Can I express any odd number with a power of two minus a prime? I have been running a computer program trying to see if I can represent any odd number in the form of
$$2^a - b$$
With b as a prime number. I have seen an earlier proof about Cohen and Selfridge regarding odd numbers that are nether a sum or a difference of a power of two and a prime, and I was curious to see if anyone has found an odd number that couldn't be represented using the above formula.
 A: Fred Cohen and J. L. Selfridge, Not Every Number is the Sum or Difference of Two Prime Powers
contains an example of such a number:
Corollary: 47867742232066880047611079 is prime and neither the sum nor
difference of a power of two and a prime.
A: Erdos's argument essentially gives that there is an infinite 
arithmetic progression of odd numbers containing 
no element of the form power of two minus a prime. 
Define 
\begin{array}{cccc}
i & a_i & m_i & p_i \\
1 & 1 & 2 & 3 \\
2 & 2   & 4   &  5  \\
3 & 4 &   8 &   17  \\
4 & 8 &  16 &  257  \\
5 & 16 & 32 & 65537 \\
6 & 0  & 64   & 641 \\
7 & 32 & 64 & 6700417 \\
\end{array}
Set $A_0=\{x:x\equiv 1 \pmod{8}\}$, 
$A_i=\{x:x\equiv 2^{a_i} \pmod{p_i}\}$ ($1\leq i\leq 7$), 
$B=A_0\cap A_1\cap\cdots\cap A_7$. 
$B$ is an AP consisting of odd numbers. 
Assume that $x\in B$ and $x=2^n-q$ for some odd prime $q$. 
For some $1\leq i\leq 7$, we have $n\equiv a_i \pmod{m_i}$, 
therefore $x\equiv 2^{a_i}-q \pmod{p_i}$. As $x\in A_i$, we 
also have $x\equiv 2^{a_i} \pmod{p_i}$, consequently $q=p_i$. 
But this is impossible as $2^n\equiv 0 \pmod{8}$ for $n\geq 3$, 
$p_i\equiv 1,3,5 \pmod{8}$ and as $x\in A_0$, $x\equiv 7 \pmod{8}$.   
