A191004 Number of ways to write n = p+q+(n mod 2)q, where p is an odd prime and q<=n/2 is a prime such that JacobiSymbol[q,n]=1 if n is odd, and JacobiSymbol[(q+1)/2,n+1]=1 if n is even.

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 3, 3, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 4, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4, 3, 5, 4, 1, 4, 1, 2, 3, 2, 2, 2, 3, 1, 4, 1, 2, 4, 2, 2, 3, 1, 2, 4, 5, 3, 3, 1, 4, 3, 2, 3, 5, 3, 4, 8, 2, 2, 7, 4, 4, 5, 2, 2, 6, 3, 3, 4, 4, 2, 4, 2, 1, 4, 4

Offset: 1

Author

Zhi-Wei Sun, Dec 30 2012

Keywords

• nonn

Comments

Conjecture: a(n)>0 for all n>5.

We have verified this for n up to 10^9. It is stronger than Goldbach's conjecture and Lemoine's conjecture.

Zhi-Wei Sun also conjectured the following refinement: Any odd number 2n+1>64 not among 105, 247, 255, 1105 can be written as p+2q, where p and q are primes, and JacobiSymbol[q,p']=1 for any prime divisor p' of 2n+1; also, any even number 2n>8 not among 32 and 152 can be written as p+q, where p and q<=n/2 are primes, and JacobiSymbol[(q+1)/2,p']=1 for any prime divisor p' of 2n+1.

Examples

			a(19)=1 since 19=5+2*7 with JacobiSymbol[7,19]=1.
a(32)=1 since 32=29+3 with JacobiSymbol[(3+1)/2,32+1]=1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..20000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Crossrefs

Cf. A002375, A046927, A185150.

Programs

Mathematica

a[n_]:=a[n]=Sum[If[(Mod[n,2]==1&&PrimeQ[n-2Prime[k]]==True&&JacobiSymbol[Prime[k],n]==1)||(Mod[n,2]==0&&n-Prime[k]>2&&PrimeQ[n-Prime[k]]==True&&JacobiSymbol[(Prime[k]+1)/2,n+1]==1),1,0],{k,1,PrimePi[n/2]}]
Do[Print[n," ",a[n]],{n,1,200}]