A372864 Numbers k such that A372720(k) = 0.

1, 500, 578, 722, 750, 1058, 1500, 1682, 1922, 2646, 2744, 3430, 3645, 4800, 5202, 5346, 5476, 5488, 5625, 6318, 6400, 6724, 7168, 7396, 8000, 8836, 10092, 10976, 11236, 11532, 11979, 12005, 13068, 13924, 14450, 14884, 15309, 16810, 16875, 16896, 18050, 18225

Offset: 1

Author

Michael De Vlieger, Jun 02 2024

Keywords

• nonn

Comments

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

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

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

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

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.

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

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}.

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.

Examples

			a(1) = 1 since tau(1) - f(1) = 1 - 1 = 0.
a(2) = 500 = 2^2 * 5*3, since tau(500) - f(500)
     = (2+1)*(3+1) - card({10,20,40,50,80,100,160,200,250,320,400,500})
     = 12 - 12 = 0.
a(3) = 578 = 2*17^2, since tau(578) - f(578)
     = (1+1)*(2+1) - card({34,68,136,272,544,578})
     = 6 - 6 = 0, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..2000

Crossrefs

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

Programs

Mathematica

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]]