Earlier I showed how to interpret a theoretical position of K+R vs. K+B as a "mate in $\omega$" on a quarter-infinite board. If I'm doing this right, it takes only an additional Black pawn on the quarter-infinite board to get $2\omega$.
Reflect the board about the diagonal, to get White Kc2, Rb3 vs. Black Ka2, Ba3, and add a Black pawn on h4:
alt text http://math.harvard.edu/%7Eelkies/2omega.png
Black to move. White can't go after the pawn at once, because then the Black King escapes. White's plan is to pin the Bishop on a3 as before, forcing the pawn to advance to h3, so that when the Rook attacks and captures it the Rook will also prevent the Black King's escape: eventually ...Bc5; Rb5, Bq19; Ra5+, Ba3; Ra4, h3; Rh4, B-any; Rxh3. But then Black will have a second chance to make the game arbitrarily long.
[In case you've not seen this before: the Rook doesn't need to be on the b-file for this to work -- e.g. Rxh3, Bc5; Rh5, Bb4; and now not Rh4?, Ka3 = draw, but (say) Rh8 and if Ka3 then Ra8+ still wins because the Bishop blocks its own King's escape!]
I think it should be possible to get $3\omega$, $4\omega$, etc. by adding more Black pawns on the same file or on multiple columns, though some care may be needed to get the counts right because the White Rook could eliminate more than one pawn at each iteration.