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
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
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
KEYWORD
easy,nonn
AUTHOR
Jonathan A. Cohen (cohenj02(AT)tartarus.uwa.edu.au), May 08 2004
STATUS
approved