How to resolve the algorithm Sequence: nth number with exactly n divisors step by step in the Go programming language

Published on 12 May 2024 09:40 PM
#Go

How to resolve the algorithm Sequence: nth number with exactly n divisors step by step in the Go programming language

Table of Contents

Problem Statement

Calculate the sequence where each term an is the nth that has n divisors. Show here, on this page, at least the first 15 terms of the sequence.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Sequence: nth number with exactly n divisors step by step in the Go programming language

Go Program Overview

This Go program generates and prints the first 33 terms of a sequence, where the terms are either calculated using modular exponentiation for prime numbers or found by searching for numbers with a specific number of divisors.

Key Concepts and Algorithm

  • Prime Numbers: The program uses Go's built-in isPrime function to identify prime numbers.
  • Modular Exponentiation: For prime numbers, the program computes p^i (p raised to the power of i) using modular exponentiation with BigInts.
  • Divisors: The countDivisors function determines the total number of divisors of a given integer.
  • Search for Numbers with Specific Divisor Count: For non-prime numbers, the program searches for integers that have a number of divisors equal to the index in the sequence (e.g., 3 divisors for the third term).

Code Structure

  • Utility Functions:
    • isPrime: Checks if a number is prime.
    • countDivisors: Calculates the number of divisors of an integer.
  • Main Sequence Generation:
    • The program initializes some constants and variables.
    • It iterates from i = 1 to 45 (the max number of terms).
    • For each i, it checks if it is prime. If so, it calculates p^i using modular exponentiation. If not, it searches for a number with i divisors using the countDivisors function.
  • Record Keeping:
    • The program uses an array records to keep track of potential numbers with specific divisor counts. This helps avoid redundant searches.

Output

The program prints the first 33 terms of the sequence, alternating between modular exponentiation results for prime numbers and integers with i divisors for non-prime numbers.

Source code in the go programming language

package main

import (
    "fmt"
    "math"
    "math/big"
)

var bi = new(big.Int)

func isPrime(n int) bool {
    bi.SetUint64(uint64(n))
    return bi.ProbablyPrime(0)
}

func generateSmallPrimes(n int) []int {
    primes := make([]int, n)
    primes[0] = 2
    for i, count := 3, 1; count < n; i += 2 {
        if isPrime(i) {
            primes[count] = i
            count++
        }
    }
    return primes
}

func countDivisors(n int) int {
    count := 1
    for n%2 == 0 {
        n >>= 1
        count++
    }
    for d := 3; d*d <= n; d += 2 {
        q, r := n/d, n%d
        if r == 0 {
            dc := 0
            for r == 0 {
                dc += count
                n = q
                q, r = n/d, n%d
            }
            count += dc
        }
    }
    if n != 1 {
        count *= 2
    }
    return count
}

func main() {
    const max = 33
    primes := generateSmallPrimes(max)
    z := new(big.Int)
    p := new(big.Int)
    fmt.Println("The first", max, "terms in the sequence are:")
    for i := 1; i <= max; i++ {
        if isPrime(i) {
            z.SetUint64(uint64(primes[i-1]))
            p.SetUint64(uint64(i - 1))
            z.Exp(z, p, nil)
            fmt.Printf("%2d : %d\n", i, z)
        } else {
            count := 0
            for j := 1; ; j++ {
                if i%2 == 1 {
                    sq := int(math.Sqrt(float64(j)))
                    if sq*sq != j {
                        continue
                    }
                }
                if countDivisors(j) == i {
                    count++
                    if count == i {
                        fmt.Printf("%2d : %d\n", i, j)
                        break
                    }
                }
            }
        }
    }
}


package main

import (
    "fmt"
    "math"
    "math/big"
)

type record struct{ num, count int }

var (
    bi     = new(big.Int)
    primes = []int{2}
)

func isPrime(n int) bool {
    bi.SetUint64(uint64(n))
    return bi.ProbablyPrime(0)
}

func sieve(limit int) {
    c := make([]bool, limit+1) // composite = true
    // no need to process even numbers
    p := 3
    for {
        p2 := p * p
        if p2 > limit {
            break
        }
        for i := p2; i <= limit; i += 2 * p {
            c[i] = true
        }
        for {
            p += 2
            if !c[p] {
                break
            }
        }
    }
    for i := 3; i <= limit; i += 2 {
        if !c[i] {
            primes = append(primes, i)
        }
    }
}

func countDivisors(n int) int {
    count := 1
    for i, p := 0, primes[0]; p*p <= n; i, p = i+1, primes[i+1] {
        if n%p != 0 {
            continue
        }
        n /= p
        count2 := 1
        for n%p == 0 {
            n /= p
            count2++
        }
        count *= (count2 + 1)
        if n == 1 {
            return count
        }
    }
    if n != 1 {
        count *= 2
    }
    return count
}

func isOdd(x int) bool {
    return x%2 == 1
}

func main() {
    sieve(22000)
    const max = 45
    records := [max + 1]record{}
    z := new(big.Int)
    p := new(big.Int)
    fmt.Println("The first", max, "terms in the sequence are:")
    for i := 1; i <= max; i++ {
        if isPrime(i) {
            z.SetUint64(uint64(primes[i-1]))
            p.SetUint64(uint64(i - 1))
            z.Exp(z, p, nil)
            fmt.Printf("%2d : %d\n", i, z)
        } else {
            count := records[i].count
            if count == i {
                fmt.Printf("%2d : %d\n", i, records[i].num)
                continue
            }
            odd := isOdd(i)
            k := records[i].num
            l := 1
            if !odd && i != 2 && i != 10 {
                l = 2
            }
            for j := k + l; ; j += l {
                if odd {
                    sq := int(math.Sqrt(float64(j)))
                    if sq*sq != j {
                        continue
                    }
                }
                cd := countDivisors(j)
                if cd == i {
                    count++
                    if count == i {
                        fmt.Printf("%2d : %d\n", i, j)
                        break
                    }
                } else if cd > i && cd <= max && records[cd].count < cd &&
                    j > records[cd].num && (l == 1 || (l == 2 && !isOdd(cd))) {
                    records[cd].num = j
                    records[cd].count++
                }
            }
        }
    }
}


  

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