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A093035
Number of triples (d1,d2,d3) where each element is a divisor of n and d1 + d2 + d3 <= n.
2
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
1,4
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
EXAMPLE
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).
PROG
(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
CROSSREFS
Cf. A233819.
Sequence in context: A072651 A209411 A369912 * A301624 A126791 A052179
KEYWORD
easy,nonn
AUTHOR
Jonathan A. Cohen (cohenj02(AT)tartarus.uwa.edu.au), May 08 2004
STATUS
approved