Rectangular six in the corner
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If there are no outside liberties:
Black can live with 5 points of territory by playing either a or b,
White can kill Black unconditionally by playing at a.
After , Black can't play at a or b because of shortage of liberties. Black is dead.
Black's throw in at is necessary: if White gets to play there she makes it a bent four in the corner. starts the ko. If Black wins this ko, by playing again at and then playing at a capturing the two White stones, the position is settled with Black living in the corner. Should White win this ko by connecting at . Black captures four stones at a and White sets up another ko for life as shown in the next diagram.
After (Black a in the previous diagram), White sets up another ko with and . She can finish the ko by connecting at , creating a bent three nakade shape. So, White has to win the ko twice to kill the corner and Black only once to live. 
The sequence is copied from the main line in the previous paragraph. Now Black can play the surprising move of (oshitsubushi) because he still has at least two liberties. White cannot connect at so Black will play there to make two eyes.
kevinwm: Aside: Even in New Zealand rules where w can play a9, it is not even a ko threat to make a ko for the corner. (That is, a9 does not make ko for the corner, and does not even make ko for the corner if b does not respond. if w a9, b tenuki, w a8, b a9, w b9, and w a7 makes oshitsubushi again.)
In this diagram, Black is dead.
No, this is not a seki. Why?
: One could argue that, this being the case, White might as well fill the outside liberty in the main line diagram instead of connecting at there; but then Black is the first to capture the second stage of the ko and here it is White (try to get first capture in a ko).