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

1,14

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.

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.

Example

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.

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}]

Crossrefs

Cf. A002375, A046927, A185150.

Sequence in context: A209156 A329325 A341149 * A191358 A204133 A342017

Adjacent sequences: A191001 A191002 A191003 * A191005 A191006 A191007

Keyword

nonn

Author

Zhi-Wei Sun, Dec 30 2012

Status

approved