OFFSET
1,1
COMMENTS
For any set S of primes whose sum of reciprocals is infinite, there are members whose prime factors are all in S. For example, by the strong form of Dirichlet's theorem this is the case for an arithmetic progression {x: x == c (mod d)} if c and d are coprime.
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
count:= 0: Res:= NULL:
for n from 1 while count < 100 do
F:= ifactors(n)[2];
if mul((t[1]^(t[2]+1)-1)^2/(t[1]^(2*t[2]+1)-1)/(t[1]-1), t = F) > 3 and andmap(s -> not(type(n/s, integer)), [Res]) then
count:= count+1; Res:= Res, n;
fi
od:
Res;
MATHEMATICA
count = 0; Res = {};
For[n = 2, count < 100, n++, F = FactorInteger[n]; If[Product[{p, e} = pe; (p^(e+1)-1)^2/((p^(2e+1)-1)(p-1)), {pe, F}] > 3 && AllTrue[Res, !IntegerQ[n/#]&], count++; AppendTo[Res, n]]
];
Res (* Jean-François Alcover, Apr 29 2019, after Robert Israel *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Israel, Jun 20 2018
STATUS
approved