Decimal to Alphanumeric

Can you give more example?

OK...
If I want to convert 10 in decimal to Alphanumeric it should be "A"
or if 20 in decimal = "L"
What if 1000?
What if 250?

OK~~~ if I can understand what you are saying... I should try ...

Let me know what is your expected result
if 1000 then ?

This is the problem I got...
I do not have the right answer to compare.
I just come out a short code like this (VB.Net)
 

Function Dec2AlphaNumeric(ByVal intDec As String) As String
Const charSeqStr = "0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,J,K,L,M,N,P,Q,R,S,T,V,W,X,Y"
Dim toFormat As String = "" 
Dim Pattern() As String = charSeqStr.Split(",")
Dim val As Double = Convert.ToInt32(intDec)
Dim i As Integer = 0
Do Until val <= 0
i = val Mod 32
toFormat = (Pattern(i) & toFormat)
val = System.Math.Floor(val / 32)
Loop
Return toFormat


End Function

My 1000 answer is Y8... is that right?.

Hope this got '1000'  to 'Y8'

        Dim strArray() As String
        strArray = New String() {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", "F", "G", "H", "J", "K", "L", "M", "N", "P", "Q", "R", "S", "T", "V", "W", "X", "Y"}
        Dim toFormat As String
        Dim val As Integer
        toFormat = ""
        val = Convert.ToInt32(TextBox1.Text)
        While (val > 0)
            Dim remain As Integer
            remain = val Mod 32
            toFormat = strArray(remain) + toFormat
            val = System.Math.Floor(val / 32)
        End While
        TextBox2.Text = toFormat

yes 1000 is Y8
to check if the result is correct or not convert base 32 number to decimal again using the following logic

Y8 = Y*(32^1) + 8 * (32 ^ 0)
^ is power function (example 32^1 = 32, 2^3 = 2*2*2 = 8)
Y=31
Y8 = 31*(32^1) + 8 * (32 ^ 0) = 31*32 + 8*1 = 992 + 8 = 1000

I'm curious. What's the motivation for this conversion?

See my previous answer here: https://www.experts-exchange.com/questions/22001532/using-text-as-counter.html?anchorAnswerId=17593224#a17593224

Hit the Button repeatedly and watch closely at the sequences it produces...

Does it "count" the way you want it to?

Public Class Form1

    Private Rev As New Revision("0123456789ABCDEFGHJKLMNPQRSTVWXY", "0")

    Private Sub Form1_Load(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles MyBase.Load
        Label1.Text = Rev.CurrentRevision
    End Sub

    Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Label1.Text = Rev.NextRevision
    End Sub

End Class

Public Class Revision

    Private chars As String
    Private values() As Char
    Private curRevision As System.Text.StringBuilder

    Public Sub New()
        Me.DefaultRevision()
    End Sub

    Public Sub New(ByVal validChars As String)
        If validChars.Length > 0 Then
            chars = validChars.ToUpper()
            values = chars.ToCharArray()
            curRevision = New System.Text.StringBuilder(values(0))
        Else
            Me.DefaultRevision()
        End If
    End Sub

    Public Sub New(ByVal validChars As String, ByVal startingRevision As String)
        Me.New(validChars)
        curRevision = New System.Text.StringBuilder(startingRevision.ToUpper())
        Dim i As Integer
        For i = 0 To curRevision.Length - 1
            If Array.IndexOf(values, curRevision.Chars(i)) = -1 Then
                curRevision = New System.Text.StringBuilder(values(0))
                MessageBox.Show("Revision has been reset." & vbCrLf & "Current Revision = " & Me.CurrentRevision, "Starting Revision contains an Invalid Character", MessageBoxButtons.OK, MessageBoxIcon.Exclamation)
                Exit For
            End If
        Next
    End Sub

    Private Sub DefaultRevision()
        chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        values = chars.ToCharArray
        curRevision = New System.Text.StringBuilder(values(0))
    End Sub

    Public ReadOnly Property ValidChars() As String
        Get
            Return chars
        End Get
    End Property

    Public ReadOnly Property CurrentRevision() As String
        Get
            Return curRevision.ToString()
        End Get
    End Property

    Public Function NextRevision(Optional ByVal numRevisions As Integer = 1) As String
        Dim forward As Boolean = (numRevisions > 0)
        numRevisions = Math.Abs(numRevisions)
        Dim i As Integer
        For i = 1 To numRevisions
            If forward Then
                Me.Increment()
            Else
                Me.Decrement()
            End If
        Next
        Return Me.CurrentRevision
    End Function

    Private Sub Increment()
        Dim curChar As Char = curRevision.Chars(curRevision.Length - 1)
        Dim index As Integer = Array.IndexOf(values, curChar)
        If index < (chars.Length - 1) Then
            index = index + 1
            curRevision.Chars(curRevision.Length - 1) = values(index)
        Else
            curRevision.Chars(curRevision.Length - 1) = values(0)
            Dim i As Integer
            Dim startPosition As Integer = curRevision.Length - 2
            For i = startPosition To 0 Step -1
                curChar = curRevision.Chars(i)
                index = Array.IndexOf(values, curChar)
                If index < (values.Length - 1) Then
                    index = index + 1
                    curRevision.Chars(i) = values(index)
                    Exit Sub
                Else
                    curRevision.Chars(i) = values(0)
                End If
            Next
            curRevision.Insert(0, values(0))
        End If
    End Sub

    Private Sub Decrement()
        Dim curChar As Char = curRevision.Chars(curRevision.Length - 1)
        Dim index As Integer = Array.IndexOf(values, curChar)
        If index > 0 Then
            index = index - 1
            curRevision.Chars(curRevision.Length - 1) = values(index)
        Else
            curRevision.Chars(curRevision.Length - 1) = values(values.Length - 1)
            Dim i As Integer
            Dim startPosition As Integer = curRevision.Length - 2
            For i = startPosition To 0 Step -1
                curChar = curRevision.Chars(i)
                index = Array.IndexOf(values, curChar)
                If index > 0 Then
                    index = index - 1
                    curRevision.Chars(i) = values(index)
                    Exit Sub
                Else
                    curRevision.Chars(i) = values(values.Length - 1)
                End If
            Next
            curRevision.Remove(0, 1)
            If curRevision.Length = 0 Then
                curRevision.Insert(0, values(0))
            End If
        End If
    End Sub

End Class

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Chaosian,
Someone is trying to be special than other.... they don't want to use Hex, Binary, or Octa.... but for really reason... I totally no clue. I never ask as well.

Alright!!! Thanks for your contribution... I manage to solve my problem ...
Appari> you have a really good explaination...
Infochandru> Thanks for your formula. That really simple than I thought.

Idle_Mind> I am going to look through your code...

Thanks again....

*chuckles*
I just thought it might be worth pointing out that, if the goal is to compress the data, this approach will actually take more space than storing numbers as numbers

really? I never know that... because if ascii?

Well, the issues amounts to this. Integers are either 32-bit (4 bytes) or 64-bit (8 bytes) depending on the hardware architecture.
Strings are 1 byte (ASCII) or 2 bytes (unicode) per character.
So, if you have a large number (say 32x32x32) your approach will require 4 bytes or 8 bytes, depending on the character set. On the other hand, this is only a 16-bit integer (2 bytes). Now, admittedly, you'll likely use the Integer type to store it, so it's going to take 4 or 8 bytes depending on the architecture. However, this is the break-even point. After this, the advantage is with the numeric type.