

A282446


Call d a recursive divisor of n iff the padic valuation of d is a recursive divisor of the padic valuation of n for any prime p dividing d; a(n) gives the number of recursive divisors of n.


5



1, 2, 2, 3, 2, 4, 2, 3, 3, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 6, 4, 4, 2, 6, 3, 4, 3, 6, 2, 8, 2, 3, 4, 4, 4, 9, 2, 4, 4, 6, 2, 8, 2, 6, 6, 4, 2, 8, 3, 6, 4, 6, 2, 6, 4, 6, 4, 4, 2, 12, 2, 4, 6, 5, 4, 8, 2, 6, 4, 8, 2, 9, 2, 4, 6, 6, 4, 8, 2, 8, 4, 4, 2, 12, 4, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

More informally, the prime tower factorization of a recursive divisor of n can be obtained by removing branches from the prime tower factorization of n (the prime tower factorization of a number is defined in A182318).
A recursive divisor of n is also a divisor of n, hence a(n)<=A000005(n) for any n, with equality iff n is cubefree (i.e. n belongs to A004709).
A recursive divisor of n is also a (1+e)divisor of n, hence a(n)<=A049599(n) for any n, with equality iff the padic valuation of n is cubefree for any prime p dividing n.
This sequence first differs from A049599 at n=256: a(256)=4 whereas A049599(256)=5; note that 256=2^(2^3), and 2^3 is not cubefree.


LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000


FORMULA

Multiplicative, with a(p^k)=1+a(k) for any prime p and k>0.
a(A014221(n))=n+1 for any n>=0.


EXAMPLE

The recursive divisors of 40 are: 1, 2, 5, 8, 10 and 40, hence a(40)=6.


PROG

(PARI) a(n) = my (f=factor(n)); return (prod(i=1, #f~, 1+a(f[i, 2])))


CROSSREFS

Cf. A000005, A004709, A049599, A014221, A182318.
Sequence in context: A106491 A073184 A073182 * A049599 A305461 A043261
Adjacent sequences: A282443 A282444 A282445 * A282447 A282448 A282449


KEYWORD

nonn,mult


AUTHOR

Rémy Sigrist, Feb 15 2017


STATUS

approved



