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

The provided Julia code deals with abundant numbers, which are positive integers for which the sum of their proper divisors (the factors excluding the number itself) exceeds the number itself. The code defines functions for finding proper divisors, checking if a number is abundant, finding abundant odd numbers, and printing the results.

Function propfact(n):

• This function calculates the proper divisors of an integer n and returns them in a sorted vector.
• It uses the factor function to obtain the prime factorization of n, then constructs a vector of proper divisors by raising each prime factor to powers from 1 to its exponent.
• It finally sorts the vector of proper divisors in ascending order.

isabundant(n):

• This function checks if a given integer n is abundant by summing its proper divisors (using the propfact function) and comparing it to n.
• It returns true if the sum of proper divisors is greater than n, and false otherwise.

prettyprintfactors(n):

• This function takes an integer n and prints information about its proper divisors.
• It calls the propfact function to get the proper divisors, calculates their sum, and then prints a formatted string containing this information.

oddabundantsfrom(startingint, needed, nprint):

• This function finds and prints abundant odd numbers starting from a given startingint, and continues until it finds needed such numbers.
• It increments the integer n by 2 in each iteration to consider only odd numbers.
• It checks if n is abundant using isabundant and increments a counter count if it is.
• If nprint is greater than 0, it prints the factors of n at the specified count (i.e., the nprint-th abundant odd number found).
• The function continues until count reaches needed.

Usage in the Code:

• The code first prints the first 25 abundant odd numbers starting from 2.
• Then, it prints the 1000th abundant odd number.
• Finally, it prints the first abundant odd number greater than one billion.

using Primes

function propfact(n)
    f = [one(n)]
    for (p, x) in factor(n)
        f = reduce(vcat, [f*p^i for i in 1:x], init=f)
    end
    pop!(f)
    sort(f)
end

isabundant(n) = sum(propfact(n)) > n
prettyprintfactors(n) = (a = propfact(n); println("$n has proper divisors $a, these sum to $(sum(a))."))

function oddabundantsfrom(startingint, needed, nprint=0)
    n = isodd(startingint) ? startingint : startingint + 1
    count = one(n)
    while count <= needed
        if isabundant(n)
            if nprint == 0
                prettyprintfactors(n)
            elseif nprint == count
                prettyprintfactors(n)
            end
            count += 1
        end
        n += 2
    end
end

println("First 25 abundant odd numbers:")
oddabundantsfrom(2, 25)

println("The thousandth abundant odd number:")
oddabundantsfrom(2, 1001, 1000)

println("The first abundant odd number greater than one billion:")
oddabundantsfrom(1000000000, 1)