Step by Step solution about How to resolve the algorithm Padovan sequence step by step in the Julia programming language

This Julia code defines and calculates Padovan sequences using different methods. Here's a detailed explanation:

1. Recursive Padovan Sequence (rPadovan):

   • The function rPadovan(n) is defined for calculating the Padovan sequence recursively.
   • For n < 4, it returns one(n), which presumably represents the first few terms of the sequence.
   • Otherwise, it calculates the nth term by summing the (n-3)th and (n-2)th terms.

2. Floor Function Calculation Padovan Sequence (fPadovan):

   • The function fPadovan(n::Int) calculates the nth term of the Padovan sequence using a floor function approximation.
   • It defines two constants p and s and uses them to calculate the result.

3. L-System Padovan Sequence (list_LsysPadovan):

   • The function list_LsysPadovan(N) generates strings representing the L-System representation of the Padovan sequence.
   • It uses rules to transform the previous string in the sequence to the next string.
   • It also keeps track of the lengths of these strings in the lens vector.

4. Constants and Data:

   • Two constants, lr and lf, are defined to store the first 64 terms of the recursive and floor function Padovan sequences, respectively.
   • Similarly, sL and lL store the first 32 strings and their lengths from the L-System representation.

5. Output:

   • The code prints the first 64 terms of the recursive, floor function, and L-System Padovan sequences, along with the corresponding strings from the L-System representation if the index is less than 11.

""" Recursive Padovan """
rPadovan(n) = (n < 4) ? one(n) : rPadovan(n - 3) + rPadovan(n - 2)

""" Floor function calculation Padovan """
function fPadovan(n)::Int
    p, s = big"1.324717957244746025960908854", big"1.0453567932525329623"
    return Int(floor(p^(n-2) / s + .5))
end

""" LSystem Padovan """
function list_LsysPadovan(N)
    rules = Dict("A" => "B", "B" => "C", "C" => "AB")
    seq, lens = ["A"], [1]
    for i in 1:N
        str = prod([rules[string(c)] for c in seq[end]])
        push!(seq, str)
        push!(lens, length(str))
    end
    return seq, lens
end

const lr, lf = [rPadovan(i) for i in 1:64], [fPadovan(i) for i in 1:64]
const sL, lL = list_LsysPadovan(32)
println("N  Recursive  Floor      LSystem      String\n=============================================")
foreach(i -> println(rpad(i, 4), rpad(lr[i], 12), rpad(lf[i], 12),
    rpad(i < 33 ? lL[i] : "", 12), (i < 11 ? sL[i] : "")), 1:64)