A218754 Number of ways to write n=p+q(3+(-1)^n)/2 with q<=n/2 and p, q, p^2+3pq+q^2 all prime.

0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 2, 1, 3, 1, 3, 1, 1, 2, 1, 0, 3, 3, 2, 3, 3, 0, 3, 0, 3, 2, 1, 1, 4, 1, 2, 2, 1, 2, 0, 2, 2, 2, 3, 0, 4, 1, 1, 2, 0, 1, 2, 3, 5, 0, 2, 1, 3, 4, 1, 1, 2, 2, 6, 2, 2, 4, 1, 2, 3, 2, 3, 3, 3, 2, 4, 1, 2, 5, 0, 3, 4, 2, 3, 4, 3, 1, 4, 3

Offset: 1

Author

Zhi-Wei Sun, Nov 04 2012

Keywords

• nonn

Comments

Conjecture: a(n)>0 for all n>=1188.

Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 23 2023

This conjecture is stronger than both Goldbach's conjecture and Lemoine's conjecture.

Zhi-Wei Sun also made the following conjecture: Given any positive odd integer d, there is a prime p(d) such that for any prime p>p(d) there is a prime q

Conjecture verified for d up to 100 and p up to 10^7. - Mauro Fiorentini, Sep 23 2023

Examples

			For n=72 we have a(72)=1 since the only primes p and q with p+q=72, q<=36 and p^2+3pq+q^2 prime are p=67 and q=5.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..20000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv preprint arXiv:1211.1588 [math.NT], 2012-2017.

Crossrefs

Cf. A002372, A046927, A218585, A218654, A218656.

Cf. A000034 = 1,2,1,2,... = (3-(-1)^n)/2. (Note: Offset shifted w.r.t. use in the definition of this sequence.) - M. F. Hasler, Nov 05 2012

Programs

Mathematica

a[n_]:=a[n]=Sum[If[PrimeQ[q]==True&&PrimeQ[n-q(3-(-1)^n)/2]&&PrimeQ[q^2+3q(n-q(3-(-1)^n)/2)+(n-q(3-(-1)^n)/2)^2]==True,1,0],{q,1,n/2}]
Do[Print[n," ",a[n]],{n,1,20000}]

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