A054743 If n = Product p_i^e_i then p_i < e_i (where e_i > 0) for all i.

1, 8, 16, 32, 64, 81, 128, 243, 256, 512, 648, 729, 1024, 1296, 1944, 2048, 2187, 2592, 3888, 4096, 5184, 5832, 6561, 7776, 8192, 10368, 11664, 15552, 15625, 16384, 17496, 19683, 20736, 23328, 31104, 32768, 34992, 41472, 46656

Offset: 1

Author

James Sellers, Apr 22 2000

Keywords

• easy
• nonn

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

Examples

			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
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10001

Crossrefs

Cf. A048103, A054744, A104126, A192135.

Cf. A207481 (same construction, but with p_i>=e_i),

Programs

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 *)

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

Formula

Sum_{n>=1} 1/a(n) = Product_{p prime} 1 + 1/((p-1)*p^p) = 1.27325025767774256043... - Amiram Eldar, Nov 24 2020

Extensions

1 prepended by Alec Jones, Jun 07 2016