A138012 a(n) = number of positive divisors, k, of n where d(k) divides n (where d(k) = number of positive divisors of k, A000005).

1, 2, 1, 2, 1, 3, 1, 3, 2, 3, 1, 6, 1, 3, 1, 3, 1, 5, 1, 4, 1, 3, 1, 8, 1, 3, 2, 4, 1, 4, 1, 3, 1, 3, 1, 9, 1, 3, 1, 6, 1, 4, 1, 4, 2, 3, 1, 8, 1, 3, 1, 4, 1, 5, 1, 6, 1, 3, 1, 11, 1, 3, 2, 3, 1, 4, 1, 4, 1, 4, 1, 12, 1, 3, 2, 4, 1, 4, 1, 8, 2, 3, 1, 11, 1, 3, 1, 6, 1, 7, 1, 4, 1, 3, 1, 10, 1, 3, 2, 4, 1, 4, 1, 6

Offset: 1

Author

Leroy Quet, Feb 27 2008

Keywords

• nonn

Comments

First occurrence of k: 1, 2, 6, 20, 18, 12, 90, 24, 36, 96, 60, 72, 5670, 972, 120, 336, 180, 420, 540, 240, 600, 2352, 360, 480, 900, 3000, 840, 1080, 1260, 720, ..., . - Robert G. Wilson v

Examples

			10 has 4 divisors (1,2,5,10). The number of divisors of each of these divisors of 10 form the sequence (1,2,2,4). Of these, three divide 10: 1,2,2. So a(10) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A138010, A138011.

Programs

Maple

with(numtheory): a:=proc(n) local div, ct, j: div:=divisors(n): ct:=0: for j to tau(n) do if `mod`(n, tau(div[j]))=0 then ct:=ct+1 else end if end do: ct end proc: seq(a(n),n=1..80); # Emeric Deutsch, Mar 14 2008

Mathematica

Table[Length[Select[Divisors[n], Mod[n, Length[Divisors[ # ]]] == 0 &]], {n,1,100}] (* Stefan Steinerberger *)
f[n_] := Count[Mod[n, DivisorSigma[0, Divisors@n]], 0]; Array[f, 104] (* Robert G. Wilson v *)

PARI

A138012(n) = sumdiv(n,d,if(!(n%numdiv(d)), 1, 0)); \\ Antti Karttunen, May 25 2017

Python

from sympy import divisors, divisor_count
def a(n): return sum([1*(n%divisor_count(d)==0) for d in divisors(n)]) # Indranil Ghosh, May 25 2017

Extensions

More terms from Stefan Steinerberger and Robert G. Wilson v, Feb 29 2008