Talk:Greatest prime dividing the n-th cubefree number

The logic of cubes_before()

Fairly obviously there are 31 multiples of 8(2^3) less than 250, and 9 multiples of 27(3^3), however of course we have to account for 8*27 = 216 being in both. I'm pretty sure it's fairly standard fare, but the logic of accounting for >=3 such clashes is eluding me. --Petelomax (talk) 02:07, 6 March 2024 (UTC)