

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).


19



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
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OFFSET

1,2


COMMENTS

1. For a squarefree 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 squarefree 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, Jan 20 2005
I have submitted comments for A000108 and A016098 that each include a combinatorial statement equivalent to the second problem and its solution.  Matthew Vandermast, Nov 22 2010


LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..10000


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.


MAPLE

with(numtheory):
a:= n> nops({seq(sort(map(x>x[2], ifactors(d)[2])), d=divisors(n))}):
seq(a(n), n=1..120); # Alois P. Heinz, Jun 12 2012


MATHEMATICA

ps[1] = {}; ps[n_] := FactorInteger[n][[All, 2]] // Sort; a[n_] := ps /@ Divisors[n] // Union // Length; Array[a, 120] (* JeanFrançois Alcover, Jun 10 2015 *)


PROG

(PARI) a(n)=my(f=vecsort(factor(n)[, 2]), v=[1], s); for(i=1, #f, s=0; v=vector(f[i]+1, i, if(i<=#v, s+=v[i]); s)); vecsum(v) \\ Charles R Greathouse IV, Feb 03 2017


CROSSREFS

Cf. A000108.
The second problem describes A076954(i). See also A006939.
Sequence in context: A073093 A222084 A088873 * A181796 A067554 A135981
Adjacent sequences: A085079 A085080 A085081 * A085083 A085084 A085085


KEYWORD

easy,nonn


AUTHOR

Amarnath Murthy, Jul 01 2003


EXTENSIONS

More terms from David Wasserman, Jan 20 2005


STATUS

approved



