Thanks buzzb. I did think something like that might happen, but that requires every entry in the whole column to be doubled, whereas, assuming that Cell 5 is the 3rd letter of 4ac, I have in the next row an unambiguous entry, from 13ac, which looks correct.
Also, I'm concerned that if all 14 of the entries in 5d are doubles, then what happens with 38,41 ac? - you'd need another 11 doubles there, and only 15 clues altogether produce doubles.
I'm puzzled!!