A383159 The sum of the maximum exponents in the prime factorizations of the unitary divisors of n.
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, 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, 3, 7, 1, 8, 1, 3, 5, 5, 3, 7, 1, 9, 4, 3, 1, 11, 3, 3, 3
Offset: 1
Examples
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. 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.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Amiram Eldar, Plot of abs((Sum_{k=1..n} a(k))/(c_1*n*log(n) + c_2*n) - 1) for n = 10^(1..10).
Crossrefs
Programs
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Mathematica
emax[n_] := If[n == 1, 0, Max[FactorInteger[n][[;; , 2]]]]; a[n_] := DivisorSum[n, emax[#] &, CoprimeQ[#, n/#] &]; Array[a, 100] (* second program: *) 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] -
PARI
emax(n) = if(n == 1, 0, vecmax(factor(n)[,2])); a(n) = sumdiv(n, d, (gcd(d, n/d) == 1) * emax(d));
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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);
Formula
a(n) = Sum_{d|n, gcd(d, n/d) = 1} A051903(d).
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).
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).
Comments