A380398 The number of unitary divisors of n that are perfect powers (A001597).
1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1
Offset: 1
Examples
a(4) = 2 since 4 have 2 unitary divisors that are perfect powers, 1 and 4 = 2^2. a(72) = 3 since 72 have 3 unitary divisors that are perfect powers, 1, 8 = 2^3, and 9 = 3^2.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
ppQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := DivisorSum[n, 1 &, CoprimeQ[#, n/#] && ppQ[#] &]; Array[a, 100]
-
PARI
a(n) = sumdiv(n, d, gcd(d, n/d) == 1 && (d == 1 || ispower(d)));
Formula
a(n) = Sum_{d|n, gcd(d, n/d) == 1} [d in A001597], where [] is the Iverson bracket.
a(n) = 1 if and only if n is squarefree (A005117).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 - Sum_{k>=2} mu(k)*(zeta(k)/zeta(k+1) - 1) = 1.49341326536904597349..., where mu is the Moebius function (A008683).
Comments