|
Search: id:A066620
|
|
|
| A066620 |
|
Number of decompositions of divisors of n into pairwise coprime triples. |
|
+0
2
|
|
| 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 1, 0, 0, 2, 0, 2, 1, 1, 0, 3, 0, 1, 0, 2, 0, 7, 0, 0, 1, 1, 1, 4, 0, 1, 1, 3, 0, 7, 0, 2, 2, 1, 0, 4, 0, 2, 1, 2, 0, 3, 1, 3, 1, 1, 0, 13, 0, 1, 2, 0, 1, 7, 0, 2, 1, 7, 0, 6, 0, 1, 2, 2, 1, 7, 0, 4, 0, 1, 0, 13, 1, 1, 1, 3, 0, 13, 1, 2, 1, 1, 1, 5, 0, 2, 2, 4, 0, 7, 0
(list; graph; listen)
|
|
|
OFFSET
|
1,12
|
|
|
COMMENT
|
a(m) = a(n) if m and n have same factorization structure.
|
|
REFERENCES
|
Amarnath Murthy, Decomposition of the divisors of a natural number into pairwise coprime sets, Smarandache Notions Journal, vol. 12, No. 1-2-3, Spring 2001.pp 303-306.
|
|
LINKS
|
M. L. Perez et al., eds., Smarandache Notions Journal
|
|
FORMULA
|
In the reference it is shown that if k is a square-free number with r prime factors and m with (r+1) prime factors then a(m) = 4*a(k) + 2^k - 1.
a(n) = (tau(n^3)-3*tau(n)+2)/6. - Vladeta Jovovic (vladeta(AT)eunet.rs), Nov 27 2004
|
|
EXAMPLE
|
a(24) = 3: the divisors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. The decompositions are (1, 2, 3), (1, 2, 9), (1, 3, 4); a(30) = 7: the triples are (1, 2, 3), (1, 2, 5), (1, 3, 5), (2, 3, 5), (1, 3, 10), (1, 5, 6), (1, 2, 15).
|
|
CROSSREFS
|
Cf. A063647.
Cf. A000005.
Sequence in context: A129561 A067742 A089233 this_sequence A025427 A091586 A116377
Adjacent sequences: A066617 A066618 A066619 this_sequence A066621 A066622 A066623
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
K. B. Subramaniam (kb_subramaniambalu(AT)yahoo.com) and Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Dec 24 2001
|
|
EXTENSIONS
|
More terms from Vladeta Jovovic (vladeta(AT)eunet.rs), Apr 03 2003
|
|
|
Search completed in 0.003 seconds
|
