I'm too noob to figure out what is bugged there? Isn't that a normal combo? 3B 236B CE CE ?
i think the input one person suggested was 3B, 236(B), G+A+K ie, hold the B while inputing G+A+K. another person suggested 3B, 236(B),8aK again while holding the B. yet another suggested "hold the 236 B and then input 98A+B+K".
if you look at the vid closely before the CE comes out, the input is something like, 3B, 1236(B)98(a)K, where you don't release B and A until very end. just look at the damage!! it's much higher than the normal 3B 236B CE CE. the extra damage is coming from the 236B inputted in the way stated above, which i think went from 40+ to 60+. the way the person on the taiwan forum explained it as (i'm translating and paraphrasing of course) "getting 236(B) out in the same amount of time as 236B using the cancel property of BE, but without actually expending gauge." another poster noted that
236(B) became 16 frames, -2 on block without expending any gauge. this is like a BE-cancel similar to QS4G where instead of the 4 cancelling the quick step into guard, the BE-like command cancel out the delay of 236(B).
i haven't got it yet because at first i thought it's just like a faster version for CE to come out. but after examining the vid and the original post further, i realized that the "glitch" is not in the CE but in the 236(B).
vid's in spoiler for your convenience.