A383156 The sum of the maximum exponents in the prime factorizations of the divisors of n.
0, 1, 1, 3, 1, 3, 1, 6, 3, 3, 1, 7, 1, 3, 3, 10, 1, 7, 1, 7, 3, 3, 1, 13, 3, 3, 6, 7, 1, 7, 1, 15, 3, 3, 3, 13, 1, 3, 3, 13, 1, 7, 1, 7, 7, 3, 1, 21, 3, 7, 3, 7, 1, 13, 3, 13, 3, 3, 1, 15, 1, 3, 7, 21, 3, 7, 1, 7, 3, 7, 1, 22, 1, 3, 7, 7, 3, 7, 1, 21, 10, 3, 1
Offset: 1
Examples
4 has 3 divisors: 1, 2 = 2^1 and 4 = 2^2. The maximum exponents in their prime factorizations are 0, 1 and 2, respectively. Therefore, a(4) = 0 + 1 + 2 = 3. 12 has 6 divisors: 1, 2 = 2^1, 3 = 3^1, 4 = 2^2, 6 = 2 * 3 and 12 = 2^2 * 3. The maximum exponents in their prime factorizations are 0, 1, 1, 2, 1 and 2, respectively. Therefore, a(12) = 0 + 1 + 1 + 2 + 1 + 2 = 7.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Amiram Eldar, Plot of 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[#] &]; Array[a, 100] (* second program: *) a[n_] := If[n == 1, 0, Module[{e = FactorInteger[n][[;; , 2]], emax, v}, emax = Max[e]; v = Table[Times @@ (Min[# + 1, k + 1] & /@ e), {k, 1, emax}]; v[[1]] + Sum[k*(v[[k]] - v[[k - 1]]), {k, 2, emax}] - 1]]; Array[a, 100]
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PARI
emax(n) = if(n == 1, 0, vecmax(factor(n)[,2])); a(n) = sumdiv(n, d, 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 ->min(x+1 , k+1), e))); v[1] + sum(k = 2, emax, k * (v[k]-v[k-1])) - 1);
Formula
a(n) = Sum_{d|n} A051903(d).
a(p^e) = e*(e+1)/2 for prime p and e >= 0. In particular, a(p) = 1 for prime p.
a(n) = tau(n, 2) - 1 + Sum_{k=2..A051903(n)} k * (tau(n, k+1) - tau(n, k)), where tau(n, k) is the number of k-free 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, tau(n, k) is a multiplicative function with tau(p^e, k) = min(e+1, k). E.g., tau(n, 2) = A034444(n), tau(n, 3) = A073184(n), and tau(n, 4) = A252505(n).
Comments