How to resolve the algorithm Abundant odd numbers step by step in the Processing programming language

Published on 12 May 2024 09:40 PM
#Processing

How to resolve the algorithm Abundant odd numbers step by step in the Processing programming language

Table of Contents

Problem Statement

An Abundant number is a number n for which the   sum of divisors   σ(n) > 2n, or,   equivalently,   the   sum of proper divisors   (or aliquot sum)       s(n) > n.

12   is abundant, it has the proper divisors     1,2,3,4 & 6     which sum to   16   ( > 12 or n);        or alternately,   has the sigma sum of   1,2,3,4,6 & 12   which sum to   28   ( > 24 or 2n).

Abundant numbers are common, though even abundant numbers seem to be much more common than odd abundant numbers. To make things more interesting, this task is specifically about finding   odd abundant numbers.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Abundant odd numbers step by step in the Processing programming language

Source code in the processing programming language

void setup() {
  println("First 25 abundant odd numbers: ");
  int abundant = 0;
  int i = 1;
  while (abundant < 25) {
    int sigma_sum = sigma(i);
    if (sigma_sum > 2 * i) {
      abundant++;
      println(i + "  Sigma sum: " + sigma_sum);
    }
    i += 2;
  }
  println("Thousandth abundant odd number: ");
  while (abundant < 1000) {
    int sigma_sum = sigma(i);
    if (sigma_sum > 2 * i) {
      abundant++;
      if (abundant == 1000) {
        println(i + "  Sigma sum: " + sigma_sum);
      }
    }
    i += 2;
  }
  println("First abundant odd number greater than 10^9: ");
  i = int(pow(10, 9)) + 1;
  while (!(sigma(i) > 2 * i)) {
    i += 2;
  }
  println(i + "  Sigma sum: " + sigma(i));
}

int sigma(int n) {
  int sum = 0;
  for (int i = 1; i < sqrt(n); i++) {
    if (n % i == 0) {
      sum += i + n / i;
    }
  }
  if (sqrt(n) % 1 == 0) {
    sum += sqrt(n);
  }
  return sum;
}

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