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The divisors of 7 are 1^1 and 7^1, which have only odd exponents (=1), so a(8) = 1+7 = 8. The divisors of 8 are 1^1, 2^1, 2^2 and 2^3; 2^2 has an even prime power and does not count, so a(8) = 1+2+8=11. The divisors of 12 are 1^1, 2^1, 3^1, 2^2, 2^1*3^1 and 2^2*3; 2^2 and 2^2*3 don't count because they have prime factors with even powers, so a(12) = 1+2+3+6 = 12.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^2/12) * Product_{p prime} (1 - 1/(p*(p+1))) = A072691 * A065463 = 0.5793804... . - Amiram Eldar, Oct 27 2022
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f[e_] := If[OddQ[e], e+2, e+1]; fun[p_, e_] := 1 + (p^f[e] - p)/(p^2 - 1); a[1] = 1; a[n_] := Times @@ (fun @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, May 14 2019 *)
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