A337360 a(n) = sigma(n) * (tau(n) + 1).

2, 9, 12, 28, 18, 60, 24, 75, 52, 90, 36, 196, 42, 120, 120, 186, 54, 273, 60, 294, 160, 180, 72, 540, 124, 210, 200, 392, 90, 648, 96, 441, 240, 270, 240, 910, 114, 300, 280, 810, 126, 864, 132, 588, 546, 360, 144, 1364, 228, 651, 360, 686, 162, 1080, 360, 1080, 400, 450, 180, 2184

Offset: 1

Author

Wesley Ivan Hurt, Aug 24 2020

Keywords

• nonn
• easy
• changed

Comments

Original name was: Sum of the coordinates of all pairs of divisors of n, (d1,d2), such that d1 <= d2.

2*a(n) is the sum of the perimeters of all distinct rectangles that can be made whose side lengths are divisors of n.

Every divisor of n occurs tau(n) + 1 times in the coordinates of divisors of n. - David A. Corneth, Aug 25 2020

Examples

			a(3) = 12; The divisors of 3 are {1,3}. The divisor pairs, (d1,d2), where d1 <= d2 are (1,1), (1,3) and (3,3). The sum of all the coordinates is then 1+1+1+3+3+3 = 12. So a(3) = 12.
a(4) = 28; The divisors of 4 are {1,2,4}. The divisor pairs, (d1,d2), where d1 <= d2 are (1,1), (1,2), (1,4), (2,2), (2,4) and (4,4). The sum of all the coordinates is then 1+1+1+2+1+4+2+2+2+4+4+4 = 28. So a(4) = 28.
a(5) = 18; The divisors of 5 are {1,5}. The divisor pairs, (d1,d2), where d1 <= d2 are (1,1), (1,5) and (5,5). The sum of all the coordinates is then 1+1+1+5+5+5 = 18. So a(5) = 18.
a(6) = 60; The divisors of 6 are {1,2,3,6}. The divisor pairs, (d1,d2), where d1 <= d2 are (1,1), (1,2), (1,3), (1,6), (2,2), (2,3), (2,6), (3,3), (3,6), (6,6). The sum of all the coordinates is then 1+1+1+2+1+3+1+6+2+2+2+3+2+6+3+3+3+6+6+6 = 60. So a(6) = 60.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005 (tau), A000203 (sigma).

Programs

Mathematica

Table[Sum[Sum[(i + k) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, n}], {n, 80}]

PARI

a(n) = sigma(n) * (numdiv(n)+1) \\ David A. Corneth, Aug 25 2020

Python

from sympy import divisor_sigma
def A337360(n): return (divisor_sigma(n,0)+1)*divisor_sigma(n) # Chai Wah Wu, Aug 07 2025

Formula

a(n) = Sum_{d1|n, d2|n, d1<=d2} (d1+d2).

a(p^k) = (p^(k+1)-1)*(k+2)/(p-1) for p prime and k >= 1. - Wesley Ivan Hurt, Aug 23 2025

Extensions

New name using formula from David A. Corneth_, Aug 25 2020