The 97 rational numbers 49/1, 49/2, 49/3, ..., 49/97 are written on a blackboard.

Two of the above numbers X and Y are chosen and replaced by X*Y-X-Y+1.

The procedure is repeated until a single number Z(say) remains on the board.

Determine the possible values of Z.

(In reply to

re: cardinality of solution set by Charlie)

That sequence cannot be the answer to this problem. 1 is in the set and if either X or Y is 1 then Z will be 0. So the number for this problem is strictly less than the corresponding term in A001147.

Also I noted that to generalize this problem would ask for the 2k-1 rational numbers {k/1, k/2, k/3, ..., k/(2k-1)}. For k=1 this is just {1}, for k=2 this is {2, 1, 1/3}. For k=3 this is {3, 3/2, 1, 3/4, 3/5}.

If we could get specific counts for k=1,2,3,4 then we may be able to calculate the answer inductively, or the OEIS might have it.