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I'm trying to find the common calculation between a series of numbers. For example:




I know that the calculation is simply the number squared. Another example would be:




In this example, I know that it is ((number * number) - 3) * number. So, with the number in the first set being 3, it would reduce down to ((3 * 3) - 3) * 3 = 18.

So, ultimately my question is trying to find the calculation between the following sets:




(for a bit of background as to why I'm asking this question, you can see my first failed attempt at asking the question here: Finding the calculation for the 24th permutation)

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Apparently, you seem to know that a polynomial is generating your values from inputs (e.g. the second one applies the polynomial $x^3-3x$ to the integers). Lagrange interpolation can determine such a polynomial. – J. M. May 4 '11 at 4:40
There are no simple answers for this kind of thing. Tell me what you want the "next" pair to be, I can find a rule that gives you all the pairs you have so far, and the one you want next. That rule will be given by a polynomial, so it won't even be ad hoc. See By the way, they aren't "sets", they aren't "sequences and series" either... – Arturo Magidin May 4 '11 at 4:47
Thanks for the clarifications. In regard to the sequences statement, I agree that I'm not stating this very well. However, couldn't the sequence that I'm looking at be: 108,405,990,? – Benjamin Powers May 4 '11 at 5:06
How do you know your second example is $(n^2-3)n$? It could just as easily be $12n^2-50n+60$, which has the advantage of involving only $n^2$, not $n^3$. – Gerry Myerson May 4 '11 at 6:16

1 Answer 1

The answer is ambiguous for example you could take $9 (200 + x (16 x -111))$ or $3 x (37 + x (5 x - 27))$ which can be found by solving the interpolation problem (those are the easy ones with low degree). There are also programs that can make some guesses for the generating sequence but they will not be happy if you feed them only 3 points.

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Thank you very much for the explanation. I am obviously coming from a different background, so I appreciate that you all even looked at my question. – Benjamin Powers May 5 '11 at 2:36

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