A340227 Number of pairs of divisors of n, (d1,d2), such that d1 < d2 and d1*d2 is squarefree.

0, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 4, 1, 4, 4, 1, 1, 4, 1, 4, 4, 4, 1, 4, 1, 4, 1, 4, 1, 13, 1, 1, 4, 4, 4, 4, 1, 4, 4, 4, 1, 13, 1, 4, 4, 4, 1, 4, 1, 4, 4, 4, 1, 4, 4, 4, 4, 4, 1, 13, 1, 4, 4, 1, 4, 13, 1, 4, 4, 13, 1, 4, 1, 4, 4, 4, 4, 13, 1, 4, 1, 4, 1, 13, 4, 4, 4, 4, 1, 13

Offset: 1

Author

Wesley Ivan Hurt, Jan 01 2021

Keywords

• nonn

Comments

If n = p where p is prime, the only pair of divisors of n such that d1 < d2 is (1,p). Since the product 1*p = p is squarefree, this satisfies the constraints. Thus, a(p) = 1 for all p. - Wesley Ivan Hurt, May 21 2021

Examples

			a(28) = 4; (1,2), (1,7), (1,14), (2,7)
a(29) = 1; (1,29)
a(30) = 13; (1,2), (1,3), (1,5), (1,6), (1,10), (1,15), (1,30), (2,3), (2,5), (2,15), (3,5), (3,10), (5,6)
a(31) = 1; (1,31)
		

Crossrefs

Cf. A008683, A074816.

Programs

Maple

with(numtheory): seq((3^nops(factorset(n))-1)/2, n=1..100); # Ridouane Oudra, Jan 27 2025

Mathematica

Table[Sum[Sum[MoebiusMu[i*k]^2 (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}]

Formula

Sum_{d1|n, d2|n, d1A008683).

a(n) = (A074816(n) - 1)/2. - Ridouane Oudra, Jan 27 2025