A054743 - OEIS

%I #47 Nov 24 2020 10:24:34

%S 1,8,16,32,64,81,128,243,256,512,648,729,1024,1296,1944,2048,2187,

%T 2592,3888,4096,5184,5832,6561,7776,8192,10368,11664,15552,15625,

%U 16384,17496,19683,20736,23328,31104,32768,34992,41472,46656

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

%C 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

%H Robert Israel, <a href="/A054743/b054743.txt">Table of n, a(n) for n = 1..10001</a>

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

%e 8 appears in the list because 8 = 2^3 and 2<3.

%e 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

%p N:= 10^10: # to get all terms <= N

%p p:= 1:

%p S:= {1}:

%p do

%p   p:= nextprime(p);

%p   if p^(p+1) > N then break fi;

%p   pp:= [seq(p^j, j=p+1 .. ilog[p](N))];

%p   S:= S union select(`<=`,{seq(seq(s*q,s=S),q=pp)},N);

%p od:

%p sort(convert(S,list)); # _Robert Israel_, Jun 07 2016

%t okQ[n_] := AllTrue[FactorInteger[n], #[[1]] < #[[2]]&];

%t Join[{1}, Select[Range[50000], okQ]] (* _Jean-François Alcover_, Jun 08 2016 *)

%o (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

%Y Cf. A048103, A054744, A104126, A192135.

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

%K easy,nonn

%O 1,2

%A _James A. Sellers_, Apr 22 2000

%E 1 prepended by _Alec Jones_, Jun 07 2016