A372972 Numbers k such that A372720(k) is negative.

162, 250, 324, 384, 486, 648, 686, 768, 972, 1152, 1250, 1296, 1372, 1458, 1536, 1728, 1875, 1944, 2058, 2250, 2304, 2430, 2500, 2560, 2592, 2662, 2738, 2916, 3000, 3072, 3362, 3402, 3456, 3698, 3750, 3840, 3888, 3993, 4050, 4116, 4374, 4394, 4418, 4500, 4608

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 A008479(k) = 1 while tau(k) = 2^omega(k).

For prime power p^m, A372720(p^m) = 1, since A008479(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.

For s > 1, an infinite number of k in R are such that g(k) is negative. For example, with s = 6, all terms k > 864 in A033845 are in this sequence.

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) = 162 = 2*3^4, since tau(162) - f(162)
     = (1+1)*(4+1) - card(A369609(162))
     = 10 - 12 = -2.
a(2) = 250 = 2*5^3, since tau(250) - f(250)
     = (1+1)*(3+1) - card(A369609(250))
     = 8 - 9 = -1.
a(3) = 324 = 2^2*3^4, since tau(324) - f(324)
     = (2+1)*(4+1) - card(A369609(324))
     = 15 - 16 = -1, etc.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001221, A007947, A008479, A126706, A372720, A372864.

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, 5000}], ?(# < 0 &)][[All, 1]]