Riddles, brain puzzles and mathematical problems

[BAndrew]

13 coins on the left, 13 on the right. If both balance, then remaining coin not weighed is the special one. If not balanced, take side that weighs less and divide it 6-6. If still no balance, the process is repeated with 3-3, and finally 1-1.

4 weighings necessary.

Great try, but there is a way to do it in 3 weightnings. Maybe I didn't make it clear, but you have to find the "best" possible answer or equivalently the minimum number of weightnings.