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