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In my research, I obtained a sequence of polynomials (I am only able to compute the first 4 of them): \begin{align} & f(2) = 1+t, \\ & f(3) = 1+4t+3t^2, \\ & f(4) = 1+6t+12t^2+7t^3, \\ & f(5) = 1+8t+20t^2+28t^3+15t^4. \end{align} Is it possible to find the general formula of $f(n)$ using these 4 polynomials? Some patterns are:

  1. The constant term is 1.
  2. The highest degree term is $(2^{n-1}-1)t^{n-1}$.
  3. $(1+t)$ is a factor of each polynomial.

Thank you very much.

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closed as unclear what you're asking by Max Alekseyev, darij grinberg, Denis Serre, Alexey Ustinov, Chris Godsil Dec 18 '17 at 11:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Without definition, your question is meaningless. The next term, $f(6)$, is obviously $42$. $\endgroup$ – FindStat Dec 10 '17 at 12:14
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    $\begingroup$ I don't know - I was joking. The serious point is that asking for the next term without definition of the sequence is meaningless. $\endgroup$ – FindStat Dec 10 '17 at 12:21
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    $\begingroup$ You need to explain how you got the first four polynomials in order for anyone to comment on finding the next one. $\endgroup$ – Stanley Yao Xiao Dec 10 '17 at 12:37
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    $\begingroup$ I don't think that the question is meaningless. I often try to understand data whose origin is irrelevant. $\endgroup$ – Richard Stanley Dec 10 '17 at 14:47
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    $\begingroup$ @RichardStanley: I would guess that the OP is trying to compute homology of some complexes (with Euler characteristic 0). If he said what exactly it is, the suggestions would not be that wild. $\endgroup$ – Mark Sapir Dec 10 '17 at 19:47
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Your data can be fitted with

$f(n)=(1+t) \left( \sum_{k=0}^{n-2} (2(2^k-1)(n-k-1)+1)t^k \right)$

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    $\begingroup$ This is consistent with fedja's guess. Your guess may be as good as anyone's. $\endgroup$ – Todd Trimble Dec 11 '17 at 3:04

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