Factorization of floor functions of fractions

[DirichletHole]

hey so

if you are taking a floor function of a fraction >1, is there any way to predict anything about it's factorization?

what about when the numerator is a factorial and the denominator is made up of factors that divide said factorial but to larger exponents then those that divide the factorial?

any info welcome

 

[Stephen Tashi]

if you are taking a floor function of a fraction >1, is there any way to predict anything about it's factorization?

It isn't clear what you are asking. I suggest you give some examples.

 

[DirichletHole]

ok. hmm...

say you are looking at floor(B/A)

based on B and/or A, is there anyway to determine the factors of floor(B/A)?

specifically B will be equal to x! (x factorial for any x that is a pos int) with all common factors of A removed. A will be x-smooth, less than B, and based on the rules for B clearly coprime to B, but aside from that there are no rules set for A. Is there perhaps an A s.t. floor(B/A) will have no prime factors less than or equal to x?

 

[mfb]

Counterexample: Let x=4.
A=x+1=5
B=4!=24, no factors to remove.
Clearly coprime.

floor(B/A)=4, which has a factor smaller than x=4.

 

[DirichletHole]

Counterexample: Let x=4.
A=x+1=5
B=4!=24, no factors to remove.
Clearly coprime.

floor(B/A)=4, which has a factor smaller than x=4.

first off, one of the requirements of A is that it only has prime factors less then or equal to x.
secondly, I already know that it is very possible for an example such as you have given to exist. What I want to know is:
Are there PARAMETERS that make it impossible for floor(B/A) to have common factors with B?

in other words
when x is any positive integer > 3
for A is divisible only by prime factors less then or equal to x
for B = x! with all the prime factors shared with A divided out so that gcd(B,A) = 1
are there any functions that satisfy the requirements of A such that the following statement always holds true:
gcd(B, floor(B/A)) = 1
?