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A383160 a(n) is the numerator of the mean of the maximum exponents in the prime factorizations of the unitary divisors of n.

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%I A383160 #9 Apr 20 2025 02:40:20
%S A383160 0,1,1,1,1,3,1,3,1,3,1,5,1,3,3,2,1,5,1,5,3,3,1,7,1,3,3,5,1,7,1,5,3,3,
%T A383160 3,3,1,3,3,7,1,7,1,5,5,3,1,9,1,5,3,5,1,7,3,7,3,3,1,11,1,3,5,3,3,7,1,5,
%U A383160 3,7,1,2,1,3,5,5,3,7,1,9,2,3,1,11,3,3,3
%N A383160 a(n) is the numerator of the mean of the maximum exponents in the prime factorizations of the unitary divisors of n.
%C A383160 First differs from A296082 at n = 27.
%C A383160 a(n) depends only on the prime signature of n (A118914).
%H A383160 Amiram Eldar, <a href="/A383160/b383160.txt">Table of n, a(n) for n = 1..10000</a>
%H A383160 Amiram Eldar, <a href="/A383160/a383160.jpg">Plot of (Sum_{k=1..n} a(k)/A383161(k))/(c_1*n - c_2*n/sqrt(log(n))) for n = 10^(1..10)</a>.
%F A383160 a(n) = numerator(Sum_{d|n, gcd(d, n/d) = 1} A051903(d) / A034444(n)) = numerator(A383159(n) / A034444(n)).
%F A383160 a(n)/A383161(n) >= A383157(n)/A383158(n), with equality if and only if n is either squarefree (A005117) or a power of prime (A000961), i.e., n is not in A126706.
%F A383160 a(n)/A383161(n) < 1 if and only if n is squarefree.
%F A383160 a(n)/A383161(n) = 1 if and only if n is a square of a prime (A001248).
%F A383160 Sum_{k=1..n} a(k)/A383161(k) ~ c_1 * n - c_2 * n / sqrt(log(n)), where c = m(2) + Sum_{k>=3} (k-1) * (m(k) - m(k-1)) = 1.36914082067166047512... is the asymptotic mean of the fractions a(k)/A383161(k), m(k) = Product_{p prime} (1 - 1/(2*p^k)) is the asymptotic mean of the ratio between the number of k-free unitary divisors and the number of unitary divisors. e.g., m(2) = A383057 and m(3) = A383058, and c_2 = A345288/sqrt(Pi) = 0.6189064491320478904... .
%e A383160 Fractions begin with 0, 1/2, 1/2, 1, 1/2, 3/4, 1/2, 3/2, 1, 3/4, 1/2, 5/4, ...
%e A383160 4 has 2 unitary divisors: 1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0 and 2, respectively. Therefore, a(4) = numerator((0 + 2)/2) = numerator(1) = 1.
%e A383160 12 has 4 unitary divisors: 1, 3 = 3^1, 4 = 2^2 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 2 and 2, respectively. Therefore, a(12) = numerator((0 + 1 + 2 + 2)/4) = numerator(5/4) = 5.
%t A383160 emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := Numerator[DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &] / 2^PrimeNu[n]]; Array[a, 100]
%o A383160 (PARI) emax(n) = if(n == 1, 0, vecmax(factor(n)[,2]));
%o A383160 a(n) = my(f = factor(n)); numerator(sumdiv(n, d, emax(d) * (gcd(d, n/d) == 1)) / (1 << omega(f)));
%Y A383160 The unitary version of A383157.
%Y A383160 Cf. A034444, A051903, A077610, A118914, A296082, A345288, A383057, A383058, A383157, A383158, A383159, A383161 (denominators).
%Y A383160 Cf. A000961, A001248, A005117, A126706.
%K A383160 nonn,easy,frac
%O A383160 1,6
%A A383160 _Amiram Eldar_, Apr 18 2025