Talk:Greatest prime dividing the n-th cubefree number

From Rosetta Code

The logic of cubes_before()

Fairly obviously there are 31 multiples of 8(2^3) less than 249, 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)

(answering my own q) it is -pairs+triples-quads... as per the recently added link. --Petelomax (talk) 15:44, 7 March 2024 (UTC)
Would something like a modified Legendre_prime_counting_function something adequate.
Instead testing with primes than only with primes cubed.Legendre_prime_counting_function#Non-recursive_partial_sieve --Horst (talk) 04:09, 7 March 2024 (UTC)


I've added a reference to "The number of integers in a given interval which are a multiple of at least one of the given numbers". One way to do this is to calculate the number of numbers between 1 and 26 not divisible by 8, add it to the number of numbers between 28 and 124 not divisible by 8 or 27 ... until you find the 10 millionth, then find the highest prime factor.--Nigel Galloway (talk) 14:43, 7 March 2024 (UTC)

Excellent, thanks. Turns out that (in the given link, ie -pairs+triples-quads, etc) was exactly what I was doing, so kudos to me, I suppose, for re-inventing (/blindly stumbling onto) an existing and well-known mathematical principle! --Petelomax (talk) 15:37, 7 March 2024 (UTC)