I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ than $a(k) = ab$, so $$\begin{align*} \sum_{k\leq x}\frac1{a(k)} &= \sum_{a b^2 \leq x} \frac{\mu^2(a)}{ab} = \sum_{a \leq x} \frac{\mu^2(a)}{a} \sum_{b^2 \leq x/a}\frac1b = \sum_{a \leq x} \frac{\mu^2(a) H_{\sqrt{x/a}}}{a}\\ &= \sum_{a \leq x} \frac{\mu^2(a) \left(\frac12\log(x) - \frac12\log(a) + \gamma + O(\sqrt{a/x})\right)}{a}\\ &= \left(\frac12\log(x) + \gamma\right) \sum_{a \leq x}\frac{\mu^2(a)}a - \frac12 \sum_{a \leq x}\frac{\mu^2(a)\log(a)}{a} + O(1). \end{align*}$$

We have $$\sum_{a \leq x}{\mu^2(a)} = \frac{6x}{\pi^2} + O(\sqrt x),$$ so by partial summation $$\sum_{a \leq x}\frac{\mu^2(a)}a = \frac6{\pi^2}\log(x) + C + O\left(\frac1 {\sqrt x}\right),$$ and $$\sum_{a \leq x}\frac{\mu^2(a) \log(a)}a = \frac{3}{\pi^2} \log^2(x) - \frac6{\pi^2} \log(x) + C_2 + O\left(\frac{\log(x)}{\sqrt x}\right),$$ which gives the rough result. The more precise result can likely be found by directly looking at the sums (without passing through $\sum_{a \leq x}\mu^2(a)$) and using more terms for the harmonic numbers.

I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ then $a(k) = ab$, so $$\begin{align*} \sum_{k\leq x}\frac1{a(k)} &= \sum_{a b^2 \leq x} \frac{\mu^2(a)}{ab} = \sum_{a \leq x} \frac{\mu^2(a)}{a} \sum_{b^2 \leq x/a}\frac1b = \sum_{a \leq x} \frac{\mu^2(a) H_{\sqrt{x/a}}}{a}\\ &= \sum_{a \leq x} \frac{\mu^2(a) \left(\frac12\log(x) - \frac12\log(a) + \gamma + O(\sqrt{a/x})\right)}{a}\\ &= \left(\frac12\log(x) + \gamma\right) \sum_{a \leq x}\frac{\mu^2(a)}a - \frac12 \sum_{a \leq x}\frac{\mu^2(a)\log(a)}{a} + O(1). \end{align*}$$

We have $$\sum_{a \leq x}{\mu^2(a)} = \frac{6x}{\pi^2} + O(\sqrt x),$$ so by partial summation $$\sum_{a \leq x}\frac{\mu^2(a)}a = \frac6{\pi^2}\log(x) + C + O\left(\frac1 {\sqrt x}\right),$$ and $$\sum_{a \leq x}\frac{\mu^2(a) \log(a)}a = \frac{3}{\pi^2} \log^2(x) - \frac6{\pi^2} \log(x) + C_2 + O\left(\frac{\log(x)}{\sqrt x}\right),$$ which gives the rough result. The more precise result can likely be found by directly looking at the sums (without passing through $\sum_{a \leq x}\mu^2(a)$) and using more terms for the harmonic numbers.

I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ than $a(k) = ab$, so $$\sum_{n\leq x}\frac1{a(k)} = \sum_{a b^2 \leq x} \frac{\mu^2(a)}{ab} = \sum_{a \leq x} \frac{\mu^2(a)}{a} \sum_{b^2 \leq \frac xa}\frac1b = \sum_{a \leq x} \frac{\mu^2(a) H_{\sqrt{\frac xa}}}{a} = \sum_{a \leq x} \frac{\mu^2(a) (\frac12\log(x) - \frac12\log(a) + \gamma + O(\sqrt{\frac{a}x}))}{a} = (\frac12\log(x) + \gamma) \sum_{a \leq x}\frac{\mu^2(a)}a - \frac12 \sum_{a \leq x}\frac{\mu^2(a)\log(a)}{a} + O(1)$$

we have $\sum_{a \leq x}{\mu^2(a)} = \frac{6x}{\pi^2} + O(\sqrt x)$, so by partial summation $\sum_{a \leq x}\frac{\mu^2(a)}a = \frac6{\pi^2}\log(x) + C + O(\frac1 {\sqrt x})$ and $\sum_{a \leq x}\frac{\mu^2(a) \log(a)}a = \frac{3}{\pi^2} \log^2(x) - \frac6{\pi^2} \log(x) + C_2 + O(\frac{\log(x)}{\sqrt x})$, which gives the rough result. The more precise result can likely be found by directly looking at the sums (without passing through $\sum_{a \leq x}{\mu^2(a)}$) and using more terms for the harmonic numbers.

I couldn't find a reference, but (as noted in the OEIS page) if we have $k = a b^2$ with squarefree $a$ than $a(k) = ab$, so $$\begin{align*} \sum_{k\leq x}\frac1{a(k)} &= \sum_{a b^2 \leq x} \frac{\mu^2(a)}{ab} = \sum_{a \leq x} \frac{\mu^2(a)}{a} \sum_{b^2 \leq x/a}\frac1b = \sum_{a \leq x} \frac{\mu^2(a) H_{\sqrt{x/a}}}{a}\\ &= \sum_{a \leq x} \frac{\mu^2(a) \left(\frac12\log(x) - \frac12\log(a) + \gamma + O(\sqrt{a/x})\right)}{a}\\ &= \left(\frac12\log(x) + \gamma\right) \sum_{a \leq x}\frac{\mu^2(a)}a - \frac12 \sum_{a \leq x}\frac{\mu^2(a)\log(a)}{a} + O(1). \end{align*}$$

We have $$\sum_{a \leq x}{\mu^2(a)} = \frac{6x}{\pi^2} + O(\sqrt x),$$ so by partial summation $$\sum_{a \leq x}\frac{\mu^2(a)}a = \frac6{\pi^2}\log(x) + C + O\left(\frac1 {\sqrt x}\right),$$ and $$\sum_{a \leq x}\frac{\mu^2(a) \log(a)}a = \frac{3}{\pi^2} \log^2(x) - \frac6{\pi^2} \log(x) + C_2 + O\left(\frac{\log(x)}{\sqrt x}\right),$$ which gives the rough result. The more precise result can likely be found by directly looking at the sums (without passing through $\sum_{a \leq x}\mu^2(a)$) and using more terms for the harmonic numbers.