# 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)
Diff to Previous Revision
--- revision 3 2013-08-10 03:27:11
+++ revision 4 2013-08-10 03:40:02
@@ -14,7 +14,10 @@
nifty if you are into working with prime numbers.
"""
factorsDict={}
- while num>1:
- factorsDict[num//p]=num//p
- num=num//p
- return sum(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)
