- #1
eljose
- 492
- 0
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)
[tex] a(n)=\pi(n+1)-\pi(n) [/tex] where [tex] pi(x) [/tex] 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:
[tex] b(n)=\pi(n+2)-2\pi(n+1)+\pi(n) [/tex] here you can check that for composite numbers and normal primes this function is always 0 except if p and p+2 are primes..
[tex] a(n)=\pi(n+1)-\pi(n) [/tex] where [tex] pi(x) [/tex] 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:
[tex] b(n)=\pi(n+2)-2\pi(n+1)+\pi(n) [/tex] here you can check that for composite numbers and normal primes this function is always 0 except if p and p+2 are primes..