Is there a way to write $\dots 2(2(2(2+1)-1)+1)-1 \dots$ in closed form?

Question

It's something like a sequence I'm working with and a part of it has coefficients, which go $3$, $5$, $11$, $21$ and so on. I was wondering if it's possible to find a (closed) formula dependent on say $n$.

The only thing I can contribute is that $(-1)^n$ for the plus and minus ones doesn't seem to lead anywhere (since one has to include the value of the preceding addition/subtraction).

With closed formula I mean a formula where you can plug in your $n$ and obtain the corresponding coefficient immediately, without having to know any of the previous ones.

Answer 1

So observe that $u_{n+1}=2u_n-(-1)^{n}$ and $u_n=2u_{n-1}+(-1)^n$

Add these to obtain $u_{n+1}+u_n=2u_n+2u_{n-1}$ or $u_{n+1}-u_n-2u_{n-1}=0$

Unsurprisingly standard methods give $u_n=A2^n+B(-1)^n$ and if $u_1=3, u_2=5$ you can compute $A=\frac 43, B=-\frac 13$

Answer 2

Try $a_0 = 1, a_n = 2^{n+1} - a_{n-1}$

Answer 3

You may find that this is OEIS seqeunce A001045. In the linked description there is a lot of information, including several formulas, mostly of recursive nature. Among the more closed formulas are $$a(n) = \sum_{k=1}^n {n\choose k}\cdot (-1)^{n+k}\cdot 3^{k-1}$$ $$a(n+1) = \lceil 2^n/3\rceil + \lfloor 2^n/3\rfloor$$