After repeating the operation $4+1=5$ times, the result is $2^{6}+1$.  After repeating the operation $4$ times, the result is $(2^{6}+1+1)\div2=2^{5}+1$.  After repeating the operation $3$ times, the result is $2^{4}+1$.  $$\cdots $$  After doing the operation for the first time, the result is $2^{2}+1$.  Before we do anything, the number is $2^{1}+1$, which is $3$. 
We are given a certain number. We double the number and then subtract $1$ from the result. We repeat this operation $4$ more times. If the end result is $2^{6}+1$, then what is the number that we started with? (Adapted from $2005$ Math Kangaroo Problems, Level $11-12$, Question \#$29$) 
Options:
[A] $$1$$ 
[B] $$2$$ 
[C] $$4$$ 
[D] $$6$$ 
[E] $$3$$ 