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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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    $\begingroup$ Without definition, your question is meaningless. The next term, $f(6)$, is obviously $42$. $\endgroup$
    – FindStat
    Commented Dec 10, 2017 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
    Commented Dec 10, 2017 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
    Commented Dec 10, 2017 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
    Commented Dec 10, 2017 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. $\endgroup$
    – user6976
    Commented Dec 10, 2017 at 19:47

1 Answer 1

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One way is

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

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    $\begingroup$ This is consistent with fedja's guess. Your guess may be as good as anyone's. $\endgroup$
    – Todd Trimble
    Commented Dec 11, 2017 at 3:04

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