000

001

010

011

100

101

110

111

But the more complex patterns are pretty, so I needed a way to generate the pattern of any given column for a function f(n). This is really very simple - I wrote a function that produces the binary digit in a given place value (2^k) of a given number (n):

([n-n%(2^k)]%[2^(k+1)])/(2^k)

Make this into a series, and substitute f(n) for n, and I have the formula for the columnar pattern in a given f(n) and place value value.

b

Σ [([f(n)-f(n)%(2^k)]%[2^(k+1)])/(2^k)]*(10^n)

n=0

Using this formula, the pattern for the 2^4 column of f(x)=x^2 from 1 to 16 is:

000110101011000

It's prettier when the trailing and leading zeros are removed:

110101011

That is my useless and probably not original (but pretty) contribution to mathematics for the day. Thank you.

EDIT: The equation will generate the proper digit, but as the series loops, the digits will not add together into a nice and tidy list. To remedy this, one must multiply the whole business by 10^n, as shown in red above, to put each one into its proper place value.

If you pick a note for 0 and a harmonising one for 1 and sing them it sounds pretty too.

ReplyDeleteI am assuming this is what you were talking about the other day in the car ^.^

ReplyDelete