Theoretical question on Fast Fourier I have a question regarding Discrete Fourier Transform and Discrete Time Fourier Transform.

Suppose that we have a sequence $x(n)= \{1,2,3,3,2,1\}$.

So is it possible to reconstruct the Discrete Time Fourier Transform $X(e^{j\omega})$ from the Discrete Fourier Transform $X(k)$ ? And how?

In my opinion , it is possible, although I am not sure if something like this would be any good. But in order to reconstruct it we probably need as much Discrete Fourier Transform samples as possible, right?

What do you people think on this matter?
This is (trivially) possible because all Fourier transform like operations are ideally invertible. So what you ask for is actually easily realized by:

$X(e^{j\Omega}) = DTFT\{ IDFT\{X(k)\} \}$

In formulae this is 

$X(e^{j\Omega}) = \sum_{l=0}^{N-1} \sum_{k=0}^{N-1} X(k)e^{2 \pi j kl/N} e^{-j\Omega l}$

which can be a bit simplified to

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \sum_{l=0}^{N-1} e^{jl (2 \pi k/N-\Omega)}$

or, finally

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \frac{e^{2 \pi j k- jN\Omega}-1}{e^{2 \pi j k/N-\Omega}-1}$.

EDIT: The question on whether you need all samples X(k) lies in the behavior of the interpolation function $I(x) = \frac{e^{j N x}-1}{e^{j x}-1}$. This is because the above result is then equal to

$X(e^{j\Omega}) = \sum_{k=0}^{N-1} X(k) \, I(2 \pi k/N - \Omega)$.

I(x) has some "sinc like" behavior, so the question on how many X(k) you need, is really when I(x) has decayed to a value "small enough".