Yes, and by adding repeated dead stones, one can create a STAR corridor for any positive odd integer.

For example for 3, we have

- Click Here To Show Diagram Code
`[go]$$ 3 STAR`

$$ --------------

$$ | X X X X X X X X X O

$$ | X O O O . O . O . O

$$ | X X X X X X X X X O[/go]

In theory we could do something for even integers, but that would require a branching tree, which doesn't really fit on a Cartesian Go board.

**4/3 attempts**I tried to do some calculations to construct a position where the first few moves were worth 4/3. However, it didn't work out.

- Click Here To Show Diagram Code
`[go]$$`

$$ | O O O . . .

$$ | O X O . . .

$$ | O X O . . .

$$ | X a O O . .

$$ | X . X O . .

$$ | X X b O O O

$$ | . O X . X O

$$ | . O . O O O

$$ | . O O O . .[/go]

Let a,b represent the sizes of those moves (only counting W's solid captures).

This has been set up so that if white spends two moves, white gets a+b+1 (with the half dead ko stone).

If white plays b (b>a), then black plays a, white gets b+1/3

If black plays b, then white plays a, white gets a

If black plays both b and a, white gets -1

Hence, the position if black plays first has a count of (a-1)/2

the position if white plays first has a count of (a+b+1+b+1/3)/2 = a/2 + b + 2/3

The overall count is the average which is (a+b)/2 + 1/3 - 1/4 = (a+b)/2 + 1/12

We require this to be the same as b+1/3

So a/2 = b/2 + 1/4 or b = a + 1/2

Also, since we want the gain to be 4/3, we need (a+b+1)-(b+1/3) = 2* (4/3)

So a = 6/3 = 2

And b = 3/2

Unfortunately, we have contradicted the assumption that b>a.

**Corridor constructions**However, at least I did manage to figure out how to construct an endgame of gain 5/2. This is b above.

Using iterative deiri counting, the technique was to gain 2 + 1/2 = 4/2 + 2/4

That is, white playing b wins them 4 points over if black plays and white responds. Next, white's response wins them 2 points over if black played there again.

This is a standard technique. Two corridors are required because if there is just one corridor, a solid entrance move compared to the defender blocking gains an odd number of points (double the number of dead stones it connects to + 1 for the less defence needed) (under territory scoring. If area scoring, then it is always even since it is double the number of intersections that have changed hands, with an extra +1 for the move the invader plays). The subsequent gains down a corridor are positive.

In addition, the sente-gote assumption implies that consecutive gains are monotone decreasing.

Hence the possible gains from a width one corridor (with no liberty issues) is the range of values the sum

S = a_0 + a_1/2 + a_2/4 + ... + a_n/2^n can take.

With the conditions:

- a_0 is an odd positive integer,

- a_i>0, and

- a_n <= a_(n-1) + a_n/2 <= a_(n-2) + a_(n-1)/2 + a_n/4 <= ... <= S

The last inequality means a_(n-1) >= a_n/2, a_(n-2) >= a_(n-1)/2 + a_n/4, etc

For example a valid sequence is 3, 5, 1 because 5>= 1/2 and 3 >= 5/2 + 1/4. In fact 3, 5, 1, 1, 1, ... fits.

It is straightforward to prove that a geometric increasing sequence of the form 1, r, r^2, ... with r>1 does not fit the inequality, whereas r<=1 does.

In particular, S<= a_0 + a_0 = 2* a_0. It is fairly clear that S can take any value in the range [a_0, 2*a_0] if we vary a_i over the reals.

Hence, a corridor can have a gain of [1,2] and [3,infinity]. Unfortunately, we can't seem to produce (2,3) with a single corridor. This is why we need an extra branch to the corridor to create 5/2 as above.

Overall, my conclusion is that I have failed at creating 4/3, and I still don't know if it is possible.

**Back to n 2/3**Here is a model.

- Click Here To Show Diagram Code
`[go]$$`

$$ | X O X . . . .

$$ | X O X . . . .

$$ | X a O O O O O

$$ | X . X b X . O

$$ | X X O O O O O[/go]

If white plays first, they get b

If black plays first (a>b) at a, then again at b, white gets -a

If black plays first at a then white at b, white gets b-a+1/3

To get the required situation, we need 2*(b-(b-a+1/3)) = b + 1/3

=> 2a - 2/3 = b + 1/3

=> 2a = b+1

This is not consistent with a>b unless a<1.

The gain is b-(b-a+1/3)=a-1/3.

We have already found a situation with 2/3 gain. In general, to get n 2/3 gain, we need a = n+1

So b = (n+1)/2 -1 = (n-1)/2.

Unfortunately, this contradicts a<1.

**Conclusion**The doubled ko was probably a special case. I didn't solve my original question, but did find something out about corridors.