A372864 - OEIS

%I #16 Jun 03 2024 18:32:32

%S 1,500,578,722,750,1058,1500,1682,1922,2646,2744,3430,3645,4800,5202,

%T 5346,5476,5488,5625,6318,6400,6724,7168,7396,8000,8836,10092,10976,

%U 11236,11532,11979,12005,13068,13924,14450,14884,15309,16810,16875,16896,18050,18225

%N Numbers k such that A372720(k) = 0.

%C Let tau = A000005, let omega = A001221, let f = A008479, and let g = A372720.

%C For squarefree k, A372720(k) >= 0, since f(k) = 1 while tau(k) = 2^omega(k).

%C For prime power p^m, A372720(p^m) = 1, since f(p^m) = m while tau(k) = m+1.

%C Therefore, apart from a(1) = 1, this sequence is a proper subset of A126706.

%C In the sequence R = {k = m*s : rad(m) | s, s > 1 in A120944}, there is a smallest term k such that g(k) <= 0 and a largest term k such that g(k) is positive. For instance, in A033845 where s = 6, only {6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 192, 216, 288, 432, 576, 864} are such that g(k) > 0.

%C Apart from terms in this sequence, all the rest of the terms k in R are such that g(k) is negative.

%C There are no 3-smooth numbers k > 1 in this sequence, however there are 3 terms {500, 6400, 8000} in A033846 (with s = rad(k) = 10). For s = 2*3*23, there are 6 terms {19044, 25392, 38088, 70656, 536544, 953856}.

%C Conjecture: proper subset of A361098, hence of A360765 and A360768. This is to say that k = a(n) is such that A003557(k) >= A119288(k), i.e., k/rad(k) >= second smallest prime factor of k, and A003557(k) > A053669(k), where A053669(k) is the smallest prime q that does not divide k.

%H Michael De Vlieger, <a href="/A372864/b372864.txt">Table of n, a(n) for n = 1..2000</a>

%e a(1) = 1 since tau(1) - f(1) = 1 - 1 = 0.

%e a(2) = 500 = 2^2 * 5*3, since tau(500) - f(500)

%e      = (2+1)*(3+1) - card({10,20,40,50,80,100,160,200,250,320,400,500})

%e      = 12 - 12 = 0.

%e a(3) = 578 = 2*17^2, since tau(578) - f(578)

%e      = (1+1)*(2+1) - card({34,68,136,272,544,578})

%e      = 6 - 6 = 0, etc.

%t rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]]; Position[Table[r = rad[n]; DivisorSigma[0, n] - Count[Range[n/r], _?(Divisible[r, rad[#]] &)], {n, 20000}], _?(# == 0 &)][[All, 1]]

%Y Cf. A000005, A001221, A007947, A008479, A010846, A372720, A372972.

%K nonn

%O 1,2

%A _Michael De Vlieger_, Jun 02 2024