# Is this a new twin prime sieve method? Any information or comments is very appreciated.

I'm studying the twin prime numbers. Instead of sieving prime numbers, I found this method to sieve $\{x: x \neq \pm 1 \text{( mod$p$)}, x \in \mathbb{N}, p \le p_i\}$, so that $(x-1,x+1)$ will be pair of twin co-primes to given prime set $\{p \le p_i\}$, and you got all twin primes when $x<p_{i+1}^2$.

Apparently this single variable sieve is easier than sieving each part of twin prime pair separately. e.g. with this it's easy to see that there're $\prod(p-2)$ twin co-primes on interval $(p_i,p_i\#)$ due to CRT.

For example, for $p_4=7$, this sieve produces $(3-2) \cdot (5-2) \cdot (7-2) = 14$ numbers on $(7,210)$ and therefore $7$ numbers (half) on $(7,105)$ due to symmetry property: $\{12, 18, 30, 42, 60, 72, 102 \}$, and this produces all twin primes $\{(11,13),(17,19),(29,31),(41,43),(59,61),(71,73),(101,103)\}$ on this interval.

Question: Is this an existing twin prime sieve method? Can you direct me to any study or application of this or similar method? Very appreciate for your help.

BTW, the same sieve method could be applied to sieve co-prime number pair of Goldbach sum of $2n$ using $\{x: x \neq \pm n \text{( mod$p$)}, x \in \mathbb{N}, p \le p_i\}$

• No it's not new. You are using the CRT and the classic sieve to make your argument. You get "all twin primes" less than $p^2$ but beyond a certain point the number of pairs may not increase. – daniel May 8 '16 at 9:17
• Halberstam's text I think skips the CRT but follows this type of idea to Brun's sieve and finally Chen's theorem. If you get through the first 60 pages you will see a variety of elementary ideas that have been tried. – daniel May 8 '16 at 9:36
• @daniel Thanks for your information. Can you add link to Halberstam's text that you referred to? – Fred Yang May 8 '16 at 13:44
• It is the Dover edition of Halberstam and Richert's Sieve Methods. store.doverpublications.com/0486479390.html – daniel May 8 '16 at 14:52
• You might consider looking at this about compound arithmetic progressions, albeit this is an unconventional strategy. math.stackexchange.com/questions/2307461/… – user1329514 Nov 20 '17 at 3:50