Is $53$ expressible in this form? It seems as if prime numbers may always be expressed in the form $a\cdot 2^b+c \cdot 3^d$ for some nonnegative integers $b,d$ and $a,c\in \{-1,0,1\}$.
Examples:
$2=1\cdot 2^1+0\cdot 3^d$
$3=0\cdot 2^b+1\cdot 3^1$
$5=1\cdot 2^1+1\cdot 3^1$
$7=1\cdot 2^2+1\cdot 3^1$
$11=1\cdot 2^3+1\cdot 3^1$
$13=1\cdot 2^2+1\cdot 3^2$
$17=1\cdot 2^3+1\cdot 3^2$
$19=1\cdot 2^4+1\cdot 3^1$
$23=1\cdot 2^5+(-1)\cdot 3^2$
$29=1\cdot 2^5+(-1)\cdot 3^1$
$31=1\cdot 2^5+(-1)\cdot 3^0$
$37=1\cdot 2^6+(-1)\cdot 3^3$
$41=1\cdot 2^5+1\cdot 3^2$
$43=1\cdot 2^4+1\cdot 3^3$
$47=1\cdot 2^7+(-1)\cdot 3^4$
We hit a brick wall at $53$. Can anyone confirm if $53$ is/isn't expressible?
What about $n\in \mathbb{N}$ in general? Is it possible to always express $n$ in this form?
Thanks.
EDIT: I invented this question on my own, there are no sources.
Note: The reason I started with prime numbers is because I found it hard to find an expression for $6$, whereas prime numbers continued to be easy to find expressions for (easy until $53$, that is).

Let's count the positive integers up to $X$ of the shape $2^b+3^d$. It is easy to see that there are at most a constant times $\log^2 X$ of them. That the same statement is true for integers of the shape $|2^b-3^d|$ is much less obvious, but follows from what are called lower bounds for linear forms in logarithms (for this exact problem, they were first used by Ellison in the 1970s). In particular, the number of positive integers in $[1,X]$ that can be written as $a \cdot 2^b + c \cdot 3^d$ is at most a constant times $\log^2 X$. It follows that, for any dense enough set of integers, most of them will not be of the form $a \cdot 2^b + c \cdot 3^d$. In the case of the primes, there are of order $X/\log X$ of them up to $X$, essentially 100% of which are not of the desired form. A like conclusion is reached if you ask the same question about, for example, squares, cubes, or Carmichael numbers.
