A222603 a(1)=1; for n>0, a(n+1) is the least practical number q>a(n) such that 2(a(n)+1)-q is practical.
1, 2, 4, 6, 8, 12, 18, 20, 24, 30, 32, 36, 42, 54, 56, 60, 66, 78, 80, 84, 90, 104, 120, 162, 176, 192, 210, 224, 234, 260, 270, 272, 276, 294, 320, 330, 342, 378, 380, 384, 390, 392, 396, 414, 416, 420, 450, 462, 464, 468, 476, 486, 510, 512, 522, 546, 594, 620, 630, 702, 704, 714, 726, 728, 744, 750, 798, 800, 810, 812, 816, 920, 924, 930, 966, 968, 972, 980, 990, 992, 1014, 1040, 1050, 1088, 1122, 1232, 1242, 1254, 1280, 1290, 1302, 1316, 1332, 1350, 1352, 1380, 1386, 1458, 1518, 1520
Offset: 1
Keywords
Examples
a(4)=6 since 2(a(3)+1)=10=6+4 with 4 and 6 both practical, and 6>a(3)=4.
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].
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) k=1 n=1 Do[If[m==1,Print[n," ",1]];If[m==k,n=n+1;Do[If[pr[2j]==True&&pr[2m+2-2j]==True,k=2j;Print[n," ",2j];Goto[aa]],{j,Ceiling[(m+1)/2],m}]]; Label[aa];Continue,{m,1,1000}]
Comments