What is the timeloss for each amount of fireworks (1,3,6)?

1 you lose nothing but not sure about 3 or 6

Does this mean that getting a 341 in 4-1 will not lose time to a 340?

No it will not.

"1 you lose nothing but not sure about 3 or 6"

My understanding was that 1 you lose 1 framerule... so a 341 is the same as a 340.

But since you only lose one framerule, it's not worth waiting for the 340 since it's the same as a 341... or do you literally lose ZERO time on a 341?

So, a 342 is TWO framerules ahead of a 341, is this correct?

1 firework loses 2 frames and you need to lose at least 1 frame to avoid it and it never saves a framerule.

imagine if there was a level where you'd have to lose exactly one frame to get the optimal framerule

"imagine if there was a level where you'd have to lose exactly one frame to get the optimal framerule"

That sounds to me like the exact situation in 8-3... Where you have to bump a block to get the optimal rule

^ do you mean 8-1?

Nope - 8-3 - I guess I should have said bump INTO a block. In 8-1 you have to hit a block, but 8-3 you have to intentionally slow yourself down by a frame BEFORE the final stairs jump so that you don't get fireworks and you still are able to hit the top of the flagpole

I mean the a situation where optimally the level would end on the last frame of XX1 on the timer, and the last frame of the frame rule would be the first frame of XX0 on the timer. In that situation, you'd have to lose exactly one frame to complete it optimally, because losing 2 or more frames would push you off the framerule and not losing any frames would get you a firework which would take up 2 frames and push you off the framerule.