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A383159 The sum of the maximum exponents in the prime factorizations of the unitary divisors of n.

%I A383159 #11 Apr 20 2025 02:40:04
%S A383159 0,1,1,2,1,3,1,3,2,3,1,5,1,3,3,4,1,5,1,5,3,3,1,7,2,3,3,5,1,7,1,5,3,3,
%T A383159 3,6,1,3,3,7,1,7,1,5,5,3,1,9,2,5,3,5,1,7,3,7,3,3,1,11,1,3,5,6,3,7,1,5,
%U A383159 3,7,1,8,1,3,5,5,3,7,1,9,4,3,1,11,3,3,3
%N A383159 The sum of the maximum exponents in the prime factorizations of the unitary divisors of n.
%C A383159 First differs from A032741 at n = 36, and from A305611 and A325770 at n = 30.
%C A383159 a(n) depends only on the prime signature of n (A118914).
%H A383159 Amiram Eldar, <a href="/A383159/b383159.txt">Table of n, a(n) for n = 1..10000</a>
%H A383159 Amiram Eldar, <a href="/A383159/a383159.jpg">Plot of abs((Sum_{k=1..n} a(k))/(c_1*n*log(n) + c_2*n) - 1) for n = 10^(1..10)</a>.
%F A383159 a(n) = Sum_{d|n, gcd(d, n/d) = 1} A051903(d).
%F A383159 a(n) = A034444(n) * A383160(n)/A383161(n).
%F A383159 a(n) <= A383156(n), with equality if and only if n is squarefree (A005117).
%F A383159 a(n) = utau(n, 2) - 1 + Sum_{k=2..A051903(n)} k * (utau(n, k+1) - utau(n, k)), where utau(n, k) is the number of k-free unitary divisors of n (k-free numbers are numbers that are not divisible by a k-th power other than 1). For a given k >= 2, utau(n, k) is a multiplicative function with utau(p^e, k) = 2 if e < k, and 1 otherwise. E.g., utau(n, 2) = A056671(n), utau(n, 3) = A365498(n), and utau(n, 4) = A365499(n).
%F A383159 Sum_{k=1..n} a(k) ~ c_1 * n * log(n) + c_2 * n, where c_1 = c(2) + Sum_{k>=3} (k-1) * (c(k) - c(k-1)) = 0.91974850283445458744..., c(k) = Product_{p prime} (1 - 1/p^2 - 1/p^k + 1/p^(k+1)), c_2 = -1 + (2*gamma - 1)*c_1 + d(2) + Sum_{k>=3} (k-1) * (d(k) - d(k-1)) = -0.50780794945146599739..., d(k) = c(k) * Sum_{p prime} (2*p^(k-1) + k*p - k - 1) * log(p) / (p^(k+1) - p^(k-1) - p + 1), and gamma is Euler's constant (A001620).
%e A383159 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) = 0 + 2 = 2.
%e A383159 12 has 4 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) = 0 + 1 + 2 + 2 = 5.
%t A383159 emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &]; Array[a, 100]
%t A383159 (* second program: *)
%t A383159 a[n_] := If[n == 1, 0, Module[{e = FactorInteger[n][[;; , 2]], emax, v}, emax = Max[e]; v = Table[Times @@ (If[# < k + 1, 2, 1] & /@ e), {k, 1, emax}]; v[[1]] + Sum[k*(v[[k]] - v[[k - 1]]), {k, 2, emax}] - 1]]; Array[a, 100]
%o A383159 (PARI) emax(n) = if(n == 1, 0, vecmax(factor(n)[,2]));
%o A383159 a(n) = sumdiv(n, d, (gcd(d, n/d) == 1) * emax(d));
%o A383159 (PARI) a(n) = if(n == 1, 0, my(e = factor(n)[, 2], emax = vecmax(e), v); v = vector(emax, k, vecprod(apply(x ->if(x < k+1, 2, 1), e))); v[1] + sum(k = 2, emax, k * (v[k]-v[k-1])) - 1);
%Y A383159 The unitary version of A383156.
%Y A383159 Cf. A005117, A034444, A051903, A056671, A077610, A365498, A365499, A383160, A383161.
%Y A383159 Cf. A032741, A305611, A325770.
%K A383159 nonn,easy
%O A383159 1,4
%A A383159 _Amiram Eldar_, Apr 18 2025