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A219185 Number of prime pairs {p,q} (p>q) with 3(p-q)-1 and 3(p-q)+1 both prime such that p+(1+(n mod 2))q=n.

%I A219185 #13 Jul 31 2016 00:06:53
%S A219185 0,0,0,0,0,0,0,1,0,1,1,1,1,0,0,2,1,1,1,2,1,0,1,3,1,1,1,1,2,1,1,1,0,1,
%T A219185 1,2,0,1,2,2,0,2,2,0,2,1,0,3,1,2,1,1,2,1,1,1,2,2,1,4,1,1,1,0,1,1,2,1,
%U A219185 1,3,1,5,2,1,2,1,0,2,0,2,3,4,2,3,3,2,2,1,3,2,1,1,2,0,0,2,1,3,2,3
%N A219185 Number of prime pairs {p,q} (p>q) with 3(p-q)-1 and 3(p-q)+1 both prime such that p+(1+(n mod 2))q=n.
%C A219185 Conjecture: a(n)>0 for all odd n>4676 and even n>30986.
%C A219185 This conjecture has been verified for n up to 5*10^7. It implies Goldbach's conjecture, Lemoine's conjecture and the twin prime conjecture.
%H A219185 Zhi-Wei Sun, <a href="/A219185/b219185.txt">Table of n, a(n) for n = 1..10000</a>
%H A219185 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, arXiv:1211.1588.
%e A219185 a(11)=1 since 11=5+2*3, and both 3(5-3)-1=5 and 3(5-3)+1=7 are prime.
%e A219185 a(16)=2 since 16=11+5=13+3, and 3(11-5)-1, 3(11-5)+1, 3(13-3)-1, 3(13-3)+1 are all prime.
%t A219185 a[n_]:=a[n]=Sum[If[PrimeQ[n-(1+Mod[n,2])Prime[k]]==True&&PrimeQ[3(n-(2+Mod[n,2])Prime[k])-1]==True&&PrimeQ[3(n-(2+Mod[n,2])Prime[k])+1]==True,1,0],
%t A219185 {k,1,PrimePi[(n-1)/(2+Mod[n,2])]}]
%t A219185 Do[Print[n," ",a[n]],{n,1,100000}]
%o A219185 (PARI) a(n)=if(n%2, aOdd(n), aEven(n))
%o A219185 aOdd(n)=my(s); forprime(q=2,(n-1)\3, my(p=n-2*q); if(isprime(n-2*q) && isprime(3*n-9*q-1) && isprime(3*n-9*q+1), s++)); s
%o A219185 aEven(n)=my(s); forprime(q=2,n/2, if(isprime(n-q) && isprime(3*n-6*q-1) && isprime(3*n-6*q+1), s++)); s
%o A219185 \\ _Charles R Greathouse IV_, Jul 31 2016
%Y A219185 Cf. A001359, A006512, A002375, A046927, A219157, A219055, A218867, A218754, A218825, A219052.
%K A219185 nonn
%O A219185 1,16
%A A219185 _Zhi-Wei Sun_, Nov 13 2012