How to resolve the algorithm Chowla numbers step by step in the Visual Basic .NET programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Chowla numbers step by step in the Visual Basic .NET programming language
Table of Contents
Problem Statement
Chowla numbers are also known as:
The chowla number of n is (as defined by Chowla's function):
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995), a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
Chowla numbers can also be expressed as:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Chowla numbers step by step in the Visual Basic .NET programming language
Source code in the visual programming language
Imports System
Module Program
Function chowla(ByVal n As Integer) As Integer
chowla = 0 : Dim j As Integer, i As Integer = 2
While i * i <= n
j = n / i : If n Mod i = 0 Then chowla += i + (If(i = j, 0, j))
i += 1
End While
End Function
Function sieve(ByVal limit As Integer) As Boolean()
Dim c As Boolean() = New Boolean(limit - 1) {}, i As Integer = 3
While i * 3 < limit
If Not c(i) AndAlso (chowla(i) = 0) Then
Dim j As Integer = 3 * i
While j < limit : c(j) = True : j += 2 * i : End While
End If : i += 2
End While
Return c
End Function
Sub Main(args As String())
For i As Integer = 1 To 37
Console.WriteLine("chowla({0}) = {1}", i, chowla(i))
Next
Dim count As Integer = 1, limit As Integer = CInt((10000000.0)), power As Integer = 100,
c As Boolean() = sieve(limit)
For i As Integer = 3 To limit - 1 Step 2
If Not c(i) Then count += 1
If i = power - 1 Then
Console.WriteLine("Count of primes up to {0,10:n0} = {1:n0}", power, count)
power = power * 10
End If
Next
count = 0 : limit = 35000000
Dim p As Integer, k As Integer = 2, kk As Integer = 3
While True
p = k * kk : If p > limit Then Exit While
If chowla(p) = p - 1 Then
Console.WriteLine("{0,10:n0} is a number that is perfect", p)
count += 1
End If
k = kk + 1 : kk += k
End While
Console.WriteLine("There are {0} perfect numbers <= 35,000,000", count)
If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()
End Sub
End Module
Imports System.Numerics
Module Program
Function chowla(n As Integer) As Integer
chowla = 0 : Dim j As Integer, i As Integer = 2
While i * i <= n
If n Mod i = 0 Then j = n / i : chowla += i : If i <> j Then chowla += j
i += 1
End While
End Function
Function chowla1(ByRef n As BigInteger, x As Integer) As BigInteger
chowla1 = 1 : Dim j As BigInteger, lim As BigInteger = BigInteger.Pow(2, x - 1)
For i As BigInteger = 2 To lim
If n Mod i = 0 Then j = n / i : chowla1 += i : If i <> j Then chowla1 += j
Next
End Function
Function sieve(ByVal limit As Integer) As Boolean()
Dim c As Boolean() = New Boolean(limit - 1) {}, i As Integer = 3
While i * 3 < limit
If Not c(i) AndAlso (chowla(i) = 0) Then
Dim j As Integer = 3 * i
While j < limit : c(j) = True : j += 2 * i : End While
End If : i += 2
End While
Return c
End Function
Sub Main(args As String())
For i As Integer = 1 To 37
Console.WriteLine("chowla({0}) = {1}", i, chowla(i))
Next
Dim count As Integer = 1, limit As Integer = CInt((10000000.0)), power As Integer = 100,
c As Boolean() = sieve(limit)
For i As Integer = 3 To limit - 1 Step 2
If Not c(i) Then count += 1
If i = power - 1 Then
Console.WriteLine("Count of primes up to {0,10:n0} = {1:n0}", power, count)
power = power * 10
End If
Next
count = 0
Dim p As BigInteger, k As BigInteger = 2, kk As BigInteger = 3
For i As Integer = 2 To 31 ' if you dare, change the 31 to 61 or 89
If {2, 3, 5, 7, 13, 17, 19, 31, 61, 89}.Contains(i) Then
p = k * kk
If chowla1(p, i) = p Then
Console.WriteLine("{0,25:n0} is a number that is perfect", p)
st = DateTime.Now
count += 1
End If
End If
k = kk + 1 : kk += k
Next
Console.WriteLine("There are {0} perfect numbers <= {1:n0}", count, 25 * BigInteger.Pow(10, 18))
If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()
End Sub
End Module
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