Find the $n^{th}$ term for this sequence I have the following sequence:

4, 7, 11, 19, 36, 69

Now, I've done the usual and found the differences, and it goes down to four levels until I get a common difference, suggesting I have to use $n^{4}$ somewhere, but I just can't find the nth term. Any help? And if you know of any easier methods to finding the nth term, I'd appreciate it.
P.S. - I know about the formulas for arithmetic progression and geometric progression, but clearly neither can be used here.
 A: The first numbers in each row of differences are 4, 3, 1, 3, 2. Assuming that the last row is all 2's then the formula is
$$
4\binom n0+ 3\binom n1 + 1\binom n2 + 3\binom n3 + 2\binom n4
$$
starting at $n=0$.
This is an instance of the Newton series.
The formula simplifies to
$$
\frac{n^4 - n^2 + 36 n + 48}{12}
$$
but this is not enlightening.
A: There are infinitely many sequences which start with that initial segment.  Since there are five terms, there will definitely be a 4th degree equation which fits it, but unless the problem is to find the simplest polynomial equation which generates that sequence, it's not clear that a polynomial would be the correct way to go (unless the minimal degree polynomial fitting the data is of lower degree than expected).
If we eyeball the sequence, we see that each term is approximately doubling each time.  This suggests that the sequence is of the form $a2^n+f(n)$ for some small $f(n)$.  
For example, if we subtract off $2^n+n$ from the $n$th term, we get the new sequence:
$$1,1,0,-1,-1,-1\ldots$$
If we define the sign function $\sigma(n)=\frac{|n|}{n}$ if $n\neq 0$ and $\sigma(0)=0$, we can thus write the sequence as 
$$2^n+n+\sigma(3-n).$$
Is this the "right" answer?  I honestly can't say.  Without more terms, there are probably quite a lot of relatively simple answers which will fit the sequence.  Having more terms (or knowledge of what kind of answer they are looking for) would help eliminate some of these possibilities.
