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Define the zeta density of the set of integers $A$ as

$$d(A) = \lim_{x \to 11+} \zeta(x) frac1{\zeta(x)} \sum_{k \in A} k^{-x}.$$

Then from results given at the OEIS, this is

$$d(A) = \lim_{x \to 11+} {1 \over \zeta(x)} \left[ {\zeta(x)^2 - \zeta(2x) \over 2\zeta(x)} \right].$$

After some simplifcation this is

$$\lim_{x \to 1} \left( {1 \over 2} - {\zeta(2x) \over 2 \zeta(x)^2} \right)$$

and recalling that $\lim_{x \to 1} \zeta(x)$ is infinity while $\zeta(2) = \pi^2/6$, this is $1/2$.

Now, if a set has a natural density, then it has a zeta density, and the two densities are equal; see for example Chapter 2 of Diaconis' PhD dissertation. So we can conclude that if your set has a natural density, then that natural density is $1/2$.

1

Define the zeta density of the set of integers $A$ as

$$d(A) = \lim_{x \to 1} \zeta(x) \sum_{k \in A} k^{-x}.$$

Then from results given at the OEIS, this is

$$d(A) = \lim_{x \to 1} {1 \over \zeta(x)} \left[ {\zeta(x)^2 - \zeta(2x) \over 2\zeta(x)} \right].$$

After some simplifcation this is

$$\lim_{x \to 1} \left( {1 \over 2} - {\zeta(2x) \over 2 \zeta(x)^2} \right)$$

and recalling that $\lim_{x \to 1} \zeta(x)$ is infinity while $\zeta(2) = \pi^2/6$, this is $1/2$.

Now, if a set has a natural density, then it has a zeta density, and the two densities are equal; see for example Chapter 2 of Diaconis' PhD dissertation. So we can conclude that if your set has a natural density, then that natural density is $1/2$.