How Can We Define a Function for Twin Primes Using Prime Counting Functions?

[eljose]

If we define a function a(n) with the next properties, a(n) is 1 iff n is prime and 0 if n is composite..then we can write the function a(n)

$$a(n)=\pi(n+1)-\pi(n)$$ where $$pi(x)$$ is the usual prime number counting function, then my question is to define a b(n) function so b(n)=1 if p and p+2 are primes (twin primes) and 0 elsewhere (no matter if p is prime or not, p and p+2 must be consecutive primes) then my question is if we somehow could write this function in the form:

$$b(n)=\pi(n+2)-2\pi(n+1)+\pi(n)$$ here you can check that for composite numbers and normal primes this function is always 0 except if p and p+2 are primes..

 

[matt grime]

b(n) is floor((pi(n)+pi(n+2))/2). But you really ought to learn what the von Mangolt function is.Oh, and you keep changing between n and p. Don't. Plus what on Earth is a 'normal' prime?

 

[eljose]

i say i don,t know if the function $$b(n)=\pi(n+2)-2\pi(n+1)+\pi(n)$$ is correct (that was my question) but you check that for a twin prime p so p+2 is also prime it gives b(p)=1 and 0 elsewhere¿? (i,m not sure) by a normal prime i mean a prime that p is prime but p+2 is not.

 

[shmoe]

pi(x) is the number of primes less than or equal to x. Your a(n) is off by one, it should be a(n)=pi(n)-pi(n-1).

Your b(n) is then just a(n+2)-a(n+1), which is clearly not correct. You could use b(n)=a(n)*a(n+2) or b(n)=floor((a(n)+a(n+2))/2) for example (the latter is what matt intended I believe)

 

[eljose]

Supposing $$\pi(x)$$ is the prime number counting function and from definition of b(n) we get $$b(n)=\pi(n+2)-2\pi(n+1)+\pi(n)$$

-if p and p+2 are primes then b(n)=1
-if p is prime b(n)=0
-if p+1 is prime then b(n)=-1
-if n is composite b(n)=0

now if we use Abel,s sum formula we would reach to the expression:

$$\sum_{p2}f(p2)+\sum_{p}[f(p-2)-f(p-1)]=(\pi(n+1)-pi(n))f(x)-\int_{2}^{\infty}dx(\pi(x+1)-\pi(x))df(x)/dx+C$$

where "´" means derivative and C is a constant, p2 is a twin prime so p and (p+2) are primes.

 

[shmoe]

Supposing $$\pi(x)$$ is the prime number counting function and from definition of b(n) we get $$b(n)=\pi(n+2)-2\pi(n+1)+\pi(n)$$

-if p and p+2 are primes then b(n)=1
-if p is prime b(n)=0
-if p+1 is prime then b(n)=-1
-if n is composite b(n)=0

by your definition of b(n), b(9)=pi(11)-2*pi(10)+pi(9)=5-2*4+4=1. Does this mean you think 9 and 11 are both prime?

Your b(n) gives a 1 when n+2 is prime and n+1 is composite, -1 when n+2 is composite and n+1 is prime, 0 when both n+2 and n+1 are prime or both are composite. It says nothing about n like you claim it does.

 

[matt grime]

I must be going blind: there is no need to explain that a ' means derivative, but it doesn't appear anywhere in your post apart from where you state that it siginfies the derivative.

 

[Zurtex]

I must be going blind: there is no need to explain that a ' means derivative, but it doesn't appear anywhere in your post apart from where you state that it siginfies the derivative.

Perhaps an entirely unrelated remark?

 

[eljose]

ups..then there is a mistake then let,s see if now i,m right:
-iff p+2 and p are primes-------->b(n)=1
-iff p is prime but p+1 and p+2 are composite b(n)=0
-iff p+1 is prime then b(n)=-1
-iff p+2 is prime then b(n)=1

I hope that now there,s all right of course p and p+1 can,t be both prime
then we have the identity:

$$\sum_{p2}f(p)+\sum_{p}[f(p-2)-f(p-1)]=a(n)f(n)+C-\int_{2}^{n}dxa(x)(df/dx)$$ (1)

where $$a(x)=\pi(x+1)-\pi(x)$$

but what,s the purpose of all that?..if the relation (1) is right we could prove some statement similars to "twin prime conjecture"..

