Getting the letter name of a column

=ADDRESS(1,n),

where n is the column number or, if this is what you want,

=MID(ADDRESS(1,n),2,IF(LEN(n)>2,2,1)),

where again n is the column number.

Aladin

I think the second formula needs a slight change to
=MID(ADDRESS(1,n),2,IF(LEN(address(1,n)>2,2,1)).

This is much better than what I was thinking. I was thinking of combining if column is greater than 26, using the Left function on the address.

Ian -- Try e.g., n=3,25,256 with both unmodified and modified versions. I think you probably overlooked the fact that LEN also accepts numbers, so there shouldn't a need for computing the address twice.
BTW, I reckon you also keep an eye on the sumproduct thing.

Aladin

Thanks to you, Ian, the second formula should be:

=MID(ADDRESS(1,n),2,IF(n>26,2,1)),

where n is or the cell of the desired column number or the column number itself.

Aladin

Yes, both do return the same with those number. When I modify my basterdize version or your slightly more to

=MID(ADDRESS(1,n),2,IF(LEN(ADDRESS(1,n,4))>2,2,1))

I get a result of AA with 27, and A with the original though?

I'll have to look at the sumproduct some more. I tend to learn through repetition. Just seems like a few months ago a asked you to explain how an array formula works!

sorry, just saw your post.

Or...

Oops! Make that...