Is it possible to express $f(n)$ and $g(n)$ in a single formula? Yes, this seems to be silly question, but please listen to my thought.

Example: $n\in\mathbb{N}$ , Let $f(n)=2n-1$ and $g(n)=2n$, Now look at this function: $$H(n)=n$$
  If We "combine" $f(n)$ and $g(n)$ we get $h(n)$

Question: İf $n\in\mathbb{N}$ and $$f(n)=\frac{2^{8n-3}-2^{2n-1}-3}{9}$$
$$g(n)=\frac{2^{8n-4}-2^{2n}-3}{9}$$
Is it possible to express $f(n)$ and $g(n)$ in a single formula?
Ps. English is not my mother language. I may not have been clear my question. For a better understanding of the problem please, edit.
Thank you!
 A: A simple example would be:
$$H(n)=\frac12((-1)^n+1)f(n/2)+\frac12(-1^{n+1}+1)g((n+1)/2)$$
For all $m\in\Bbb{N}$, we have:
$$H(2m)=\frac12(1+1)f(m)+\frac12(-1+1)g((2m+1)/2)=f(m)$$
$$H(2m-1)=\frac12(-1+1)f((2m-1)/2)+\frac12(1+1)g(m)=g(m)$$
A: I think the word you want might be "interpolate." What you're essentially doing with $f(n)$ and $g(n)$ is letting
$$H(1)=f(1),\ H(2)=g(1)$$
$$H(3)=f(2),\ H(4)=g(2)$$
$$H(5)=f(3),\ H(6)=g(3)$$
$$\vdots$$
so you're defining $H(2n-1)=f(n)=2n-1$ and $H(2n)=g(n)=2n$, so $H(x)=x$ for all integers $x$. You can do the exact same thing with your new functions and get
$$H(x)=\bigg\{\begin{array}{cc}\frac{1}{9}\left(2^{4x+1}-2^x-3\right) & \mathrm{if\ }x\ \mathrm{is\ odd} \\ \frac{1}{9}\left(2^{4x-4}-2^x-3\right) & \mathrm{otherwise}\end{array}$$
It doesn't look as nice as $H(x)=x$, but it's a function nonetheless. 
A: Hint. that what wolfie says, when i try to put first values of that functions: $g(n)$, $f(n)$ [https://www.wolframalpha.com/input/?i=1,+3,+453,+909,+116501,+233013,+29826133,+59652309,+7635497301,+15270994773,+...]
