|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
|
|