The prime counting function

[R2357]

Hello,

I've thought about the prime counting function pi(n), and I thought that when n is a primorial : 17#, 19#, 23# and so on, there may exist a way to find the exact value pi(n) that could be simpler than computing all the primes (whitch would, of course need to be tested)

Example : We know that pi(210)=46, let's suppose we didn't know it, we could start by calculating the number of possible positions doing 1-1*1/2=1/2, 1/2-1/2*1/3=2/6, 2/6-2/6*1/5=8/30, 8/30-8/30*1/7=48/210,
Then, we would add 4, the number of primes up to x#, (in that case 7#), giving us 52, then we take away 1, because the first number in any cycle x# is always a possible prime position but 1 isn't prime, so we get to 51,
And, (probably the most time consuming part for this method) : we take away, for each of the primes above x (for x#), whose square is below x#, the number of primes that they can multiply by to give less than x#, in the case 7#, 11 can be multiplied up to 19, giving us 4 taken away (11*11, 13, 17, 19), and 13² takes away another number,

We have thus 51-(4+1)=46.

Maybe an interesting idea...

[CRGreathouse]

There's nothing in this method that limits it to primorials.

If you push a little further you'll get the Legendre sieve and hence the beginnings of sieve theory.

[Batalov]

Originally Posted by R2357

And, (probably the most time consuming part for this method) : we take away, for each of the primes above x (for x#), whose square is below x#, the number of primes that they can multiply by to give less than x#, in the case 7#, 11 can be multiplied up to 19, giving us 4 taken away (11*11, 13, 17, 19), and 13² takes away another number,

Maybe an interesting idea...

It is an interesting idea indeed ...and a nice reinvention (not in a bad sense; people reinvent things all the time while they learn - which is very much like embryological parallelism. While one grows as a mathematical being, one repeats what the past evolution has already done).

I think this idea is ~188 years old. you see - for squareful numbers, whose moebius function is zero, you subtract them back because you've already added them in the first pass. But if you used moebius function, you would have "added them with weight 0" (i.e. skipped).