prime counting function I made

Hi,

I forgot to mention:

n = nth prime a = value to count primes up to (nth prime)^2 b = my prime counting function c = mathematica primepi d = error between b and c

For example for nth prime = 10 (prime(10)=29)

count of primes = (EulerPhi(Pn(n))/Pn(n)*(prime(n)^2))+n

count of primes = (EulerPhi(Pn(n))/Pn(n)*(prime(n)^2))+n

count of primes = 142.834

This gives the estimate of how many primes there are up to 29^2 (841)

Mathematica PrimePi gives 146

So the error is 0.978312 (142.834/146)

see below for n=10:

cheers, Jamie

Try it for 37^2=24649, my formula is more accurate over certain wide ranges of numbers.

2728.31 = (EulerPhi(Pn(37))/Pn(37)*(prime(37)^2))+n

Mathematica PrimePi = 2729

cheers, Jamie

Can the 'where' of the 'certain ranges' be predicted I wonder?

I'm slightly reminded of the twelve hour analog clock that is absolutely accurate twice a day. If I knew exactly 'when' it was going to be absolutely accurate I could use it as a master timer.

What's the function "Pn" in this context?

Right. A list of primes is even more accurate, though over smaller ranges.

Exactly.

Hi,

Pn(1) = 2 Pn(2) = 6 Pn(3) = 30

see

cheers, Jamie

Hi,

I think so, the formula is less than 24 hours old, and I'm still testing to see what the error looks like, right now it seems to be a converging error, but it may oscillate.

I made a new formula in the meantime that is more accurate I think, but also a bit less nice looking:

Here it is:

An example prime counting formula:

to estimate the primes up to prime(n)^2 where prime(n) is prime and n > 1

count of primes = prime(n)^2 - (sum(((((prime(n)^2)-1) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1), x=1 to n-1) + 1)

example for n=4, prime(n)=7, prime(n)^2=49

count of primes = prime(n)^2 - (sum(((((prime(n)^2)-1) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1), x=1 to n-1) + 1)

count of primes = 49 - (sum((((48) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1), x=1 to 3) + 1)

count of primes = 49 - ((((48) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1) + (((48) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1) + (((48) / (Pn(x)/EulerPhi(Pn(x-1)))) - 1) + 1)

count of primes = 49 - ((((48) / (Pn(3)/EulerPhi(Pn(3-1)))) - 1) + (((48) / (Pn(2)/EulerPhi(Pn(2-1)))) - 1) + (((48) / (Pn(1)/EulerPhi(Pn(1-1)))) - 1) + 1)

count of primes = 49 - ((((48) / (30/2))) - 1) + (((48) / (6/1))) - 1) + (((48) / (2/1))) - 1) + 1)

count of primes = 49 - (((3.2 - 1) + (8 - 1) + (24 - 1)) + 1) count of primes = 49 - ((2.2 + 7 + 23) + 1) count of primes = 49 - (32.2 + 1) count of primes = 49 - 33.2 count of primes = 49 - 33.2 count of primes = 15.8 actual count of primes up to prime(n)^2 = 15

So the formula got 15.8 and there are actually 15 primes. I have to test it more, but it should remain fairly accurate.

cheers, Jamie

seems to do the job, and it dates from 1896. Apparently there is a mathematical proof that it is right.

You can't use it to prove the Legendre Conjecture

which is a pity.

Jamie seems to have re-invented the wheel ....