A338713 Number of numbers of the form i*n with 1 <= i <= n and tau(i*n) = 4.

0, 0, 1, 1, 2, 1, 3, 1, 1, 1, 4, 0, 5, 1, 1, 0, 6, 0, 7, 0, 1, 1, 8, 0, 1, 1, 1, 0, 9, 0, 10, 0, 1, 1, 1, 0, 11, 1, 1, 0, 12, 0, 13, 0, 0, 1, 14, 0, 1, 0, 1, 0, 15, 0, 1, 0, 1, 1, 16, 0, 17, 1, 0, 0, 1, 0, 18, 0, 1, 0, 19, 0, 20, 1, 0, 0, 1, 0, 21, 0, 0, 1, 22, 0, 1, 1, 1, 0, 23

Offset: 1

Author

N. J. A. Sloane, Nov 11 2020

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

tau is A000005.

For tau(i*n) = 2, 3, 6, see A010051, A296084, A338714.

Inspired by A333995.

Programs

Mathematica

Table[Count[Table[n i,{i,n}],?(DivisorSigma[0,#]==4&)],{n,90}] (* _Harvey P. Dale, Jun 14 2022 *)

PARI

a(n) = sum(i=1, n, numdiv(i*n)==4); \\ Michel Marcus, Nov 11 2020

Python

from collections import Counter
from sympy import factorint
def A338713(n):
    f = Counter(factorint(n))
    return sum(1 for i in range(1,n+1) if (l:=tuple((f+Counter(factorint(i))).values()))==(1,1) or l==(3,)) # Chai Wah Wu, Oct 14 2023