A209253 Number of ways to write 2n-1 as the sum of a Sophie Germain prime and a practical number.

0, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 3, 4, 3, 5, 2, 3, 4, 4, 4, 5, 2, 3, 5, 2, 4, 7, 4, 2, 6, 2, 5, 6, 2, 2, 6, 1, 3, 7, 4, 3, 7, 4, 5, 8, 2, 3, 8, 3, 3, 8, 4, 4, 7, 4, 5, 8, 3, 4, 7, 1, 5, 9, 5, 3, 9, 3, 4, 8, 4, 3, 9, 3, 5, 8, 2, 2, 9, 4, 3, 8, 4, 4, 10, 1, 3, 10, 5, 4, 10, 4, 3, 9, 5, 5, 10, 4

Offset: 1

Author

Zhi-Wei Sun, Jan 14 2013

Keywords

• nonn

Comments

Conjecture: a(n)>0 for all n>1.

This has been verified for n up to 5*10^6.

Examples

			a(40)=1 since 2*40-1=23+56 with 23 a Sophie Germain prime and 56 a practical number.
		

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 [math.NT], 2012-2017

Crossrefs

Cf. A005384, A005153, A208243, A208244, A208246, A208249.

Programs

Mathematica

f[n_]:=f[n]=FactorInteger[n]
Pow[n_,i_]:=Pow[n,i]=Part[Part[f[n],i],1]^(Part[Part[f[n],i],2])
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n],s+1],1]<=DivisorSigma[1,Product[Pow[n,i],{i,1,s}]]+1,0,1],{s,1,Length[f[n]]-1}]
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n,2]+Con[n]==0)
a[n_]:=a[n]=Sum[If[PrimeQ[2Prime[k]+1]==True&&pr[2n-1-Prime[k]]==True,1,0],{k,1,PrimePi[2n-1]}]
Do[Print[n," ",a[n]],{n,1,100}]