human
Sum of GCD of all numbers upto N with N itself | Function to Find Sum of GCD of each numbers.
Consider all prime factors of no . and subtract their multiples from result.
Check if p is a prime factor.
If yes , then update no and result.
If no has a prime factor greater than Math . sqrt ( n ) then at - most one such prime factor exists.
Return the result.
Finding GCD of pairs.
Calculate the divisors.
Return count of numbers from 1 to N with GCD d with N.
Check if d1 and d2 are equal then skip this.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to sum of GCD of all numbers upto N with N itself and the other steps required.
 def getCount ( d , n ) :
	no = n // d ;
	result = no ;
	for p in range ( 2 , int ( pow ( no , 1 / 2 ) ) + 1 ) :
		if ( no % p == 0 ) :
			while ( no % p == 0 ) :
				no //= p ;
			result -= result // p ;
	if ( no > 1 ) :
		result -= result // no ;
	return result ;
def sumOfGCDofPairs ( n ) :
	res = 0 ;
	for i in range ( 1 , int ( pow ( n , 1 / 2 ) ) + 1 ) :
		if ( n % i == 0 ) :
			d1 = i ;
			d2 = n // i ;
			res += d1 * getCount ( d1 , n ) ;
			if ( d1 != d2 ) :
				res += d2 * getCount ( d2 , n ) ;
	return res ;
if __name__ == ' _ _ main _ _ ' :
	n = 12 ;
	print ( sumOfGCDofPairs ( n ) ) ;