A209320 Number of ways to write 2n = p+q with p and q both prime, p+1 and q-1 both practical.
0, 0, 1, 2, 3, 2, 2, 2, 2, 3, 4, 5, 3, 2, 3, 3, 5, 7, 3, 3, 4, 4, 5, 8, 4, 3, 5, 2, 4, 8, 3, 4, 6, 2, 4, 7, 3, 4, 7, 2, 4, 9, 4, 4, 9, 5, 3, 9, 3, 5, 8, 3, 4, 10, 4, 6, 8, 5, 4, 14, 2, 4, 8, 2, 6, 10, 4, 4, 7, 4, 4, 10, 5, 4, 8, 3, 4, 9, 5, 5, 7, 3, 3, 13, 6, 5, 7, 4, 2, 11, 5, 5, 9, 4, 2, 9, 3, 6, 10, 7
Offset: 1
Keywords
Examples
a(8) = 2 since 2*8 = 3+13 = 11+5 with 3, 5, 11, 13 all prime and 3+1, 13-1, 11+1, 5-1 all practical.
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
Programs
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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[2n-Prime[k]]==True&&pr[Prime[k]+1]==True&&pr[2n-Prime[k]-1]==True,1,0],{k,1,PrimePi[2n-2]}] Do[Print[n," ",a[n]],{n,1,100}]
Comments