What is the pattern to this sequence? $$0, 1, 3, 13, 51, 205$$
More specifically, $$(0,0)\quad(1,1)\quad (2,3)\quad (3,13) \quad(4,51)\quad (5,205)$$
I have tried using the interpolation feature in Grapher.app and Wolfram Alpha, but cannot seem to get what I'm looking for which is a quadratic function that hands me the numbers from the top part. 
UPDATE/EDIT
For those of you interested, I encountered the above sequence after factoring $\frac{2}{5}$ from the sum $$\sum_{n=1}^\infty \frac{2^n}{2^{2n} + (-1)^{n+1}} = \frac{2}{5} +\frac{4}{15}+\frac{8}{65}+\frac{16}{255}+\frac{32}{1025} ...$$
$$=\frac{2}{5} * \left(1+ \frac{2}{3}+\frac{4}{13}+\frac{8}{51}+\frac{16}{205} ...\right)$$
The numerators in the brackets are powers of $2$, while the denominators were the ones in question. 
Thanks for the help!
 A: There is no quadratic function for that sequence. One possible equation is:
$ a(n) = 3a(n-1) + 4a(n-2)$
This sequence represents:


*

*The inverse binomial transform of powers of 5.

*The number of walks of length n between any two distinct vertices of the complete graph $K_5$.

*The number of segments (sides) per iteration of the space-filling Peano-Hilbert curve.

*For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 3's along the central diagonal, and 2's along the subdiagonal and the superdiagonal.


Source: the online encyclopedia of integer sequences.
A: For a question like this we can come up with our own pattern. The next number can be anything we want. We can come up with a pattern for each answer.
A: a(n)=a(n-1)*4 then check the value of n position, if it is odd then add one,if it is even then substrata one  
A: In cases like this, it is recommended to try and visit the OEIS.
A: Of course, with such pattern recognition questions, you can let the next value be anything.
My suspicion is that $u_{n+1} = 4 u_{n} + (-1)^n$.
A: Using the recurrence $a_n = 4a_{n-2} +3a_{n-1}$, which this sequence appears to satisfy, you can derive an explicit formula for $a_n$, namely:
$a_n = \frac{4^n - (-1)^n}{5}$
It's exponential - it grows too fast to be quadratic. You can find a quadratic function that passes through any three of those points, but it won't pass through the others at the same time.
A: It is clear from the (edited) question that the pattern is $a_n=\dfrac{2^{2n}+(−1)^{n+1}}{5}$. Is that answer acceptable? 
