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SIN
8 : 8
M(x)

Plotting m(x) reveals a clear fractal pattern

Mocha's Sequence is generated like so:

  • 1,...
  • 1,2,... (next power of 2)
  • 1,2,4,... (next power of 2)
  • 1,2,4,3,... (3 is missing. Put the mean of the leftmost empty region (4 and 2).)
  • 1,2,4,3,8,... (next power of 2)
  • 1,2,4,3,8,6,... (5, 6, and 7 are missing. Put the mean of the leftmost empty region (8 and 4).)
  • 1,2,4,3,8,6,5,... (5 and 7 are missing. Put the mean of the leftmost empty region (6 and 4).)
  • 1,2,4,3,8,6,5,7... (7 is missing. Put the mean of the leftmost empty region (8 and 6).)
  • 1,2,4,3,8,6,5,7,16... (next power of 2)

And so on.

It was inspired by the order in which I mark lines on my graphs:

Mm

By using the third iteration, you too can closely approximate 1/5 as 0.203125!

  1. Let n be the number of divisions needed on a graph
  2. If n mod 2 = 0, divide the axis. Go to 5.
  3. If n can be written as 2^n+1, move hand as if to divide at 2^n. Move hand opposite direction as if to divide new hypothetical region by 2^(2n). Repeat forever, or until desired accuracy, and then draw line. Go to 5.
  4. Guess where the line goes.
  5. Move to the leftmost region that needs lines. If one exists, go to 2. Otherwise, you are DONE!

m(n)

1,2,4,3,8,6,5,7,16,12,10,9,11,14,13,15,32,24,20,18,17,19,22,21,23,28,26,25,27,30,29,31,64,48,40,36,34,33...

m-1(n)

1,2,4,3,7,6,8,5,12,11,13,10,...

Properties

  • The the average difference between m(n) and n approaches 4 as n->inf.
  • There are an infinite number of n such that m(n)=n. Such an n mod 10 is even for n>1.
  • The series is fractal.
  • m(2^n)=2^n-1 for n>1
  • m(2^n+1)=2^(n+1)
  • m(2^n+2)=3*2^(n-1) for n>0

See Also

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