Let $f(n,k)$ be the number of times A picks up the $k$th coin when the game starts with $n$ coins. Note that the total number of ways an $n$ coin game can go is $2^{n-1}$, since each player has 2 options at each move, except the last move, which is forced. Then by the reasoning in the answer by piyush and my comment, we have the following equations. $\eqalign{f(n+2,1)&=2^n+f(n,1)\cr f(n+2,2)&=2f(n,1)+f(n,2)\cr f(n+2,k)&=f(n,k)+2f(n,k-1)+f(n,k-2)\cr}$ The third equation holds for $3\le k\le n-2$. Also, $f(n+2,n+1)=f(n+2,2)$, and $f(n+2,n+2)=f(n+2,1)$.
This gives a formula for the numbers in the table, but of course it's a recursive formula. Whether this system of equations can be solved in closed form, I don't know.
EDIT: For any fixed $k$ it shouldn't be too hard to get a formula for $f(n,k)$ as a function of $n$. Let's just take the case $n$ odd. From $f(n+2,1)=2^n+f(n,1)$ we get $f(n,1)=2^{n-2}+2^{n-4}+\cdots+2^1+f(1,1)$ and summing the geometric series we get $f(n,1)={2^n+1\over3}$ Then from $f(n+2,2)=2f(n,1)+f(n,2)$ we get $f(n,2)=(2/3)(2^{n-2}+1+2^{n-4}+1+\cdots+2^1+1)$ and there's another geometric series to sum. Now you can play the same game with $f(n+2,3)=f(n,3)+2f(n,2)+f(n,1)$ and so on. Perhaps a pattern emerges if you do enough of these.