-the sum of the inverse of twin primes {p2| p and p+2 are primes} converges whereas its equivalent to normal prime diverge..note that if we set f(x)=1/x+1 then f(x-2)-f(x-1) sum over all prime would diverge as:

$$ln[ln(x-2)/ln(x-1)]$$ for x---->oo this is 0

-there are infinite number of "twin primes" set f(x)=x and you can check that the integral on the right is O(n) for n big this is divergent.

-Unfortunately for the case f(x)=1 we can,t obtain anything as...

$$\pi2(x)=\sum_{n=2}^{x}(\pi(x+2)-2\pi(x+1)+\pi(x))$$

but we don,t know "a priori" if there are infinite number of twin primes that contribute with 1 to the series, taking the difference operator twice for b(n) and letting n--->oo we get that the probability of finding a twin prime is $$P=1/(ln(n))^{2}$$

of course the title is wrong..the formula b(n) couldn,t be considered as a charasteristic function for twin primes...

 

[shmoe]

ups..then there is a mistake then let,s see if now i,m right:
-iff p+2 and p are primes-------->b(n)=1
-iff p is prime but p+1 and p+2 are composite b(n)=0
-iff p+1 is prime then b(n)=-1
-iff p+2 is prime then b(n)=1

I hope that now there,s all right of course p and p+1 can,t be both prime
then we have the identity:

n and n+1 are prime when n=2.

b(n)=1 has nothing at all to do with n being prime or not. This b(n) has nothing at all to do with twin primes.

$$\sum_{p2}f(p)+\sum_{p}[f(p-2)-f(p-1)]=a(n)f(n)+C-\int_{2}^{n}dxa(x)(df/dx)$$ (1)

where $$a(x)=\pi(x+1)-\pi(x)$$

What is this supposed to be? Are these sums supposed to be over an infinite range? Over twin primes and primes respectively I guess? I don't see how this is supposed to follow from Abel summation, if your sum is over primes or twin primes then you should end up with their counting functions on the right.

Your a(n) is still not the characteristic function of the primes, in case you are still hoping that it is.

-Unfortunately for the case f(x)=1 we can,t obtain anything as...

$$\pi2(x)=\sum_{n=2}^{x}(\pi(x+2)-2\pi(x+1)+\pi(x))$$

This is false. If n=[x] the sum on the right is pi(n+2)-pi(n+1)-pi(3)+pi(2), which is not at all the counting function for twin pimes.

 

[eljose]

-Sorry i forgot...the detail that n and n+1 are primes if n=2
-if you see the function $$b(n)=\Delta\Delta{\pi(n)}$$ where this delta is the forward difference operator then using Abel sum formula:

$$\sum_{n}\Delta^{2}{\pi(x)}f(n)=\Delta{\pi(n)}f(n)-\int_{2}^{n}dx\Delta{\pi(x)}df(x)$$

 

[shmoe]

-if you see the function $$b(n)=\Delta\Delta{\pi(n)}$$ where this delta is the forward difference operator then using Abel sum formula:

$$\sum_{n}\Delta^{2}{\pi(x)}f(n)=\Delta{\pi(n)}f(n)-\int_{2}^{n}dx\Delta{\pi(x)}df(x)$$

Fine, but you haven't answered my question as to what that first sum was supposed to be over. It looks like you mean p2 to be a twin prime, but that isn't what the sum $\sum_{n}\Delta^{2}{\pi(x)}f(n)$ gives (or at least what it looks like you mean, you still can't be bothered to give the range of your summation, and you're apparently using "n" to mean more than one thing), it will just be your second sum, the sum with f(p-2)-f(p-1) in it

 

[eljose]

Sorry...i was wrong i don,t mean that the relation between the sum of b(n)f(n) by Abel,s sum formula relation is wrong but i think i didn,t consider it well in fact if we redefine:

$$b(n)=\pi(n+1)-2\pi(n)+\pi(n-1)=[\pi(n+1)-\pi(n)]-[\pi(n)-\pi(n-1)$$

so taking the product b(n)f(n) and summing it for every n we get:

$$\sum_{n=0}^{\infty}(f(n-1)-f(n))(\pi(n)-\pi(n-1))=\sum_{p}(f(p-1)-f(p))$$

but from here i don,t know how to associate it to a sum over twin primes...