So on a n *
n grid (n rows and n columns respectively) the entries (in the horizontal rows)
represent the product terms for the Riemann zeta function ζ(s), (s ≥ 1),
whereas the corresponding entries (in the vertical columns) entries represent
the reciprocals of the product entries (1/x > 1) based on our derived
formula,

^{2})(1 – x

^{3})(1 – x

^{4}) … = 1 – 1/{(1/x – 1)} + 1/{(1/x – 1)(1/x

^{2}– 1)} –

1/{(1/x – 1)(1/x

^{2}– 1)(1/x

^{3}– 1)} + 1/{(1/x – 1)(1/x

^{2}– 1)(1/x

^{3}– 1)(1/x

^{4}– 1)} – …

Now the Riemann zeta function provides important information regarding the collective relationship of the primes to the natural number system.

Therefore in complementary manner the derived formula should provide important information on the individual nature of each prime.

With this in mind I then set about looking for a way of using the formula to test for primality (of individual numbers).

Now when we start with x = 1/ 2 (with 1/x = 2) the formula generates the terms 1, 3, 7, 15, … i.e. 2

^{n}– 1, for n = 1, 2, 3, 4, …

I had earlier shown how this number sequence is in turn related to the unique number sequence associated with (x – 1)(x – 2) = x

^{2 }– 3x + 2 which is 1, 3, 7, 15, …

So in this case we multiply the unique terms of this sequence * 1 to obtain the corresponding numbers that the formula generates for 1/x = 2.

I had also shown that when x = 1/3 (with 1/x = 3) the formula generates the terms 2, 8, 26, 80 i.e. 3

^{n}– 1, for n = 1, 2, 3, 4, …

And this in turn is related to the unique number sequence associated with (x – 1)(x – 3) = x

^{2 }– 4x + 3, which is 1, 4, 13, 40, …

So in this case, we multiply the unique terms of the sequence * 2 to obtain the corresponding numbers that the formula generates for 1/x = 3.

And in general when we start with x = 1/k (1/x = k), the formula generates the terms

k

^{n}– 1, for n = 1, 2, 3, 4, …

And this in turn is related to the unique number sequence associated with (x – 1)(x – k) = where we multiply the unique terms of the sequence * (k – 1) to obtain the corresponding numbers that the formula generates for 1/x = k.

It then struck me that if we subtract k – 1 from k

^{n}– 1 i.e. k

^{n}– 1 – k + 1 = k

^{n}– k that

(k

^{n}– k)/n would offer a good test of primality.

Fro example in the simplest case where k = 2, we have (2

^{n }– 2)/n which invariably results in an integer when n is prime.

For example, when n = 2, (2

^{2 }– 2)/2 = 1

n = 3, (2

^{3 }– 2)/3 = 2

n = 5, (2

^{5 }– 2)/5 = 6

n = 7, (2

^{7 }– 2)/7 = 18

Furthermore when n is not prime, (2

^{n }– 2)/n is not an integer.

For example when n = 4, (2

^{4 }– 2)/4 = 14/4 (which is not an integer)

when n = 6, (2

^{6 }– 2)/6 = 62/6 (which is not an integer)

when n = 8, (2

^{8 }– 2)/8 = 254/8 (which is not an integer)

when n = 9, (2

^{9 }– 2)/9 = 510/9 (which is not an integer).

So in this case, where k = 2, (2

^{n }– 2)/n offers a clear cut test for primality

Thus for any integer value of n if (2

^{n }– 2)/n is perfectly divisible by n (so that n is thereby a factor) then n is prime; if not perfectly divisible then n is not prime.

When k > 3, results are not quite so clearcut.

It still is true that when n is prime, and not a factor of k, (k

^{n}– k)/n will result in an integer. Unfortunately however n will no longer be exclusively confined to primes.

For example when k = 3, (3

^{6 }– 3)/6 = 726/6 = 121. However n = 6 is not prime.

However for all other values where (3

^{n }– 3)/n results in an integer up to but not including 66, n is prime.

Then when k = 4 many non-prime values for n will occur.

For example (4

^{4}– 4)/4 = 252/4 = 63 (but 4 is not prime)

And (4

^{8}– 8)/8 = 65,528/8 = 8191 (but 8 is not prime).
Then when k = 5 again many non-prime values will occur. For
example in the first 20, n = 4, 5, 10, 15 are all non-prime.

There are close connections here with Fermat’s Little Theorem.
However what is important is the context in which my own investigation has
arisen.

## No comments:

## Post a Comment