Surprising behaviour of polynomial that generates the series 1,2,4,8,...2^(k-1) Consider the generating function $f(n)$ that produces the following values:
$$f(1) = 1$$
$$f(2) = 2$$
$$f(3) = 4$$
Obviously these values can be generated by $f(n)= 2^{n-1}$.
These values can equally well be generated by $f(n) = (n^2-n+2)/2$, a second order polynomial.
Many (all?) integer series $f(k)$, where $k = 1,2,3,...,K-1,K$ can be generated by a polynomial of order $K-1$.
The integer series $2^{n-1}$, where $n = 1,2,3,...,K$ can also be generated by a polynomial of order $K-1$.
The following interesting thing happens.
If we describe the series $1,2,4$ by $f(n) = \frac{n^2-n+2}{2}$ then $f(4) = 7$
For $1,2,4,8$, $f(n) = \frac{n^3-3n^2+8n}{6}$ and $f(5) = 15$
For $1,2,4,8,16$, $f(n) = \frac{n^4-6n^3+23n^2-18n}{24}$ and $f(6)=31$
For $1,2,4,8,16,32$, $f(n) = \frac{n^5-10n^4+55n^3-110n^2+184n}{120}$ and $f(7)=63$
For $1,2,4,8,16,32,64$, $f(n) = \frac{n^6-15n^5+115n^4-405n^3+964n^2-660n+720}{720}$ and $f(8)=127$ 
I have verified this till order 14.
Lets add the series "1" and "1,2" for completeness:
For $1$, $f(n) = 1$ and $f(2) =1$. $f(2) = 2 \cdot f(1)-1$
For $1,2$, $f(n) = n$ and $f(3) = 3$. $f(3) = 2 \cdot f(2)-1$
This suggests that $f(k+1) = 2 \cdot f(k) -1$ when $f(n)$ is the $k-1$ th order polynomial function that generates the values $1,2,4,...2^{k-1}$.
This raises the question if this is true for all integer series of this type and if so, how to prove it. I hope that I am not overlooking the obvious.
Another observation is that if you write the polynomials that describe the series $1,2,4,8,...$ in a fractional form where all coefficients of $n^k$ in the numerator are integers, then the denominator always seems to be $(K-1)!$ ($1,1,2,6,24,120,$ etc.)
Can anybody shine some light on these observations please?

The second observation is true of all polynomials which interpolate an integer sequence.  This is the subject of the method of finite differences, the "main theorem" of which is this: if we define $\Delta f(n) = f(n+1) - f(n)$, then the unique polynomial of degree $n$ which interpolates the sequence $f(0), f(1), ... f(n)$ is
$$f(x) = \sum_{i=0}^{n} \Delta^i f(0) {x \choose i}.$$
(You should think of this as analogous to Taylor expansion.  The proof uses the identity $\Delta {x \choose i} = {x \choose i-1}$.)  In particular, the $\Delta^i f(0)$ are all integer if and only if $f(0), f(1), ... f(n)$ are all integers, which is the second pattern you observe.
For the powers of $2$ we have $\Delta^i f(0) = 1$ for $1 \le i \le n$, which gives
$$f(x) = \sum_{i=0}^{n} {x \choose i}.$$
It follows that $f(n+1) = 2^{n+1} - 1$, which I think is the first pattern you observe.
