[1] Start with this commutator to swap three 1x1x(n-1) blocks,
U2 l U2 r' U2 r U2 l'
Why this one and not its inverse? (As an obvious counterexample).
Because it doesn't discolor all three of the 1x(n-2) center blocks it affects. That is, it doesn't discolor the 1x(n-2) center block in the left half of the U layer.
[2] Add F2 to have all three wings in the same slice
U2 l U2 r' U2 r U2 l'
F2
[3] Add r, the quarter turn to solve back one 1x1x(n-1) block and induce the odd permutation.
U2 l U2 r' U2 r U2 l'
F2
r
[4] Add F2 to restore the cube somewhat
U2 l U2 r' U2 r U2 l'
F2
r
F2
[5] Add the following piece to pair up the remaining edges
U2 l U2 r' U2 r U2 l'
F2
r
F2
r' F2 r
[6] Add F2 to restore the cube somewhat
U2 l U2 r' U2 r U2 l'
F2
r
F2
r' F2 r
F2
[7] Add the non-center preserving PLL parity algorithm
U2 l U2 r' U2 r U2 l'
F2
r
F2
r' F2 r
F2
r2 F2 r2 F2 r2
This is a 20 btm algorithm. However, we can make an obvious change to promote a move cancellation to 19 btm.
U2 l U2 r' U2 r U2 l'
F2
r
F2
r' F2 r
r2 F2 r2 F2 r2
F2
Using the commutator
F' r' F U2 F' r F U2, we can achieve an 18 btm which is nearly <U, r>:
r2 U2 r2 U2 r U2 r U2 r' U2 B2 U' r' U B2 U' r U'