Hardy Ramanujam theorem states that the number of prime factors of n will approximately be log(log(n)) for most natural numbers n
Examples :
5192 has 2 distinct prime factors and log(log(5192)) = 2.1615
51242183 has 3 distinct prime facts and log(log(51242183)) = 2.8765
As the statement quotes, it is only an approximation. There are counter examples such as
510510 has 7 distinct prime factors but log(log(510510)) = 2.5759
1048576 has 1 prime factor but log(log(1048576)) = 2.62922
This theorem is mainly used in approximation algorithms and its proof lead to bigger concepts in probability theory.
// CPP program to count all prime factors
#include <bits/stdc++.h>
using namespace std;
// A function to count prime factors of
// a given number n
int exactPrimeFactorCount(int n)
{
int count = 0;
if (n % 2 == 0) {
count++;
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= sqrt(n); i = i + 2) {
if (n % i == 0) {
count++;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 2
if (n > 2)
count++;
return count;
}
// driver function
int main()
{
int n = 51242183;
cout << "The number of distinct prime factors is/are "
<< exactPrimeFactorCount(n) << endl;
cout << "The value of log(log(n)) is "
<< log(log(n)) << endl;
return 0;
}
// Java program to count all prime factors
import java.io.*;
class GFG {
// A function to count prime factors of
// a given number n
static int exactPrimeFactorCount(int n)
{
int count = 0;
if (n % 2 == 0) {
count++;
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i = i + 2)
{
if (n % i == 0) {
count++;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
count++;
return count;
}
// driver function
public static void main (String[] args)
{
int n = 51242183;
System.out.println( "The number of distinct "
+ "prime factors is/are "
+ exactPrimeFactorCount(n));
System.out.println( "The value of log(log(n))"
+ " is " + Math.log(Math.log(n))) ;
}
}
// This code is contributed by anuj_67.
# Python3 program to count all
# prime factors
import math
# A function to count
# prime factors of
# a given number n
def exactPrimeFactorCount(n) :
count = 0
if (n % 2 == 0) :
count = count + 1
while (n % 2 == 0) :
n = int(n / 2)
# n must be odd at this
# point. So we can skip
# one element (Note i = i +2)
i = 3
while (i <= int(math.sqrt(n))) :
if (n % i == 0) :
count = count + 1
while (n % i == 0) :
n = int(n / i)
i = i + 2
# This condition is to
# handle the case when n
# is a prime number greater
# than 2
if (n > 2) :
count = count + 1
return count
# Driver Code
n = 51242183
print ("The number of distinct prime factors is/are {}".
format(exactPrimeFactorCount(n), end = "\n"))
print ("The value of log(log(n)) is {0:.4f}"
.format(math.log(math.log(n))))
# This code is contributed by Manish Shaw
# (manishshaw1)
// C# program to count all prime factors
using System;
class GFG {
// A function to count prime factors of
// a given number n
static int exactPrimeFactorCount(int n)
{
int count = 0;
if (n % 2 == 0) {
count++;
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this point. So
// we can skip one element
// (Note i = i +2)
for (int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
if (n % i == 0) {
count++;
while (n % i == 0)
n = n / i;
}
}
// This condition is to handle the
// case when n is a prime number
// greater than 2
if (n > 2)
count++;
return count;
}
// Driver function
public static void Main ()
{
int n = 51242183;
Console.WriteLine( "The number of"
+ " distinct prime factors is/are "
+ exactPrimeFactorCount(n));
Console.WriteLine( "The value of "
+ "log(log(n)) is "
+ Math.Log(Math.Log(n))) ;
}
}
// This code is contributed by anuj_67.
<?php
// PHP program to count all prime factors
// A function to count
// prime factors of
// a given number n
function exactPrimeFactorCount($n)
{
$count = 0;
if ($n % 2 == 0)
{
$count++;
while ($n % 2 == 0)
$n = $n / 2;
}
// n must be odd at this
// point. So we can skip
// one element (Note i = i +2)
for($i = 3; $i <= sqrt($n); $i = $i + 2)
{
if ($n % $i == 0)
{
$count++;
while ($n % $i == 0)
$n = $n / $i;
}
}
// This condition is to
// handle the case when n
// is a prime number greater
// than 2
if ($n > 2)
$count++;
return $count;
}
// Driver Code
$n = 51242183;
echo "The number of distinct prime".
" factors is/are ",exactPrimeFactorCount($n),"\n";
echo "The value of log(log(n)) ".
"is ",log(log($n)),"\n";
// This code is contributed by m_kit
?>
<script>
// Javascript program to count all prime factors
// A function to count
// prime factors of
// a given number n
function exactPrimeFactorCount(n)
{
let count = 0;
if (n % 2 == 0)
{
count++;
while (n % 2 == 0)
n = n / 2;
}
// n must be odd at this
// point. So we can skip
// one element (Note i = i +2)
for(let i = 3; i <= Math.sqrt(n); i = i + 2)
{
if (n % i == 0)
{
count++;
while (n % i == 0)
n = n / i;
}
}
// This condition is to
// handle the case when n
// is a prime number greater
// than 2
if (n > 2)
count++;
return count;
}
// Driver Code
let n = 51242183;
document.write("The number of distinct prime factors is/are " +
exactPrimeFactorCount(n) + "<br>");
document.write("The value of log(log(n)) is "+ Math.log(Math.log(n)) + "<br>");
// This code is contributed by _saurabh_jaiswal
</script>
Output:
The number of distinct prime factors is/are 3 The value of log(log(n)) is 2.8765
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
