Argthtjtr

We can use the excel spreadsheet itself to help increment the letters - the increment will work from A to XFC

First create the cell reference: INDIRECT(A1&"1")

Then find the address of the next column over: ADDRESS(1,COLUMN(INDIRECT(A10&"1"))+1)

Then from the $??$1 we extract the letters: 2 ways:

1. 
   

   Look for the second $, and snip the text out between them

   
   
   

   =MID(ADDRESS(1,COLUMN(INDIRECT(A1&"1"))+1),2,FIND("$",ADDRESS(1,COLUMN(INDIRECT(A1&"1"))+1),2)-2)
   
   

   
   


2. 
   

   Replace the 1 and $ with nothing in the string

   
   
   

   =SUBSTITUTE(SUBSTITUTE(ADDRESS(1,COLUMN(INDIRECT(A1&"1"))+1),"$",""),"1","")
   
   

   
   

Choose which one works best for you

Another example: Type this into cell A1, then copy the formula to any or all cells.

=CHAR(MOD(ROW(A1)-1;26)+65)&CHAR(MOD(COLUMN(A1)-1;26)+65)

Intended as an example of how one may think about the problem.

This will reset to "A" when it reaches "Z"

=IF(A1="Z", "A", CHAR(CODE(A1)+1))

You can build out beyond that with more if statements.

OR

I just wrote this formula for something similar:

IF(RIGHT(C2,1)="Z",IF(RIGHT(C2,2)="ZZ","A",CHAR(CODE(MID(C2,5,1))+1)),MID(C2,5,1))&IF(RIGHT(C2,1)="Z","A",CHAR(CODE(MID(C2,6,1))+1))

As another solution for it with using ADDRESS() is:

=MID(ADDRESS(1,26+ROW()),2,2)

Above formula will return AA in first row and AB in second row and so on.

Also with using math the formula is:

=CONCATENATE(CHAR(INT(ROW()/26)+65),CHAR(MOD(ROW()-1,26)+65))

I used this code to obtain Cell address

=ADDRESS(ROW($AT$17),COLUMN($AT$17)+Increment,4,1)

This example is for AT17 cell address.

Increment you define how many numbers you will increment

I wrote this code in cell Ax17.

Later I obtained value of AT17 with

=INDIRECT(AX17)

Done !!!

Now you can increment columns instead of rows !!!

Here is my solution (cell A1 contains "AA"):

=IF(CHAR(CODE(RIGHT(A2;1))+1)="[";CHAR(CODE(LEFT(A2;1))+1);CHAR(CODE(LEFT(A2;1))))&IF(CHAR(CODE(RIGHT(A2;1))+1)="[";CHAR(65);CHAR(CODE(RIGHT(A2;1))+1))

Fill Column A (from row 1) with consecutive numbers starting with 0 to 100 [or till requirement]

Fill Cell B1 with below formula

=CONCATENATE(CHAR(MOD(QUOTIENT(A1,26*26),26)+65),CHAR(MOD(QUOTIENT(A1,26),26)+65),CHAR(MOD(A1,26)+65))

Copy down the formula from B1 to other rows in Column B [till the row you have filled Column A]

This works for 3 characters AAA to ZZZ

Formula needs to be modified as per no. of characters required (AA to ZZ / AAAAA to ZZZZZ / etc)

I know this is slightly off the main question, but I think it answers the fuller question...
If you have a letter in A1, and you wish it to be stepped by a number in B1, the following formula combo will achieve it from single letters to ZZ.

 =IF(LEN(A1)>1,IF((CODE(RIGHT(A1,1))+$B$1)>CODE("Z"),CHAR(CODE(LEFT(A1,1))+1)&CHAR(CODE(RIGHT(A1,1))-21),LEFT(A1,1)&CHAR(CODE(RIGHT(A1,1))+$B$1)),IF((CODE(A1)+$B$1)>CODE("Z"),"A"&CHAR(CODE(A1)-21),CHAR(CODE(A1)+$B$1)))

Copy it down the column and the results are there. Change the B5 number and the results change.

=IF(AND(LEN(G1)=1,G1="Z"), "AA", IF(LEN(G1)=1, CHAR(CODE(G1)+1), IF(RIGHT(G1,1)<>"Z", LEFT(G1,1)&CHAR(CODE(RIGHT(G1,1))+1), CHAR(CODE(G1)+1)&"A")))

For Columns, the below is the right solution.

=IF(LEN(ADDRESS(1,COLUMN()))=4,MID(ADDRESS(1,COLUMN()),2,1),MID(ADDRESS(1,COLUMN()),2,2))