A093035 Number of triples (d1,d2,d3) where each element is a divisor of n and d1 + d2 + d3 <= n.

0, 0, 1, 4, 1, 17, 1, 20, 8, 20, 1, 103, 1, 20, 27, 54, 1, 109, 1, 112, 27, 20, 1, 315, 8, 20, 27, 112, 1, 315, 1, 112, 27, 20, 27, 481, 1, 20, 27, 324, 1, 321, 1, 112, 125, 20, 1, 695, 8, 112, 27, 112, 1, 321, 27, 324, 27, 20, 1, 1285, 1, 20

Offset: 1

Author

Jonathan A. Cohen (cohenj02(AT)tartarus.uwa.edu.au), May 08 2004

Keywords

• easy
• nonn

Comments

It appears that a(n) depends on both parity of n and its prime signature. For instance a(odd prime)=1, a(even semiprime)=20, a(odd semiprime)=27, a(odd prime cube)=27, a(odd prime fourth power)=64. Maybe it is possible to find a formula for a(n). Similar sequences with pairs, quadruples, ... instead of triples can be envisioned. - Michel Marcus, Aug 21 2013

There's more to the story above. It seems that a(A233819(n)) gives the largest possible value per prime signature. Some prime signatures may have more than two possible values for a(n). - David A. Corneth, May 19 2020

Examples

			a(9) = 8 because the divisors of 9 are {1,3,9} making the valid triples (1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Crossrefs

Cf. A233819.

Programs

PARI

a(n) = {nb = 0; d = divisors(n); for (i = 1, #d, for (j = 1, #d, for (k = 1, #d, if (d[i]+d[j]+d[k] <= n, nb++);););); nb;} \\ Michel Marcus, Aug 21 2013