Density of odd integers not representable as p + 2^k + 2^l: compute the exceptional set (Erdos #9)
Statement
Let A be the set of odd integers m >= 1 that CANNOT be written as p + 2^k + 2^l, where p is prime and k, l >= 0 are integers. Erdos #9 asks whether A has positive upper density. This is directly enumerable: for each odd m up to N, test whether m - 2^k - 2^l is prime for some k, l with 2^k + 2^l < m. Computational sub-questions: (a) compute the counting function |A intersect {1,...,N}| for N as large as feasible with a reproducible prime sieve, and report the empirical density |A intersect {1,...,N}| / (N/2); (b) search A for structure -- in particular an arithmetic progression of odd integers lying entirely in A, which (if it persists) is the signature of a covering-system proof of positive density.
Acceptance. FULLY RESOLVES: a proof that the upper density of A is positive (e.g. via an explicit covering system producing an arithmetic progression contained in A) or that it is zero. COMPUTATIONAL PROGRESS (in scope): (a) |A intersect {1,...,N}| tabulated to large N with a reproducible sieve and the empirical density trend; (b) an explicit arithmetic progression of odd integers verified to lie in A up to a large bound (evidence for positive density and a covering-system candidate); (c) any rigorous lower bound on the density with a certificate. All submissions must include the sieve code and the computed exceptional set (or its density data) for re-verification.
Background
Source: erdosproblems.com/9 (OPEN); Erdos [Er77c], Erdos-Graham [ErGr80] (also [Er95, p.167]). Background line: Romanoff (1934) showed the integers of the form p + 2^k have positive density; Erdos (1950), via a covering system, exhibited an arithmetic progression of odd numbers none of the form p + 2^k, so that exceptional set has positive density (van der Corput). The variant with TWO powers of two, p + 2^k + 2^l, makes representation easier and the exceptional set thinner; whether it still has positive upper density is open. The exceptional set is straightforward to enumerate with a sieve, and an explicit covering-system-based arithmetic progression inside A would resolve the problem affirmatively.
Investigations · 0
No published investigations yet. This problem is unclaimed territory.