How to resolve the algorithm Curzon numbers step by step in the Wren programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Curzon numbers step by step in the Wren programming language
Table of Contents
Problem Statement
A Curzon number is defined to be a positive integer n for which 2n + 1 is evenly divisible by 2 × n + 1. Generalized Curzon numbers are those where the positive integer n, using a base integer k, satisfy the condition that kn + 1 is evenly divisible by k × n + 1. Base here does not imply the radix of the counting system; rather the integer the equation is based on. All calculations should be done in base 10. Generalized Curzon numbers only exist for even base integers.
and even though it is not specifically mentioned that they are Curzon numbers:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Curzon numbers step by step in the Wren programming language
Source code in the wren programming language
/* Curzon_numbers.wren */
import "./gmp" for Mpz
import "./fmt" for Fmt
for (k in [2, 4, 6, 8, 10]) {
System.print("The first 50 Curzon numbers using a base of %(k):")
var count = 0
var n = 1
var pow = Mpz.from(k)
var curzon50 = []
while (true) {
var z = pow + Mpz.one
var d = k*n + 1
if (z.isDivisibleUi(d)) {
if (count < 50) curzon50.add(n)
count = count + 1
if (count == 50) {
Fmt.tprint("$4d", curzon50, 10)
System.write("\nOne thousandth: ")
}
if (count == 1000) {
Fmt.print("$,d", n)
break
}
}
n = n + 1
pow.mul(k)
}
System.print()
}
import "./math" for Int
import "./fmt" for Fmt
var isCurzon = Fn.new { |n, k|
var r = k * n
return Int.modPow(k, n, r+1) == r
}
var k = 2
while (k <= 10) {
System.print("Curzon numbers with base %(k):")
var n = 1
var count = 0
while (count < 50) {
if (isCurzon.call(n, k)) {
Fmt.write("$4d ", n)
count = count + 1
if (count % 10 == 0) System.print()
}
n = n + 1
}
while (true) {
if (isCurzon.call(n, k)) count = count + 1
if (count == 1000) break
n = n + 1
}
Fmt.print("1,000th Curzon number with base $d: $,d\n", k, n)
k = k + 2
}
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