Convert column index into corresponding column letter

I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):

function columnToLetter(column){  var temp, letter = '';  while (column > 0)  {    temp = (column - 1) % 26;    letter = String.fromCharCode(temp + 65) + letter;    column = (column - temp - 1) / 26;  }  return letter;}function letterToColumn(letter){  var column = 0, length = letter.length;  for (var i = 0; i < length; i++)  {    column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);  }  return column;}

This works good

=REGEXEXTRACT(ADDRESS(ROW(); COLUMN()); "[A-Z]+")

even for columns beyond Z.

Simply replace COLUMN() with your column number. The value of ROW() doesn't matter.

=SUBSTITUTE(ADDRESS(1,COLUMN(),4), "1", "")

This takes your cell, gets it's address as e.g. C1, and removes the "1".

How it works

• COLUMN() gives the number of the column of the cell.
• ADDRESS(1, ..., <format>) gives an address of a cell, in format speficied by <format> parameter. 4 means the address you know - e.g. C1.
  • The row doesn't matter here, so we use 1.
  • See ADDRESS docs
• Finally, SUBSTITUTE(..., "1", "") replaces the 1 in the address C1, so you're left with the column letter.