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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 (list; graph; refs; listen; history; text; internal format)
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] (* Jean-Fran├žois 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

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Last modified May 23 18:42 EDT 2017. Contains 286926 sequences.