|
Search: id:A085082
|
|
|
| A085082 |
|
Number of distinct prime signatures arising among the divisors of n. i.e. among several divisors of n with the same prime signature only one contributes to the count. Let this function be called tau'(n). |
|
+0
3
|
|
| 1, 2, 2, 3, 2, 3, 2, 4, 3, 3, 2, 5, 2, 3, 3, 5, 2, 5, 2, 5, 3, 3, 2, 7, 3, 3, 4, 5, 2, 4, 2, 6, 3, 3, 3, 6, 2, 3, 3, 7, 2, 4, 2, 5, 5, 3, 2, 9, 3, 5, 3, 5, 2, 7, 3, 7, 3, 3, 2, 7, 2, 3, 5, 7, 3, 4, 2, 5, 3, 4, 2, 9, 2, 3, 5, 5, 3, 4, 2, 9, 5, 3, 2, 7, 3, 3, 3, 7, 2, 7, 3, 5, 3, 3, 3, 11, 2, 5, 5, 6, 2, 4, 2, 7, 4
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
1. For a square-free number with n distinct prime divisors a(n) = n+1. 2. If n = p^r then a(n) =tau'(n)= tau(n)= r+1. Question: Find tau'(n) in the following cases: 1. n = m^k where m is a square-free number with r distinct prime divisors. 2. n = product {(p_i)^i}, n has r distinct prime divisors p_i., i = 1 to r.
Answers: 1. (r+k)!/(r!k!). 2. A000108(r+1). - David Wasserman (wasserma(AT)spawar.navy.mil), Jan 20 2005
|
|
EXAMPLE
|
a(30) = 4 and the divisors with distinct prime signatures are 1,2,6,30. The divisors 3 and 5 with the same prime signature as of 2 and the divisors 10 and 15 with the same prime signature as that of 6 are not counted.
The divisors of 36 are 1, 2, 3, 4, 6, 9, 12 and 36. We can group them as (1), (2, 3), (6), (4, 9), (12, 18), (36) so that every group contains divisors with the same prime signature and we have a(36) = 6.
|
|
CROSSREFS
|
Cf. A000108.
Sequence in context: A065151 A073093 A088873 this_sequence A067554 A135981 A135615
Adjacent sequences: A085079 A085080 A085081 this_sequence A085083 A085084 A085085
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jul 01 2003
|
|
EXTENSIONS
|
More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Jan 20 2005
|
|
|
Search completed in 0.002 seconds
|
