A210681 Number of ways to write 2n = p+q+r (p<=q) with p, q, r-1, r+1 all prime and p-1, p+1, q-1, q+1, r all practical.
0, 0, 0, 0, 1, 2, 3, 3, 3, 2, 2, 3, 5, 6, 4, 3, 2, 4, 7, 10, 8, 5, 3, 5, 8, 12, 10, 6, 2, 3, 6, 11, 12, 6, 4, 3, 5, 9, 10, 6, 5, 4, 5, 8, 8, 5, 7, 7, 6, 8, 7, 6, 6, 8, 6, 7, 8, 5, 7, 8, 6, 7, 7, 4, 6, 7, 5, 6, 8, 4, 8, 6, 4, 5, 7, 5, 5, 8, 5, 6, 8, 6, 4, 7, 6, 6, 7, 5, 3, 7, 3, 4, 8, 6, 8, 5, 4, 3, 7, 6
Offset: 1
Keywords
Examples
a(5)=1 since 2*5=3+3+4 with 3 and 5 both prime, and 2 and 4 both practical. a(6)=2 since 2*6=3+3+6=3+5+4 with 3,5,7 all prime and 2,4,6 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.
- Zhi-Wei Sun, Sandwiches with primes and practical numbers, a message to Number Theory List, Jan. 13, 2013.
- Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II: CANT, New York, NY, USA, 2015 and 2016, Springer Proc. in Math. & Stat., Vol. 220, Springer, New York, 2017, pp. 279-310. (See also 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) pp[k_]:=pp[k]=pr[Prime[k]-1]==True&&pr[Prime[k]+1]==True pq[n_]:=pq[n]=PrimeQ[n-1]==True&&PrimeQ[n+1]==True&&pr[n]==True a[n_]:=a[n]=Sum[If[pp[j]==True&&pp[k]==True&&pq[2n-Prime[j]-Prime[k]]==True,1,0],{j,1,PrimePi[n-1]},{k,j,PrimePi[2n-Prime[j]]}] Do[Print[n," ",a[n]],{n,1,100}]
Comments