OFFSET

1,2

COMMENTS

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}.

LINKS

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

EXAMPLE

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.

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]]

CROSSREFS

KEYWORD

nonn

AUTHOR

Michael De Vlieger, Jun 02 2024

STATUS

approved