A064372 Additive function a(n) defined by the recursive formula a(1)=1 and a(p^k)=a(k) for any prime p.

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3

Offset: 1

Author

Steven Finch, Sep 26 2001

Keywords

• nonn
• easy
• nice

Comments

That is, if i, j, k, ... are relatively prime, then a(i*j*k*...) = a(i) + a(j) + a(k) + ... - N. J. A. Sloane, Nov 20 2007

Starts almost the same as A001221 (the number of distinct primes dividing n): the first twelve terms which are different are a(1), a(64), a(192), a(320), a(448), a(576), a(704), a(729), a(832), a(960), a(1024) and a(1088), since the first non-unitary values of n are a(6) and(10). - Henry Bottomley, Sep 23 2002

a(A164336(n)) = 1. - Reinhard Zumkeller, Aug 27 2011

Examples

			a(30) = a(5^1 * 3^1 * 2^1) = a(1) + a(1) + a(1) = 3.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A001221, A079553, A106444, A106490, A106491, A106495, A124010, A164336.

Programs

Haskell

a064372 1 = 1
a064372 n = sum $ map a064372 $ a124010_row n
-- Reinhard Zumkeller, Aug 27 2011

Maple

a:= proc(n) option remember; `if`(n=1, 1,
      add(a(i[2]), i=ifactors(n)[2]))
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Aug 23 2020

Mathematica

a[1] = 1; a[n_] := a[n] = Plus @@ a /@ FactorInteger[n][[All, 2]]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Sep 19 2012 *)

Formula

a(n) = A106491(n) - A106490(n) = A106495(A106444(n)). - Antti Karttunen, May 09 2005

a(1) = 1, a(n) = Sum_{k=1..A001221(n)} a(A124010(n,k)) for n > 1. - Reinhard Zumkeller, Aug 27 2011