A checked finding · covering systems
Can repeating patterns catch every integer?
Give each odd number a remainder — 0 mod 3, 1 mod 5, and so on — and every integer it hits lights up. Try to leave no gaps. With 3, 5, 7, 9, 11 the best you can ever cover is 65.37%, and the leftover holes never close. That small, exact ceiling is a doorway to a problem that's still open after 75 years.
LCM(3,5,7,9,11) = 3465, so this one panel is the entire number line, forever.3465 cells — one full period. Green = covered, red = a hole that repeats forever. Columns run mod 77 (=7·11), rows mod 45 (=9·5), so the leftover holes fall into clean arithmetic-progression stripes.
a₉ mod 3 = 1 ≠ a₃ = 0, so the 9-pattern lands on fresh numbers the 3-pattern missed — it pulls its full weight. This is the winning move.n removes exactly 1/n, full stop. The single real decision in this whole puzzle is whether 9 reinforces 3 or dodges it.1⁄3+1⁄5+1⁄7+1⁄9+1⁄11 ≈ 0.878 < 1 — there simply isn't enough coverage to go around, even before overlaps eat into it.2, is the one you're barred from.Past 11 — keep adding steps, watch the gaps stall.
Eleven was only where the picture stopped fitting on screen, not where the math stops. Add bigger odd steps and the gaps shrink — but slower and slower, even after the coverage budget sails past 1. That stall is the clearest face of the open problem.
Σ1/n shoot past 1 while the gaps stubbornly refuse to close — that stall is the whole difficulty.Uncovered % after adding each stamp (in increasing order) — see how each new step buys less than the last:
A look at the first 231 integers (green = covered, red = a gap):
The overlap-free trick: 1 − 1/√c
Take only the odd steps in a band (x/c, x] — none divides another, so the nested-overlap waste vanishes and the coverage has an exact closed form that doesn't depend on x. It climbs toward 100% as the band widens, but only reaches it in the limit — and getting there drags the small overlapping steps back in.
x/c and x. None divides another, so coverage has a clean formula that doesn't care how big x is:At x = 1,000,000 that's the odd numbers from 11,933 up to 1,000,000.
c → ∞. And a bigger c means reaching down to smaller steps... all the way back to 3, 5, 7, the very numbers that divide each other and bring back the overlap waste this band was built to dodge. Clean coverage tops out; closing the last sliver drags the whole hard problem back in.1 − 1/√c is an average over random starting points — it says nothing about whether some hand-picked alignment leaves zero gaps, which is what covering actually demands. Second, it only describes one family (steps in a single band, none dividing another). Real coverings roam every scale and use divisible steps on purpose — the very thing this band throws away. A clean ceiling for one strategy is not a limit on all of them.≈ ½ ln(c) no matter the scale, so coverage ≈ 1 − e^(−½ln c) = 1 − 1/√c is a pure function of the band ratio. A beautiful, exact ceiling — and still not a covering.Where the doodle actually leads: a prime, forced
Push the band idea to its edge and the survivors turn out to be exactly the primes — so the question becomes whether the small-prime multiples can ever cover the whole interval. They can't, with room to spare. The margin even grows. And yet it's still not a new proof — here is the honest reason why.
(x/3, x] by every prime up to √x. Every survivor is prime, so a prime is guaranteed the moment the interval outruns the largest gap between survivors — the Jacobsthal number j(P(√x)). The entire idea becomes one inequality:Safety margin = (interval length 2x/3) ÷ (max survivor gap j). Above 1 means a prime is forced. Click a point.
j(P(√x)) is about x itself — the same size as 2x/3 — with no controlled constant below 2/3. Turning that 20× empirical cushion into a theorem is the open Jacobsthal problem. The idea is real; the last step is a genuine research frontier, not a missing trick.Covering systems were introduced by Erdős in 1950. The version above — all moduli odd, distinct, and > 1 — is Erdős Problem #7, and whether any such covering exists at all is still unknown. Hough's 2015 theorem showed the smallest modulus in any covering is bounded (later sharpened to ≤ 616,000 by Balister–Bollobás– Morris–Sahasrabudhe–Tiba), and a 2026 Lean-verified result pushed the odd case past lcm 10,000. The exact 65.37% here isn't in the literature — it's just a finite computation nobody bothered to tabulate. It rhymes with the rest of the site: the worst-case hunter asks where Goldbach is weakest; this asks where odd remainders run out of room.