LemmyLaLibre@feddit.ch to Memes@lemmy.ml · 1 year agoI say double it.feddit.chimagemessage-square134fedilinkarrow-up1775
arrow-up1753imageI say double it.feddit.chLemmyLaLibre@feddit.ch to Memes@lemmy.ml · 1 year agomessage-square134fedilink
minus-squarealerternate@lemmy.worldlinkfedilinkarrow-up22·1 year agomath checks out. log2(8 billion) ~= 32.9
minus-squareghariksforge@lemmy.worldlinkfedilinkarrow-up9arrow-down2·1 year agoIt’s a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.
minus-squareSabazius@lemmy.worldlinkfedilinkarrow-up7·edit-21 year agoIt’ll just be one fewer junctions. 2^n is always one more than the sum of 21+…2(n-1)
minus-squareMagikjak@lemmy.worldlinkfedilinkarrow-up1·1 year agoI think you have to include 2^0 for that to be true? e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.
math checks out. log2(8 billion) ~= 32.9
It’s a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.
It’ll just be one fewer junctions. 2^n is always one more than the sum of 21+…2(n-1)
I think you have to include 2^0 for that to be true?
e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.