## Saturday, October 24, 2009

### Connecting prime distribution to a continuous function

Going back into my archives to copy from prior posts I have a result from 2002 where a simple progression of formulas shows how the prime distribution is connected to a continuous function:

1. Prime counting function sieve form:

With natural numbers where p_j is the j_th prime:

P(x,n) = x - 1 - sum for j=1 to n of {P([x/p_j],j-1) - (j-1)}

where if n is greater than the count of primes up to and including sqrt(x) then n is reset to that count.

The [] is the floor function and is actually redundant as you're to use natural numbers.

The sqrt() is the integer square root, for instance sqrt(10) = 3, as that's floor(sqrt(10)), and it's automatically positive because natural numbers are only the positive integers greater than 0.

2. Prime counting function difference equation form:

With natural numbers

P(x,y) = x - 1 - sum for j=2 to y of {(P([x/j],j-1) - P(j-1, sqrt(j-1)))(P(j, sqrt(j)) - P(j-1, sqrt(j-1)))}

where if y>sqrt(x), then P(x,y) = P(x,sqrt(x)).

And there is no longer need to tell the equation of any numbers that are prime. If you program it, you'll notice it plucks out primes on its own as it counts, as only when j is prime does:

P(j, sqrt(j)) - P(j-1, sqrt(j-1)) = 0

3. Continuous function:

In the complex plane

P'y(x,y) = -(P(x/y,y) - P(y, sqrt(y))) P'(y, sqrt(y))

and a continuous function is found easily enough from the difference equation from above. To see the full derivation click here.

The partial differential equation form is rather succinct like the sieve form, and I just gave the PDE versus talking about summation as here there is no actual prime count involved.

And that is why the prime distribution is connected to a continuous function: a sieve form as P(x,n) equation can go to a difference equation form, which leads to a partial differential equation.

James Harris