I'm stuck on a problem. I have a term $3^m$ where $m$ is an integer $> 0$. and I want to represent it as $2^m + a$ however I don't want to keep the $a$. I am looking for a formula to represent $a$ using numbers and the variable $m$. Is this possible to do?
Thanks
This is more a collection of information than a formula. It seems tough to get an elementary function on this.
We can get a generating function for this.
There's a lot of information on this on Sloan's Online Encyclopedia of Integer Sequences, A001047.
It looks like you have:
$$3^m = 2^m + a$$
Also, I believe you want to represent $a$ as a formula of $m$.
This can be done by a generating function. The way this works is that we use induction. We consider what we'd get with $3^{m-1}$, and then calculate what we'd get with $3^m$.
First, we want to figure out the difference between $3^m$ and $2^m$. We realize that we can visualize $3^m$ as a "hypercube"; that is, an $m$-dimensional cube with sides of length 3. Now $2^m$ is an $m$ dimensional cube with sides of length 2. So the difference between the larger cube and the smaller cube is adding length one to each side of the smaller cube.
We do this by first adding $3^{m-1}$. This gives us part of what we need. Then we can add twice the previous case (remember, we're doing this by induction). That will give us the required difference.
Now, we can construct a generating function to give us the answer. If you're interested in a very, very simple introduction, you can download a free copy of Wilf's Generatingfunctionology here.
THE GENERATING FUNCTION
We state this as recursively as:
$$a_{m+1} = 2 a_m + 3^{m-1}$$
This translates into the generating function:
$$ \begin{align} \sum_{m=1}^\infty{a_{m+1}x^m} &= 2 \cdot \sum_{m=1}^\infty{a_{m}x^m} + \frac{3}{(1-3x)} \\ \frac{A(m)-1}{x} &= 2A(m) + \frac{3}{(1-3x)}\\ A(m) &= \frac{1}{1-5x+6} \end{align} $$
The series proceeds like:
$$1, 5x, 19x^2, 65x^3, 211x^4, \dots$$