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
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 thecountDivisors
function.
- Record Keeping:
- The program uses an array
records
to keep track of potential numbers with specific divisor counts. This helps avoid redundant searches.
- The program uses an array
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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