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The number of unitary divisors of n that are perfect powers (A001597).
3

%I #10 Jan 25 2025 08:29:39

%S 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,

%T 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,

%U 1,1,1,3,1,1,2,2,1,1,1,2,2,1,1,2,1,1,1

%N The number of unitary divisors of n that are perfect powers (A001597).

%C First differs from A368978 at n = 32, from A007424 and A369163 at n = 36, from A278908, A307848, A358260 and A365549 at n = 64, and from A323308 at n = 72.

%C a(n) depends only on the prime signature of n (A118914).

%C The record values are 2^k, for k = 0, 1, 2, ..., and they are attained at A061742(k).

%C The sum of unitary divisors of n that are perfect powers is A380400(n).

%H Amiram Eldar, <a href="/A380398/b380398.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n, gcd(d, n/d) == 1} [d in A001597], where [] is the Iverson bracket.

%F a(n) = A091050(n) - A380399(n).

%F a(n) = 1 if and only if n is squarefree (A005117).

%F 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).

%e a(4) = 2 since 4 have 2 unitary divisors that are perfect powers, 1 and 4 = 2^2.

%e a(72) = 3 since 72 have 3 unitary divisors that are perfect powers, 1, 8 = 2^3, and 9 = 3^2.

%t ppQ[n_] := n == 1 || GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := DivisorSum[n, 1 &, CoprimeQ[#, n/#] && ppQ[#] &]; Array[a, 100]

%o (PARI) a(n) = sumdiv(n, d, gcd(d, n/d) == 1 && (d == 1 || ispower(d)));

%Y Cf. A001597, A005117, A008683, A061742, A077610, A091050, A118914, A323308, A380399, A380400.

%Y Cf. A007424, A278908, A307848, A323308, A358260, A365549, A368978, A369163.

%K nonn,easy

%O 1,4

%A _Amiram Eldar_, Jan 23 2025