A210480
Number of primes p
0, 0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 3, 3, 4, 4, 4, 5, 4, 4, 3, 3, 2, 2, 4, 4, 4, 5, 5, 7, 6, 6, 3, 4, 3, 3, 5, 5, 4, 5, 5, 7, 7, 6, 4, 3, 3, 4, 4, 3, 2, 4, 4, 7, 6, 6, 3, 3, 4, 4, 4, 4, 2, 4, 4, 6, 5, 5, 3, 2, 4, 4, 6, 3, 3, 4, 4, 7, 5, 6, 4, 4, 4, 4, 7, 6, 5, 4, 3, 8, 5, 7, 3, 3, 5
Offset: 1
Keywords
Examples
a(1846)=1 since 1846=1289+557 with 1289 and 557 both prime, and 1288 and 1290 both practical. a(15675)=1 since 15675=919+14756 with 919 prime, and 918, 920, 14756 all practical.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..60000
- G. Melfi, On two conjectures about practical numbers, J. Number Theory 56 (1996) 205-210 [MR96i:11106].
- Zhi-Wei Sun, Sandwiches with primes and practical numbers, a message to Number Theory List, Jan. 13, 2013.
- 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[pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True&&(PrimeQ[n-Prime[k]]==True||pr[n-Prime[k]]==True),1,0],{k,1,PrimePi[n-1]}] Do[Print[n," ",a[n]],{n,1,100}]
Comments