#### Amicable Number

A
b
```package com.thealgorithms.maths;

/**
* Amicable numbers are two different numbers so related that the sum of the
* proper divisors of each is equal to the other number. (A proper divisor of a
* number is a positive factor of that number other than the number itself. For
* example, the proper divisors of 6 are 1, 2, and 3.) A pair of amicable
* numbers constitutes an aliquot sequence of period 2. It is unknown if there
* are infinitely many pairs of amicable numbers. *
*
* <p>
*
* <p>
* Simple Example : (220,284) 220 is divisible by {1,2,4,5,10,11,20,22,44,55,110
* } <- Sum = 284
* 284 is divisible by -> 1,2,4,71,142 and the Sum of that is. Yes right you
* probably expected it 220
*/
public class AmicableNumber {

public static void main(String[] args) {

AmicableNumber.findAllInRange(1, 3000);
/* Res -> Int Range of 1 till 3000there are 3Amicable_numbers These are  1: = ( 220,284)	2: = ( 1184,1210)
3: = ( 2620,2924) So it worked	*/

}

/**
* @param startValue
* @param stopValue
* @return
*/
static void findAllInRange(int startValue, int stopValue) {

/* the 2 for loops are to avoid to double check tuple. For example (200,100) and (100,200) is the same calculation
* also to avoid is to check the number with it self. a number with itself is always a AmicableNumber
* */
StringBuilder res = new StringBuilder();
int countofRes = 0;

for (int i = startValue; i < stopValue; i++) {
for (int j = i + 1; j <= stopValue; j++) {
if (isAmicableNumber(i, j)) {
countofRes++;
res.append("" + countofRes + ": = ( " + i + "," + j + ")" + "\t");
}
}
}
res.insert(
0,
"Int Range of "
+ startValue
+ " till "
+ stopValue
+ " there are "
+ countofRes
+ " Amicable_numbers.These are \n ");
System.out.println(res.toString());
}

/**
* Check if {@code numberOne and numberTwo } are AmicableNumbers or not
*
* @param numberOne numberTwo
* @return {@code true} if {@code numberOne numberTwo} isAmicableNumbers
* otherwise false
*/
static boolean isAmicableNumber(int numberOne, int numberTwo) {

return ((recursiveCalcOfDividerSum(numberOne, numberOne) == numberTwo
&& numberOne == recursiveCalcOfDividerSum(numberTwo, numberTwo)));
}

/**
* calculated in recursive calls the Sum of all the Dividers beside it self
*
* @param number div = the next to test dividely by using the modulo
* operator
* @return sum of all the dividers
*/
static int recursiveCalcOfDividerSum(int number, int div) {

if (div == 1) {
return 0;
} else if (number % --div == 0) {
return recursiveCalcOfDividerSum(number, div) + div;
} else {
return recursiveCalcOfDividerSum(number, div);
}
}
}
```  