User blog comment:Wythagoras/Very strong variant of BEAF/@comment-11227630-20131005034440/@comment-10429372-20131005061520

Under my definition, {{X,X/2}&X/2}&n reaches θ(ΩΩ+Ωθ(Ω Ω) ).

Then {{{X,X/2}&X/2}&X/2}&n reaches θ(ΩΩ+Ωθ(Ω Ω+Ωθ(Ω Ω) ) )

And {X,X/3} is equal to {{{{{...}&X/2}&X/2}&X/2}&X/2}&n, so this {X,X/3} reaches θ(ΩΩ2)