• ghariksforge@lemmy.world
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      1 year ago

      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.

      • Sabazius@lemmy.world
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        1 year ago

        It’ll just be one fewer junctions. 2^n is always one more than the sum of 21+…2(n-1)

        • Magikjak@lemmy.world
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          1 year ago

          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.