%I #25 Apr 12 2020 10:06:24
%S 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,
%T 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,
%U 4,8,2,9,2,4,6,6,4,8,2,8,4,4,2,12,4,4
%N Call d a recursive divisor of n iff the p-adic valuation of d is a recursive divisor of the p-adic valuation of n for any prime p dividing d; a(n) gives the number of recursive divisors of n.
%C 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).
%C 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).
%C 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 p-adic valuation of n is cubefree for any prime p dividing n.
%C 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.
%H Rémy Sigrist, <a href="/A282446/b282446.txt">Table of n, a(n) for n = 1..10000</a>
%F Multiplicative, with a(p^k)=1+a(k) for any prime p and k>0.
%F a(A014221(n))=n+1 for any n>=0.
%e The recursive divisors of 40 are: 1, 2, 5, 8, 10 and 40, hence a(40)=6.
%t a[1] = 1; a[n_] := a[n] = Times @@ (1 + a/@ (Last /@ FactorInteger[n])); Array[a, 100] (* _Amiram Eldar_, Apr 12 2020 *)
%o (PARI) a(n) = my (f=factor(n)); return (prod(i=1, #f~, 1+a(f[i,2])))
%Y Cf. A000005, A004709, A049599, A014221, A182318.
%K nonn,mult
%O 1,2
%A _Rémy Sigrist_, Feb 15 2017