A085082 Number of distinct prime signatures arising among the divisors of n.

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

Offset: 1

Author

Amarnath Murthy, Jul 01 2003

Keywords

• easy
• nonn

Comments

For a squarefree number n with k distinct prime divisors, a(n) = k+1.

If n = p^r then a(n) = tau(n) = r+1.

Question: Find a(n) in the following cases:

1. n = m^k where m is a squarefree number with r distinct prime divisors.

2. n = Product_{i=1..r} (p_i)^i, where p_i is the i-th distinct prime divisor of n.

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

Examples

			a(30) = 4 and the divisors with distinct prime signatures are 1, 2, 6 and 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.
		

Links

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

Crossrefs

Cf. A000108.

The second problem describes A076954(i). See also A006939.

Programs

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-François Alcover, Jun 10 2015 *)

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

Extensions

More terms from David Wasserman, Jan 20 2005