How to resolve the algorithm Mertens function step by step in the Dyalect programming language

Published on 12 May 2024 09:40 PM
#Dyalect

How to resolve the algorithm Mertens function step by step in the Dyalect programming language

Table of Contents

Problem Statement

The Mertens function M(x) is the count of square-free integers up to x that have an even number of prime factors, minus the count of those that have an odd number. It is an extension of the Möbius function. Given the Möbius function μ(n), the Mertens function M(x) is the sum of the Möbius numbers from n == 1 through n == x.

This is not code golf.   The stackexchange link is provided as an algorithm reference, not as a guide.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Mertens function step by step in the Dyalect programming language

Source code in the dyalect programming language

func mertensNumbers(max) {
    let mertens = Array.Empty(max + 1, 1)
    for n in 2..max {
        for k in 2..n {
            mertens[n] -= mertens[n / k]
        }
    }
    mertens
}
 
let max = 1000
let mertens = mertensNumbers(max)
 
let count = 200
let columns = 20
print("First \(count - 1) Mertens numbers:")
 
for i in 0..
    if i % columns > 0 {
        print(" ", terminator: "")
    }
    print(i == 0 ? "   " : mertens[i].ToString().PadLeft(2, ' ') + " ", terminator: "")
    if (i + 1) % columns == 0 {
        print()
    }
}
 
var (zero, cross, previous) = (0, 0, 0)
for i in 1..max {
    let m = mertens[i]
    if m == 0 {
        zero += 1
        if previous != 0 {
            cross += 1
        }
    }
    previous = m
}
 
print("M(n) is zero \(zero) times for 1 <= n <= \(max).")
print("M(n) crosses zero \(cross) times for 1 <= n <= \(max).")

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