How to count rational numbers
- From: kunzmilan <kunzmilan@xxxxxxxx>
- Date: Wed, 31 Oct 2007 06:27:51 -0700
I already discredited myself by my miscarried attempt to count sums of
prime numbers. Nevertheless, I try it again.
In the rational matrix R, r(i,j) = j/i
1) n(n - 1)/2 elements are lesser than 1 in the matrix T,
2) n elements are equal to 1, and
3) n(n - 1)/2 elements are greater than 1.
The value 1/2 repeats in every even row, thus this fraction repeats n/
2 times, supposing that n is even. We must subtract [(n/2) - 1] from
the value of T to eliminated these repeated values 1/2. We will call
these orrective elements as c(i).
The values 1/3 and 2/3 repeat in every third row, thus these 2
rational numbers repeat 2n/3 times in the matrix T. We must subtract
[(2n/3) - 2] from the value of T to eliminated repeated values 1/3 and
2/3, again supposing that n is divisible by 3.
1/4, 2/4, 3/4 repeat in every 4. row, but 2/4 was already counted as
1/2. This leaves 2 uncounted elements in every 4. row. We must
subtract [(n/2) - 2] from the value of T to eliminated repeated values
1/4 and 3/4.
The prime 5 gives repeatings as 4n/5, and the term [(4n/5) - 4],
generally [({p - 1}n/p) - p + 1].
respectively. This leaves fractions 1/6 and 5/6 to be counted.From j/6, 3/6 corresponds to 1/3, and 2/6 and 4/6 to 1/3 and 2/3,
Carefully continuing, and finding ways to eliminate possible rounding
errors, it were possible to find the counting function for any n.
It can be conjectured, that for none n
T - \sum c(i) is not equal to n.
kunzmilan
.
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