Task: Chowla numbers
Java: public class Chowla {

    public static void main(String[] args) {
        int[] chowlaNumbers = findChowlaNumbers(37);
        for (int i = 0; i < chowlaNumbers.length; i++) {
            System.out.printf("chowla(%d) = %d%n", (i+1), chowlaNumbers[i]);
        }
        System.out.println();

        int[][] primes = countPrimes(100, 10_000_000);
        for (int i = 0; i < primes.length; i++) {
            System.out.printf(Locale.US, "There is %,d primes up to %,d%n", primes[i][1], primes[i][0]);
        }
        System.out.println();

        int[] perfectNumbers = findPerfectNumbers(35_000_000);
        for (int i = 0; i < perfectNumbers.length; i++) {
            System.out.printf("%d is a perfect number%n", perfectNumbers[i]);
        }
        System.out.printf(Locale.US, "There are %d perfect numbers < %,d%n", perfectNumbers.length, 35_000_000);
    }

    public static int chowla(int n) {
        if (n < 0) throw new IllegalArgumentException("n is not positive");
        int sum = 0;
        for (int i = 2, j; i * i <= n; i++)
            if (n % i == 0) sum += i + (i == (j = n / i) ? 0 : j);
        return sum;
    }

    protected static int[][] countPrimes(int power, int limit) {
        int count = 0;
        int[][] num = new int[countMultiplicity(limit, power)][2];
        for (int n = 2, i=0;  n <= limit; n++) {
            if (chowla(n) == 0) count++;
            if (n % power == 0) {
                num[i][0] = power;
                num[i][1] = count;
                i++;
                power *= 10;
            }
        }
        return num;
    }

    protected static int countMultiplicity(int limit, int start) {
        int count = 0;
        int cur = limit;
        while(cur >= start) {
            count++;
            cur = cur/10;
        }
        return count;
    }

    protected static int[] findChowlaNumbers(int limit) {
        int[] num = new int[limit];
        for (int i = 0; i < limit; i++) {
            num[i] = chowla(i+1);
        }
        return num;
    }

    protected static int[] findPerfectNumbers(int limit) {
        int count = 0;
        int[] num = new int[count];

        int k = 2, kk = 3, p;
        while ((p = k * kk) < limit) {
            if (chowla(p) == p - 1) {
                num = increaseArr(num);
                num[count++] = p;
            }
            k = kk + 1;
            kk += k;
        }
        return num;
    }

    private static int[] increaseArr(int[] arr) {
        int[] tmp = new int[arr.length + 1];
        System.arraycopy(arr, 0, tmp, 0, arr.length);
        return tmp;
    }
}

VB: Option Explicit

Private Declare Function AllocConsole Lib "kernel32.dll" () As Long
Private Declare Function FreeConsole Lib "kernel32.dll" () As Long
Dim mStdOut As Scripting.TextStream

Function chowla(ByVal n As Long) As Long
Dim j As Long, i As Long
  i = 2
  Do While i * i <= n
    j = n \ i
    If n Mod i = 0 Then
    chowla = chowla + i
      If i <> j Then
      chowla = chowla + j
      End If
    End If
    i = i + 1
  Loop
End Function
 
Function sieve(ByVal limit As Long) As Boolean()
Dim c() As Boolean
Dim i As Long
Dim j As Long
  i = 3
  ReDim c(limit - 1)
    Do While i * 3 < limit
      If Not c(i) Then
        If (chowla(i) = 0) Then
          j = 3 * i
          Do While j < limit
            c(j) = True
            j = j + 2 * i
          Loop
        End If
    End If
    i = i + 2
    Loop
  sieve = c()
End Function

Sub Display(ByVal s As String)
  Debug.Print s
  mStdOut.Write s & vbNewLine
End Sub
 
Sub Main()
Dim i As Long
Dim count As Long
Dim limit As Long
Dim power As Long
Dim c() As Boolean
Dim p As Long
Dim k As Long
Dim kk As Long
Dim s As String * 30
Dim mFSO As Scripting.FileSystemObject
Dim mStdIn As Scripting.TextStream
    
  AllocConsole
  Set mFSO = New Scripting.FileSystemObject
  Set mStdIn = mFSO.GetStandardStream(StdIn)
  Set mStdOut = mFSO.GetStandardStream(StdOut)
  
  For i = 1 To 37
    Display "chowla(" & i & ")=" & chowla(i)
  Next i
  
  count = 1
  limit = 10000000
  power = 100
  c = sieve(limit)
  
  For i = 3 To limit - 1 Step 2
    If Not c(i) Then
      count = count + 1
    End If
    If i = power - 1 Then
      RSet s = FormatNumber(power, 0, vbUseDefault, vbUseDefault, True)
      Display "Count of primes up to " & s & " = " & FormatNumber(count, 0, vbUseDefault, vbUseDefault, True)
      power = power * 10
    End If
  Next i

  count = 0: limit = 35000000
  k = 2:     kk = 3

  Do
    p = k * kk
    If p > limit Then
      Exit Do
    End If

    If chowla(p) = p - 1 Then
      RSet s = FormatNumber(p, 0, vbUseDefault, vbUseDefault, True)
      Display s & " is a number that is perfect"
      count = count + 1
    End If
    k = kk + 1
    kk = kk + k
  Loop
        
  Display "There are " & CStr(count) & " perfect numbers <= 35.000.000"

  mStdOut.Write "press enter to quit program."
  mStdIn.Read 1

  FreeConsole

End Sub