explicit recurrence formula

I can advise you two methods:

(Edit) Method of generating functions:

We are going to obtain a particular solution, which is often the most difficult part.

Let us consider the case where:

$$a_0=1, a_1=2, a_2=3$$

with the following terms :

$$a_3=11, \ a_4=36, \ a_5=96, \ a_6=239, \ a_7=579, \ a_8=1364, \ \text{etc.}$$

In this case, one gets the following formula, very coherent to yours (it suffices to add the general solution of the associated homogeneous recurrence):

$$a_n=(29/50) \cos(\pi n/2)+(53/50) \sin(\pi n/2)-(27/25) \times 2^n+(3/2)+(4/5) \times n 2^n\tag{0'}$$

Let us establish it using the technique of generating functions.

Let the formal series $g(x)=\sum_{n=0}^{\infty}a_nx^n$ be the generating function of the sequence $a_n$.

It is not difficult to verify that, when multiplying initial relationship:

$$a_{n}-2a_{n-1}+a_{n-2}-2a_{n-3}=2^n-3\tag{1'}$$

by $x^n$ and then suming for all $x^n$, we get:

$$g(x)(1-2x+x^2-2x^3)-\underbrace{(3+x-x^2)}_{p(x)}=\frac{1}{1-2x}-\frac{3}{1-x}\tag{2'}$$

where the fractions on the RHS are the generating functions of sequence $2^n$ and constant sequence $3$ resp. The presence of polynomial $p(x)$ is justified by the fact that the recurrence between the $a_n$ uses four terms, therefore cannot take place before degree $3$.

From (2'), one can deduce the partial fraction decomposition:

$$g(x)=\dfrac{53x+29}{50(x^2+1)}-\dfrac{3}{2(x-1)}+\dfrac{47}{25(2x-1)}+\dfrac{4}{5(2x-1)^2}\tag{3'}$$

(3') gives (0').

Second method using "auxiliary sequences" (in the initial version of this answer, it was placed in front) with the objective to get rid progressively from the "spurious" terms using :

Let us write the initial relationship

$$a_{n}-2a_{n-1}+a_{n-2}-2a_{n-3}=2^n-3\tag{1}$$

and its equivalent "shifted version":

$$a_{n+3}-2a_{n+2}+a_{n+1}-2a_{n}=2^{n+3}-3\tag{2}$$

Take the combination $(3) = 1 \times (2) - 8 \times (1)$ giving

$$(a_{n+3}-8a_n)-2(a_{n+2}-8a_{n-1})+(a_{n+1}-8a_{n-2})-2(a_{n}-8a_{n-3})=21\tag{3}$$

Otherwise said, by defining

$$b_{n}:=a_{n+3}-8a_n \tag{4}$$

$$b_n-2b_{n-1}+b_{n-2}-2b_{n-3}=21\tag{3'}$$

Now, we must get rid of constant $21$. This is done by writing (3') under the form:

$$(b_n+k)-2(b_{n-1}+k)+(b_{n-2}+k)-2(b_{n-3}+k)=21\tag{5}$$

for a certain unknown $k$ that must be such that $-2k=21 \iff k=-\dfrac{21}{2}$

Setting

$$c_n:=b_n-\dfrac{21}{2}\tag{6}$$

relationship (5) has become:

$$c_n-2c_{n-1}+c_{n-2}-2c_{n-3}=0\tag{7}$$

Now, and only now, you are "authorized" to use the characteristic polynomial technique giving you the general expression of the $c_n$, from which you will get, using (6) under the form $b_n=c_n+\dfrac{21}{2}$ the general expression of $b_n$, then, using (4), you will get a relationship that you will write 3 times:

$$a_{n+3}-8a_n=b_n, \ a_{n+2}-8a_{n-1}=b_{n-1}, \ a_{n+1}-8a_{n-2}=b_{n-2}$$

and by combinations of these equations, get rid, once more, of the "spurious" RHS.