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 A282446 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. 9
 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 p-adic 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. MATHEMATICA a = 1; a[n_] := a[n] = Times @@ (1 + a/@ (Last /@ FactorInteger[n])); Array[a, 100] (* Amiram Eldar, Apr 12 2020 *) 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 A334762 A305461 Adjacent sequences:  A282443 A282444 A282445 * A282447 A282448 A282449 KEYWORD nonn,mult AUTHOR Rémy Sigrist, Feb 15 2017 STATUS approved

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Last modified May 6 05:18 EDT 2021. Contains 343580 sequences. (Running on oeis4.)