# Factorization of floor functions of fractions

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.

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
Mentor
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.

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
?