To start consider the x^2 series starting with 0...

0, 1, 4

I noticed you can always find the next term by additions and subtractions of the initial 3 terms as follows:

4 + (4-1) + (4-1)-(1-0) = 9

This simplifies to the recurrence relationship:

A(n) = 3*(A(n-1) - A(n-2)) + A(n-3)

For example:

25 = 3*(16-9) + 4

And then I wondered what happens if you start it with other values than squares...

3, 8, 15

24 = 3*(15-8) - 3

I found the significance of this number is if you find the quadratic equation that fits these three points with any x values each a distance 1 apart, like here I randomly chose 4,5,6 as the x coordinates:

This function evaluated at the next integer 7 is 24.

The y values don't have to be positive or integers:

4.5, -3.2, 7

35.1 = 3*(7+3.2)+4.5

This function evaluated at 7 = 35.1

I also found that 3*(A(n+1)-A(n+2))+A(n+3) works equally well for finding A(n) going backwards knowing the 3 values to the right of A(n).

A similar formula can be found for cubic polynomials and it is:

Now to show this one for cubes works as well:

2, -5, 7.5, 16

-3 = 4*(16)-6*(7.5)+4*(-5)-(2)

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