101, 103, X, 115, 131
Form the first differences and simplify:
103-101, X-103, 115-X, 131-115
2, X-103, 115-X, 16
Since the first and last are first and fourth powers of 2, the most likely
sequence is
2, 4, 8, 16
Let's see if that works for the first differences. If so, then this would
need to be a consistent though dependent system:
They both simplify to the same equation
X = 107
That's the most likely solution for X
Checking:
101, 103, 107, 115, 131
Edwin