View source for 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 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. --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 02:07, 6 March 2024 (UTC)
::: (answering my own q) it is -pairs+triples-quads... as per the recently added link. --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 15:44, 7 March 2024 (UTC)
:Would something like a modified [[Legendre_prime_counting_function]] something adequate.<br>Instead testing with primes than only with primes cubed.[[Legendre_prime_counting_function#Non-recursive_partial_sieve]] --[[User:Horst|Horst]] ([[User talk:Horst|talk]]) 04:09, 7 March 2024 (UTC)

==Combinatronics==
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.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk: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! --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 15:37, 7 March 2024 (UTC)