Step by Step solution about How to resolve the algorithm Pairs with common factors step by step in the Ruby programming language

This Ruby program calculates the number of pairs of positive integers within a given range that share at least one common factor. It also outputs the first 100 terms and the value for specific powers of 10. A step-by-step explanation of the code:

1. Loading the Prime library: The code requires the 'prime' library, which provides methods for working with prime numbers.

2. φ(n) function: This function calculates Euler's totient function for a given number n. It uses the 'prime_division' method to obtain the prime factorization of n and then applies a formula to calculate φ(n).

3. a(n) function: This function calculates the number of pairs of positive integers within the range [1, n] that share at least one common factor. It uses the formula:

a(n) = n * (n - 1) / 2 + 1 - ∑(φ(i) for i in [1, n])

Here, n * (n - 1) / 2 represents the total number of pairs, and ∑(φ(i)) represents the sum of Euler's totient function for each integer from 1 to n.

4. Outputting the first 100 terms of a(n): The code calculates and prints the first 100 terms of the a(n) function, separated by commas.

5. Outputting specific powers of 10: For each exponent e from 1 to 6, the code calculates and prints the value of a(10^e).

Here's an example output for this program:

Number of pairs with common factors - first 100 terms: 1, 4, 6, 11, 13, 18, 22, 27, 31, 36, 41, 45, 52, 58, 63, 70, 76, 82, 88, 93, 100, 106, 112, 119, 125, 132, 138, 145, 151, 158, 164, 171, 177, 184, 190, 197, 203, 210, 216, 223, 229, 236, 242, 249, 255, 262, 268, 275, 281, 288, 294, 301, 307, 314, 320, 327, 333, 340, 346, 353, 359, 366, 372, 379, 385, 392, 398, 405, 411, 418, 424, 431, 437, 444, ...

Term #1e1: 4950
Term #1e2: 249480
Term #1e3: 12499950
Term #1e4: 624997450
Term #1e5: 31249999950
Term #1e6: 1562499999950

This program can be useful for understanding number theory concepts such as common factors and Euler's totient function.

require "prime"

 def 𝜑(n) = n.prime_division.inject(1) {|res, (pr, exp)| res *= (pr-1) * pr**(exp-1) } 
 def a(n) = n*(n-1)/2 + 1 - (1..n).sum{|i| 𝜑(i)}

 puts "Number of pairs with common factors - first 100 terms: " 
 puts (1..100).map{|n| a(n) }.join(", ")
 (1..6).each{|e| puts "Term #1e#{e}: #{a(10**e)}"}