login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Numbers of the form m*p^3, where m > 1 is squarefree and prime p does not divide m.
3

%I #33 Aug 02 2024 03:03:19

%S 24,40,54,56,88,104,120,135,136,152,168,184,189,232,248,250,264,270,

%T 280,296,297,312,328,344,351,375,376,378,408,424,440,456,459,472,488,

%U 513,520,536,552,568,584,594,616,621,632,664,680,686,696,702,712,728,744,750

%N Numbers of the form m*p^3, where m > 1 is squarefree and prime p does not divide m.

%C Can be regarded as the cube version of A072357. Subsequence of A036537 (the number of divisors of any term is a power of 2). Also a subsequence of A048109.

%C Subsequence of A126706. - _Michael De Vlieger_, Feb 22 2023

%C The asymptotic density of this sequence is Sum_{p prime} d(p) = 0.074177741367259601921..., where d(p) = 1/(p^2*(p+1)*zeta(2)) is the density of the subset of terms that are divisible by p^3. - _Amiram Eldar_, Aug 01 2024

%H Amiram Eldar, <a href="/A360793/b360793.txt">Table of n, a(n) for n = 1..10000</a>

%F For a(n) = m*p^3, A000005(a(n)) = 2^k, where k = 2 + A001221(m).

%F Equals A048109 \ A030078.

%e 1608 = 2^3*201 is in this sequence (p = 2; m = 201 is odd and squarefree).

%e A001221(201) = 2, therefore 1608 has 2^(2+2) = 16 divisors.

%p filter:= proc(n) local F;

%p F:= sort(ifactors(n)[2][..,2]);

%p nops(F) >= 2 and F[-1] = 3 and F[-2] = 1

%p end proc:

%p select(filter, [$1..1000]); # _Robert Israel_, Mar 01 2023

%t Select[Range[1000], (e = Sort[FactorInteger[#][[;; , 2]]])[[-1]] == 3 && Length[e] > 1 && e[[-2]] == 1 &] (* _Amiram Eldar_, Feb 21 2023 *)

%o (PARI) isok(k) = if (k>1, my(f=factor(k), v=f[,2], m); if (vecmax(v)==3, w=select(x->(x==3), v, 1); if (#w == 1, m = k/f[w[1],1]^3; (m>1) && issquarefree(m)))); \\ _Michel Marcus_, Feb 21 2023

%o (Python)

%o from itertools import count, islice

%o from sympy import factorint

%o def A360793_gen(startvalue=1): # generator of terms >= startvalue

%o return filter(lambda n:len(f:=sorted(factorint(n).values(),reverse=True))>1 and f[0]==3 and f[1] == 1,count(max(startvalue,1)))

%o A360793_list = list(islice(A360793_gen(),20)) # _Chai Wah Wu_, Feb 28 2023

%Y Cf. A000005, A001221, A001222, A013661, A030078, A036537, A048109, A072357, A126706.

%Y Subsequence of A190641.

%K nonn

%O 1,1

%A _David James Sycamore_, Feb 21 2023