human
Highly Composite Numbers | Function to count the number of divisors of the N.
Sieve method for prime calculation.
Traversing through all prime numbers.
Calculate number of divisor with formula total div = ( p1 + 1 ) * ( p2 + 1 ) * ... . . * ( pn + 1 ) where n = ( a1 ^ p1 ) * ( a2 ^ p2 ) . ... * ( an ^ pn ) ai being prime divisor for n and pi are their respective power in factorization.
Function to check if a number is a highly composite number.
Count number of factors of N.
Loop to count number of factors of every number less than N.
If any number less than N has more factors than N , then return false.
Given Number N.
Function Call"
Solution in Python:
 ... Can you provide a code to highly Composite Numbers and the other steps required?
assistant
#Here is code to highly Composite Numbers and the other steps required.
 def divCount ( n ) :
	Hash = [ True for i in range ( n + 1 ) ]
	p = 2
	while ( ( p * p ) < n ) :
		if bool ( Hash [ p ] ) :
			i = p * 2
			while i < n :
				Hash [ i ] = False
				i += p
		p += 1
	total = 1
	for P in range ( 2 , n + 1 ) :
		if ( bool ( Hash [ P ] ) ) :
			count = 0
			if ( n % P == 0 ) :
				while ( n % P == 0 ) :
					n = n // P
					count += 1
				total = total * ( count + 1 )
	return total
def isHighlyCompositeNumber ( N ) :
	NdivCount = divCount ( N )
	for i in range ( N ) :
		idivCount = divCount ( i )
		if ( idivCount >= NdivCount ) :
			return bool ( False )
	return bool ( True )
N = 12
if ( bool ( isHighlyCompositeNumber ( N ) ) ) :
	print ( " Yes " )
else :
	print ( " No " )