Least amount of steps to get over 1000 I am wanting to type out 1000 characters onto a message, I start off with one character and I have that character in my clipboard. I then paste it to get two characters.
From here I can either select all, copy, navigate to the end of message and paste it, or I can paste the single one I have. the former option requires 4 steps to increase it to 4 with 2 characters in my clipboard, but the latter requires 1 step to increase it to 3 with 1 character in my clipboard. I could then make this choice at each iteration. Either double the amount of characters in the clipboard and add it to the end in 4 steps or keep the amount the same and add it onto the end.
What is the least amount of time I could take to get $1000$ or more characters assuming that each step takes 1 second. i.e. doubling the increase and pasting takes 4 seconds or simply pasting the existing increase takes 1 second.
my second question expands off of this, how would I work out how many steps it takes to get over $n$ characters assuming $n$ is any real integer.
would I be right in assuming it would be wise to double the increase every chance except the last iteration which would be 2 single pastes without doubling?
 A: Two more pastes get you to $4$ characters, which is two steps faster than copy.  Now to get to 8 is a toss-up; four steps either way.  If you do the copy, you then have four on the clipboard, so two more pastes will get to $16$.  I think the long run most efficient is copy then two pastes, which gets a factor of $4$ with six steps instead of $8$.  My best is then $$\begin {array} {l r r} \text{Action} & \text{message} & \text {clipboard}\\3 \text{ pastes} & 4&1\\\text{copy}&8&4\\2\text{ pastes} &16&8\\\text{copy}&32&16\\2\text{ pastes}&64&16\\\text{copy}&128&64\\2\text{ pastes}&256&64\\\text{copy}&512&256\\2\text{ pastes}&1024&256\end{array}$$ for a total of $27$ actions, meaning $27$ seconds  
Added: for long term operations, we can think of three choices, one paste then copy, two pastes then copy, or three pastes then copy.  As a figure of merit, I suggest the log of the expansion factor divided by the number of actions.  We get $$\begin {array} \\ \text{pattern}&\text{expansion}&\text{actions}&\text{log(expansion)/actions}\\\text{paste+copy}&3&5&.2197\\ \text{2 paste+copy}&4&6&.2310\\ \text{3 paste+copy}&5&7&.2299\end{array}$$
so 2 pastes seems to win.
