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well, using f(n)=1*2^(n-1); f(1)=1 breaks scratch | ScratchStats
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well, using f(n)=1*2^(n-1); f(1)=1 breaks scratch
_D
_Duplicate_
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Shared September 16, 2026
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that basically means the function starts at 1, you multiply 2 times v_faster lwk
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1381902828
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Created
September 16, 2026
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September 17, 2026
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September 16, 2026
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