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.

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, 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, 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

Offset

1,16

Comments

Conjecture: a(n)>0 for all odd n>4676 and even n>30986.

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.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Example

a(11)=1 since 11=5+2*3, and both 3(5-3)-1=5 and 3(5-3)+1=7 are prime.
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.

Mathematica

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],
{k, 1, PrimePi[(n-1)/(2+Mod[n, 2])]}]
Do[Print[n, " ", a[n]], {n, 1, 100000}]

Prog

(PARI) a(n)=if(n%2, aOdd(n), aEven(n))
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
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
\\ Charles R Greathouse IV, Jul 31 2016

Crossrefs

Cf. A001359, A006512, A002375, A046927, A219157, A219055, A218867, A218754, A218825, A219052.

Sequence in context: A367405 A112400 A316523 * A365658 A116861 A340032

Adjacent sequences: A219182 A219183 A219184 * A219186 A219187 A219188

Keyword

nonn

Author

Zhi-Wei Sun, Nov 13 2012

Status

approved