RosettaCodeData/Task/M-bius-function/Python/m-bius-function-2.py

65 lines
1.1 KiB
Python

# Python Program to evaluate
# Mobius def M(N) = 1 if N = 1
# M(N) = 0 if any prime factor
# of N is contained twice
# M(N) = (-1)^(no of distinct
# prime factors)
import math
# def to check if n
# is prime or not
def isPrime(n) :
if (n < 2) :
return False
for i in range(2, n + 1) :
if (n % i == 0) :
return False
i = i * i
return True
def mobius(n) :
p = 0
# Handling 2 separately
if (n % 2 == 0) :
n = int(n / 2)
p = p + 1
# If 2^2 also
# divides N
if (n % 2 == 0) :
return 0
# Check for all
# other prime factors
for i in range(3, int(math.sqrt(n)) + 1) :
# If i divides n
if (n % i == 0) :
n = int(n / i)
p = p + 1
# If i^2 also
# divides N
if (n % i == 0) :
return 0
i = i + 2
if(p % 2 == 0) :
return -1
else :
return 1
# Driver Code
print("Mobius numbers from 1..99:")
for i in range(1, 100):
print(f"{mobius(i):>4}", end = '')
# This code is contributed by
# Manish Shaw(manishshaw1)