Obtains the total number of factors of p in n! for any prime p.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | # Author: Samuel J Erickson
# Date: 8/9/2013
# Description: de Polignac's formula gives all factors of p in n! for any prime
# p.
def polignac(num,p):
"""
input: num can be any positive integer and p and prime number.
output: Gives the total number of factors of p in num! (num factorial).
Stated another way, this function returns the total number of factors
of p of all numbers between 1 and num; de Polignac's formula is pretty
nifty if you are into working with prime numbers.
"""
factorsDict={}
if num>1 and p==1:
print("Invalid entry since 1 is the identity.")
else:
while num>1:
factorsDict[num//p]=num//p
num=num//p
return sum(factorsDict)
|
I forgot to mention that you will want to call getPLargest(num,p) in order to actually obtain the output from polignac's formula.