

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.


2



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
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OFFSET

1,2


COMMENTS

By a result of Melfi, each positive even number can be written as the sum of two practical numbers.
For a practical number p, define h(p) as the least practical number q>p such that 2(p+1)q is practical. Construct a simple (undirected) graph H as follows: The vertex set of H is the set of all practical numbers, and for two vertices p and q>p there is an edge connecting p and q if and only if h(p)=q. Clearly H contains no cycle.
Conjecture: The graph H constructed above is connected and hence it is a tree.


LINKS



EXAMPLE

a(4)=6 since 2(a(3)+1)=10=6+4 with 4 and 6 both practical, and 6>a(3)=4.


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<3Mod[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+22j]==True, k=2j; Print[n, " ", 2j]; Goto[aa]], {j, Ceiling[(m+1)/2], m}]];
Label[aa]; Continue, {m, 1, 1000}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



