Need a formula
There are many types of sequences that could be forced to create the numbers you specify.
The simplest type of sequence that could conceivably obtain your numbers is a finite difference sequence.
A finite difference series is one where successive differences between the terms eventually result in a constant. For example, consider the 1 through n terms N1, N2, N3, N4,........Nn:
n...................1......2......3......4......5......6......7......8......9......10
N..................2......6.....13.....23....36....52....71....93....118....136
1st Diff..............4......7......10...13.....16....19....22....25......28
2nd Diff.................3......3......3......3......3......3......3......3
An expression can be derived enabling the definition the nth term of any finite difference series. The expression is a function of the number of successive differences required to reach the constant difference. If the first differences are constant, the expression is of the first order, i.e., N = an + b. If the second differences are constant, the expression is of the second order, i.e., N = an^2 + bn + c. Similarly, constant third differences derive from N = an^3 + bn^2 + cn + d.
Take the following example:
n.................1......2......3......4......5......6................n
N................3......9.....19.....35....59....93...............Nn
1st Diff............6.....10.....16....24....34
2nd Diff...............4.......6......8.....10
3rd Diff....................2......2......2
Using the data points (n1, N1), (n2,N2), (n3,N3), etc., we substitute them into N = an^3 + bn^2 + cn + d as follows:
(n1,N1) = (1,3) produces a(1^3) + b(1^2) + c(1) + d = 3 or a + b + c + d = 3
(n2,N2) = (2,9) produces a(2^3) + b(2^2) + c(2) + d = 9 or 8a + 4b + 2c + d = 9
(n3,N3) = (3,19) produces a(3^3) + b(3^2) + c(3) + d = 19 or 27a + 9b + 3c + d = 19
(n4,N4) = (4,35) produces a(4^3) + b(4^2) + c(4) + d = 35 or 64a + 16b + 4c + d = 35
Subtracting each successive pair yields
7a + 3b + c = 6
19a + 5b + c = 10
37a + 7b + c = 16
Again, subtracting each successive pair yields
12a + 2b = 4
18a + 2b = 6
Subtracting these yields 6a = 2 making a = 1/3, b = 0, c = 11/3, and d = -1 resulting in our final expression for the nth term of this series Nn = (n^3)/3 + (11n)/3 - 1 = (n^3 + 11n - 3)/3.
Checking it out for the 6th term we have [(6^3) + (66) - 3]/3 = [216 + 66 - 3]/3 = 279/3 = 93.
Considering your numbers, lets create the sequence
n......1......2......3
N.....18....72....432
Diff......54...350
Diff.........306
Now this limited information can be assumed to be the beginning of a finite difference sequence with second differences equal to 306. The next term in the sequence would derive from
n......1......2......3......4
N.....18....72...432..1098
Diff......54....350..666
Diff.........306...306
With the second differences assumed to be constant, the general expression for any term would be of the form N = an^2 + bn + c.
Substituting the data already derived, we get
a(1^2) + b(1) + c = 18 or a + b + c = 18
a(2^2) + b(2) + c = 72 or 4a + 2b + c = 72
a(3^2) + b(3) + c = 432 or 9a + 3b + c = 432
Solving, a = 153, b = -405 and c = 270.
Therefore, in terms of n = 1, 2, 3, ...n, N = 153n^2 - 405n + 270.
But, you want the derived terms to be based on 2, 3, 4,...n, or one more than the n used to derive this expression.
We therefore substitute (n + 1) for n and get N = 153(n + 1)^2 - 405(n + 1) + 270 or N = 153n^2 - 711n + 828.
Substitute n = 2, 3 and 4 and you will get 18, 72 and 432. Higher values of n will produce consistant values of N for the sequence with second differences of 306.
