package 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. * * *

* link: https://en.wikipedia.org/wiki/Amicable_numbers * * *

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); } } }