The first differences are 3-1, 7-3, 15-7, 29-15, 51-29
or 2, 4, 8, 14, 22

The second differences are 4-2, 8-4, 14-8, 22-14
or 2, 4, 6, 8

The third differences are 4-2, 6-4, 8-6
or 2, 2, 2

Since the third differences are all the same, the equation is a
third degree polynomial equation.  Suppose the third dregree 
polynomial equation is: 

y = Ax³+Bx²+Cx+D

Substituting the first four points:

Substituting x=1, y=1, 
1 = A(1)³+B(1)²+C(1)+D or  A+B+C+D = 1

Substituting x=2, y=3, 
3 = A(2)³+B(2)²+C(2)+D or  8A+4B+2C+D = 3

Substituting x=3, y=7, 
7 = A(3)³+B(3)²+C(3)+D or  27A+9B+3C+D = 7

Substituting x=4, y=15, 
15 = A(4)³+B(4)²+C(4)+D or  64A+16B+4C+D = 15

Solve the system of 4 equations in 4 unknowns:



and get (A,B,C,D) = (1/3,-1,8/3,-1)

So the equation is



[If you are taking calculus of finite differences using
Stirling numbers, then this can be done that way, but I
assume you are not in that course.] 

Edwin