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Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals

$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$

where $*$ denotes Dirichlet convolution. We may write rewrite the given asymptotic as

$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$

noticing that the right hand side this is of the form

$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$

Mobius inversion tells us that

$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$

where $\alpha^{-1}$ is the multiplicative inverse of $\alpha$ with respect to Dirichlet convolution. Applying Mobius inversion to our sum, we have that

$$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$

which equals

$$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$

$$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$

Notice that the Dirichlet inverse to the function $g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.

2 deleted 44 characters in body

Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals

$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$

where $*$ denotes Dirichlet convolution. We may write

$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$

noticing that the right hand side is of the form

$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$

Mobius inversion tells us that

$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$

where $\alpha^{-1}$ is the multiplicative function satisfying $\alpha^{-1}*\alpha=\epsilon$ where $\epsilon(1)=1$, and all other values are inverse of $0$. \alpha$with respect to Dirichlet convolution. Applying Mobius inversion to our sum, we have that $$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$ which equals $$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$ $$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$ Notice that the Dirichlet inverse to the function$g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers. 1 Let$g(n)$denote the indicator function for the fourth powers. Then your sum equals $$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$ where$*$denotes Dirichlet convolution. We may write $$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$ noticing that the right hand side is of the form $$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$ Mobius inversion tells us that $$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$ where$\alpha^{-1}$is the multiplicative function satisfying$\alpha^{-1}*\alpha=\epsilon$where$\epsilon(1)=1$, and all other values are$0$. Applying Mobius inversion to our sum, we have that $$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$ which equals $$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$ $$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$ Notice that the Dirichlet inverse to the function$g(n)\$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.