Step by Step solution about How to resolve the algorithm Sequence: smallest number greater than previous term with exactly n divisors step by step in the Ruby programming language

This Ruby code generates a sequence of numbers, where each number in the sequence has one more divisor than the previous number. The code uses the Enumerator class to create a lazy sequence, and then takes the first 15 numbers from the sequence.

Here's a breakdown of the code:

1. The require 'prime' line includes the prime module, which provides methods for working with prime numbers.

2. The num_divisors method takes a number n and returns the number of divisors of n. It uses the prime_division method to compute the prime factorization of n, and then applies the formula (n + 1) to each prime factor to count the number of divisors for that factor.

3. The seq variable is an Enumerator object that represents the sequence of numbers with increasing numbers of divisors. The Enumerator.new block defines the logic for generating the sequence.

4. The cur variable keeps track of the current number of divisors, starting from 0.

5. The (1..).each loop generates an infinite sequence of positive integers.

6. Inside the loop, the if statement checks if the current number i has one more divisor than cur. If it does, then i is added to the sequence using the y block argument, and cur is incremented by 1.

7. The seq.take(15) line takes the first 15 numbers from the sequence and prints them out.

require 'prime'
 
def num_divisors(n)
  n.prime_division.inject(1){|prod, (_p,n)| prod *= (n + 1) } 
end

seq = Enumerator.new do |y|
  cur = 0
  (1..).each do |i|
    if num_divisors(i) == cur + 1 then
      y << i
      cur += 1
    end
  end
end

p seq.take(15)