OFFSET
1,2
COMMENTS
Closed under multiplication. Use A104126 to construct A192135 by putting A104126(n) * prime(n)^k in a list up to some chosen bound. Create this sequence by multiplying any k elements of A192135 with distinct prime factors in a list (k>1). The last list along with A192135 is this sequence when sorted. - David A. Corneth, Jun 07 2016
LINKS
Robert Israel, Table of n, a(n) for n = 1..10001
FORMULA
Sum_{n>=1} 1/a(n) = Product_{p prime} 1 + 1/((p-1)*p^p) = 1.27325025767774256043... - Amiram Eldar, Nov 24 2020
EXAMPLE
8 appears in the list because 8 = 2^3 and 2<3.
Construction of elements up to 1000: 1. Put 2^3 and 3^5 in a list; {8, 81} (The terms of A104126 up to 1000.) 2. For each element, put products the last list with their distinct prime factors up to 1000. Gives: {8, 16, 32, 64, 128, 256, 512, 81, 243, 729} (Terms from A192135 up to 1000). 3. Put products of k powers of distinct primes in a new list up to 1000: {648} (k>1). Unite {648} with {8, 16, 32, 64, 128, 256, 512, 81, 243, 729}. {8, 16, 32, 64, 128, 256, 512, 81, 243, 729, 648}. Sort the list. This gives: {8, 16, 32, 64, 81, 128, 243, 256, 512, 648, 729}, which are the elements below 1000 in this sequence. - David A. Corneth, Jun 07 2016
MAPLE
N:= 10^10: # to get all terms <= N
p:= 1:
S:= {1}:
do
p:= nextprime(p);
if p^(p+1) > N then break fi;
pp:= [seq(p^j, j=p+1 .. ilog[p](N))];
S:= S union select(`<=`, {seq(seq(s*q, s=S), q=pp)}, N);
od:
sort(convert(S, list)); # Robert Israel, Jun 07 2016
MATHEMATICA
okQ[n_] := AllTrue[FactorInteger[n], #[[1]] < #[[2]]&];
Join[{1}, Select[Range[50000], okQ]] (* Jean-François Alcover, Jun 08 2016 *)
PROG
(PARI) lista(nn) = {for (n=1, nn, f = factor(n); ok = 1; for (i=1, #f~, if (f[i, 1] >= f[i, 2], ok = 0; break; ); ); if (ok, print1(n, ", ")); ); } \\ Michel Marcus, Jun 15 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
James A. Sellers, Apr 22 2000
EXTENSIONS
1 prepended by Alec Jones, Jun 07 2016
STATUS
approved