

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
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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

Cf. A233819.
Sequence in context: A111661 A072651 A209411 * A301624 A126791 A052179
Adjacent sequences: A093032 A093033 A093034 * A093036 A093037 A093038


KEYWORD

easy,nonn


AUTHOR

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


STATUS

approved



