) ( mod p ) we mean
n ≡ r ( m o d p ) n\equiv r\pmod{p} n ≡ r ( mod p ) for some integer
with
.
References
[EGRS75] Erdős, P., and Graham, R. L. and Ruzsa, I. Z. and Straus, E. G., On the prime factors of ( \sp 2 n \sb n ) (\sp{2n}\sb{n}) ( \sp 2 n \sb n ) . Math. Comp. (1975), 83-92. Two ways to claim this Each is a separate task with its own bundle and its own bounty. Pick the one your proof argues for.
Bounty
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3ddd79b5dceb · 5 hours ago
Lean type
True ↔
Asymptotics.IsEquivalent Filter.atTop
(fun n => ∑ p ∈ Finset.range (n + 1) with Nat.Prime p ∧ ↑p / 2 < ↑n % ↑p, 1 / ↑p) fun n =>
Real.log (Real.log ↑n) / 2What you must prove
import FormalConjectures.ErdosProblems.«726»
import TaskSupport
namespace Bounty
theorem target : ¬ (fcTypeOfName% "Erdos726.erdos_726") := by
sorry
end Bounty
Pinned source: FormalConjectures/ErdosProblems/726.lean
Source type SHA-256 sha256:48a37b8b57bd535899e5cf8ee92fafd7e186700826b378d3155894413c941648
Task id fc-379fc029-erdos726-erdos-726-21ddd3c4de-counterexample-v1
Task commitment sha256:741ea699ff1831f7e96fe36ddac06072e5beb12c4c1b62132d7fd9243ddddda7 Something wrong with this formalization?
A statement that does not faithfully capture the original conjecture is the one real risk here, so we would rather hear about it early - before someone spends weeks on it.
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Erdős problem 726 · Conjectures.io