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Folding EulerTotient function

©Linus Jarbo 29 December 2009, compilation of texts for skissa.se.

If you divide (or fold) a square in 1/2, 1/3, 1/4, 1/5 etc. and then pile the patterns onto each other. They do not space evenly. Worth noting is that if the pattern doesn't fit at all(while piling), not even partly, it is a prime number and not dividable to a positive integer other than by itself and 1. (for example: 4 fits partly in 2, so 4 is not a prime)



Worth nothing, while stacking each new additional pattern onto all the previously created and stacked pattern(counting), each new pattern/number - adds a specific amount of new lines, dividing the already divided square further.

-If the amount of new lines created with a stacking - equals the number itself minus one, then the number is a prime, because it is not divisible by any other number than itself or one. thus, not fitting with any other lines. (i.e divisions.)

There exist a function that estimates the additional lines for each stacking. However, the function is just an estimate, or else, we would have a mean to calculate prime numbers, all you had to do was compare the amount of added lines with the number you currently where counting with - minus one. (minus one, since the representation of, for example 3, only has 2 lines)

1, 2, 4, 6, 10, 12, 18, 22, 28, 32, 42,

a(n) = (3/Pi^2)*n^2 + O(n*(log(n))^(2/3)*(log(log(n)))^(4/3));

a(n+1) - a(n) = n - 1;
n = prime;

turns out this stacking of patterns is the sum of Eulers totient function.

and one property of the Eulers totient function sequence is that:
For prime n, totient(n) = n - 1

...and if you for any given number want to know how many new lines that would appear in the square compared to the one you have already stacked - you can use this formula.

So - the conclusion is
- it is possible to use a piece of paper to calculate the totient function (if you can fold it)

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