Jacobsthal's function and the quadratic bound

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Jacobsthal's function and the quadratic bound

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Take any integer with at most k prime factors. How long can a run of consecutive integers get when every one of them shares a factor with it? This class shows you how to bound that length, and how the new proof that it grows at most quadratically fits together.

From covering to counting

Jacobsthal's function j(n) is the least m such that every m consecutive integers include one coprime to n. h(k) is the largest j(n) over all n with at most k prime divisors, so h(k) − 1 is the longest run that k primes can cover [0:05]. For example, 2, 3 and 5 cover 24–28, and h(3) = 6. The first k primes are the worst case up to k = 23, but not at k = 24.
Jacobsthal asked whether h(k) ≪ k². Iwaniec proved k² log² k, and the best lower bound is about k(log k)². The new result gives h(k) ≤ C k²/(log log 3k)² [0:54].
Turn it around and count survivors. Forbid one class aₚ at each prime p ≤ z. By the CRT, this is the same as being coprime to ∏p after a shift. Theorem 1.2 says at least c·YV₀/B² integers in [1, Y] survive [1:32].

Why a count, not positivity

If no choice can cover Y integers, then no run of length Y is covered by the primes ≤ z. But n can also have prime divisors above z, and each one strikes out survivors [2:21]. Take z ≍ k log k / log log k. Then Y ≍ k²/(log log k)², and the at most k large primes remove only O(k(1 + X/log X)) survivors, where X = Y/z. That is a small fraction of the guaranteed count, so an integer coprime to n remains [2:59].

The engine

Decreasing-prime tree: S_d = N_d − Σ S_dp. Signs alternate with depth, as in Brun's sieve. An admission rule keeps the progressions long, and replacing N_d by μ_d = (Y/d)V₀ gives the reference tree [3:36].
Parameter 2: the lower linear-sieve function f vanishes at s = 2, which is where the exponent 2 comes from. A boundary term of size B⁻² survives. Derivative weights of f and F turn paths into a Markov chain with regenerations, and the key renewal theorem evaluates the term [4:19].
Comparison: the fundamental lemma handles long progressions. On short ones, an inverse estimate forces alignment with a rational A/D. Variance bounds and stops at even nodes keep the losses inside the positive margin [5:12].

Next time you read a sieve bound for gaps or coverings, ask: does it prove positivity, or a count that pays for the primes it left out?

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