A209254 Number of ways to write 2n-1 = p+q with q practical, p and p^4+q^4 both prime.

0, 1, 1, 2, 2, 2, 3, 1, 2, 2, 3, 2, 4, 3, 1, 3, 1, 1, 4, 2, 5, 5, 1, 4, 1, 2, 4, 3, 1, 6, 3, 4, 4, 5, 1, 6, 7, 2, 4, 3, 4, 2, 4, 5, 1, 2, 3, 7, 5, 2, 4, 8, 4, 6, 5, 1, 2, 2, 3, 8, 3, 1, 5, 6, 2, 4, 7, 4, 8, 4, 2, 7, 6, 3, 4, 3, 1, 6, 6, 1, 7, 6, 2, 8, 9, 5, 7, 3, 3, 10, 7, 3, 9, 14, 1, 9, 4, 3, 4, 6

Offset: 1

Author

Zhi-Wei Sun, Jan 14 2013

Keywords

• nonn

Comments

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

Zhi-Wei Sun also conjectured that any odd integer greater than one can be written as p+q with q practical, and p and p^2+q^2 both prime. This is a refinement of Ming-Zhi Zhang's problem related to A036468.

Examples

			a(8)=1 since 2*8-1=11+4 with 4 practical, 11 and 11^4+4^4=14897 both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• G. Melfi, On two conjectures about practical numbers, J. Number Theory 56 (1996) 205-210 [MR96i:11106].
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arxiv:1211.1588 [math.NT], 2012-2017.

Crossrefs

Cf. A000040, A005153, A208243, A208244, A208246, A209253.

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[pr[2n-1-Prime[k]]==True&&PrimeQ[Prime[k]^4+(2n-1-Prime[k])^4]==True,1,0],{k,1,PrimePi[2n-1]}]
Do[Print[n," ",a[n]],{n,1,100}]