A337174 Number of pairs of divisors of n (d1,d2) such that d1 <= d2 and d1*d2 >= n.

1, 2, 2, 4, 2, 6, 2, 6, 4, 6, 2, 12, 2, 6, 6, 9, 2, 12, 2, 12, 6, 6, 2, 20, 4, 6, 6, 12, 2, 20, 2, 12, 6, 6, 6, 25, 2, 6, 6, 20, 2, 20, 2, 12, 12, 6, 2, 30, 4, 12, 6, 12, 2, 20, 6, 20, 6, 6, 2, 42, 2, 6, 12, 16, 6, 20, 2, 12, 6, 20, 2, 42, 2, 6, 12, 12, 6, 20, 2, 30, 9, 6, 2

Offset: 1

Author

Wesley Ivan Hurt, Jan 28 2021

Keywords

• nonn
• easy

Examples

			a(4) = 4; (1,4), (2,2), (2,4), (4,4).
a(5) = 2; (1,5), (5,5).
a(6) = 6; (1,6), (2,3), (2,6), (3,3), (3,6), (6,6).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A002620, A337175, A038548.

Programs

Mathematica

Table[Sum[Sum[Sign[Floor[i*k/n]] (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 100}]
a[n_] := Floor[(DivisorSigma[0, n]+1)^2/4]; Array[a, 100] (* Amiram Eldar, Feb 02 2025 *)

PARI

a(n) = (numdiv(n)+1)^2\4; \\ Amiram Eldar, Feb 02 2025

Python

from sympy import divisor_count
def A337174(n):
    return (divisor_count(n)+1)**2//4 # Chai Wah Wu, Jan 29 2021

Formula

a(n) = Sum_{d1|n, d2|n} sign(floor(d1*d2/n)).

a(n) = tau*(tau+2)/4 if tau is even and a(n) = (tau+1)^2/4 if tau is odd, where tau = A000005(n) is the number of divisors of n, i.e., a(n) = A002620(A000005(n)+1) = floor((A000005(n)+1)^2/4). - Chai Wah Wu, Jan 29 2021

a(n) = A337175(n) + A038548(n). - Ridouane Oudra, May 24 2025

a(n) = (2*d^2 + 4*d + 1 - (-1)^d)/8, where d = A000005(n). (see Wu formula)