Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat. 
What about: $2, 3, 6, 7, 14, 15, 30,... $?
Again the sequence has a group of three terms it repeats: Add one to first term to get second term, then double second term to get third term. 
How do you compute the $n$th term of such sequences directly, without iterating through all preceding terms?

Denote the $n$-th term by $x_{n}$ and have a look at the terms with
odd index $n$. Based on what you remarked yourself:
$x_{1}=1$
$x_{3}=2x_{1}+3=2+3$
$x_{5}=2x_{3}+3=2\left(2+3\right)+3=2^{2}+2.3+3$
$x_{7}=2x_{5}+3=2\left(2^{2}+2.3+3\right)+3=2^{3}+2^{2}.3+2.3+3$
et cetera.
This starts to look like 
$x_{2n+1}=2^{n}+\left(2^{n-1}+2^{n-2}+\cdots+1\right)3=2^{n}+\left(2^{n}-1\right)3=2^{n+2}-3$
This can be proved by induction and its easy now to find that $x_{2n}=x_{2n+1}-3=2^{n+2}-6$. 
On a sortlike way you come to $y_{2n+1}=2^{n+2}-2$ and $y_{2n}=2^{n+1}-1$ where $y_{n}$
denotes the $n$-th term of the second sequence.
